Explanation:
Okay, so I need to find the range of the function $ f(x) = (\arccos x)^2 + (\arcsin x)^2 $. Hmm, let me start by recalling some basic properties of inverse trigonometric functions.

First, I know that the domain of both $ \arccos x $ and $ \arcsin x $ is $ [-1, 1] $. So, the domain of $ f(x) $ is also $ [-1, 1] $. That's straightforward. Now, the question is about the range, so I need to figure out all the possible values that $ f(x) $ can take as $ x $ varies from -1 to 1.

Let me think about the relationship between $ \arccos x $ and $ \arcsin x $. I remember there's an identity that connects them. Specifically, $ \arccos x + \arcsin x = \frac{\pi}{2} $ for all $ x $ in $ [-1, 1] $. Is that right? Let me verify with a quick example. If $ x = 0 $, then $ \arccos 0 = \frac{\pi}{2} $ and $ \arcsin 0 = 0 $, so their sum is $ \frac{\pi}{2} $. If $ x = 1 $, $ \arccos 1 = 0 $ and $ \arcsin 1 = \frac{\pi}{2} $, so again the sum is $ \frac{\pi}{2} $. Similarly for $ x = -1 $, $ \arccos(-1) = \pi $ and $ \arcsin(-1) = -\frac{\pi}{2} $, so their sum is $ \pi - \frac{\pi}{2} = \frac{\pi}{2} $. Okay, so that identity seems to hold.

So, if I let $ \theta = \arcsin x $, then $ \arccos x = \frac{\pi}{2} - \theta $. Therefore, substituting into $ f(x) $, we get:

$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2
$$

Let me expand that:

$$
\left( \frac{\pi}{2} - \theta \right)^2 + \theta^2 = \left( \frac{\pi^2}{4} - \pi \theta + \theta^2 \right) + \theta^2 = \frac{\pi^2}{4} - \pi \theta + 2\theta^2
$$

So, $ f(x) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. But since $ \theta = \arcsin x $, and $ x \in [-1, 1] $, $ \theta $ ranges from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $. Wait, but $ \arcsin x $ actually gives values in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, right? So $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

But maybe I can express $ f(x) $ in terms of $ \theta $, and then find its range over $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Alternatively, since $ \theta $ is $ \arcsin x $, maybe I can also think of $ f(x) $ as a function of $ \theta $, and then find its minimum and maximum values.

Alternatively, maybe I can consider $ f(x) $ as a quadratic function in terms of $ \theta $, and then find its minimum or maximum. Let me try that.

So, $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. This is a quadratic function in $ \theta $, opening upwards because the coefficient of $ \theta^2 $ is positive (2). Therefore, its minimum occurs at the vertex, and the maximum will occur at one of the endpoints of the interval $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

First, let me find the vertex of this quadratic. The vertex occurs at $ \theta = \frac{\pi}{4} $, since for a quadratic $ ax^2 + bx + c $, the vertex is at $ x = -\frac{b}{2a} $. Here, $ a = 2 $, $ b = -\pi $, so:

$$
\theta = \frac{\pi}{2 \times 2} = \frac{\pi}{4}
$$

Wait, no. Wait, the formula is $ -b/(2a) $. So here, $ b = -\pi $, so:

$$
\theta = -(-\pi)/(2 \times 2) = \pi / 4
$$

Yes, that's correct. So the vertex is at $ \theta = \pi/4 $. Since the parabola opens upwards, this is the minimum point. Therefore, the minimum value of $ f(\theta) $ is at $ \theta = \pi/4 $, and the maximum occurs at one of the endpoints $ \theta = -\pi/2 $ or $ \theta = \pi/2 $.

So let me compute the value of $ f(\theta) $ at these points.

First, at $ \theta = \pi/4 $:

$$
f(\pi/4) = 2\left( \frac{\pi}{4} \right)^2 - \pi \left( \frac{\pi}{4} \right) + \frac{\pi^2}{4}
$$

Let me compute each term:

First term: $ 2 \times \frac{\pi^2}{16} = \frac{\pi^2}{8} $

Second term: $ -\pi \times \frac{\pi}{4} = -\frac{\pi^2}{4} $

Third term: $ \frac{\pi^2}{4} $

Adding them up:

$ \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8} $

So the minimum value of $ f(x) $ is $ \frac{\pi^2}{8} $.

Now, let's check the endpoints. First, $ \theta = \pi/2 $:

$$
f(\pi/2) = 2\left( \frac{\pi}{2} \right)^2 - \pi \left( \frac{\pi}{2} \right) + \frac{\pi^2}{4}
$$

Calculating each term:

First term: $ 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $

Second term: $ -\pi \times \frac{\pi}{2} = -\frac{\pi^2}{2} $

Third term: $ \frac{\pi^2}{4} $

Adding them up:

$ \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4} $

Similarly, for $ \theta = -\pi/2 $:

$$
f(-\pi/2) = 2\left( -\frac{\pi}{2} \right)^2 - \pi \left( -\frac{\pi}{2} \right) + \frac{\pi^2}{4}
$$

Calculating each term:

First term: $ 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $

Second term: $ -\pi \times (-\frac{\pi}{2}) = \frac{\pi^2}{2} $

Third term: $ \frac{\pi^2}{4} $

Adding them up:

$ \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4} $

Wait, hold on. Wait, that seems like a big jump. Let me check again.

Wait, $ f(-\pi/2) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. So substituting $ \theta = -\pi/2 $:

First term: $ 2 \times (-\pi/2)^2 = 2 \times \pi^2 / 4 = \pi^2 / 2 $

Second term: $ -\pi \times (-\pi/2) = \pi^2 / 2 $

Third term: $ \pi^2 / 4 $

So total is $ \pi^2 / 2 + \pi^2 / 2 + \pi^2 / 4 = (\pi^2 + \pi^2 / 4) = (5\pi^2)/4 $. Yes, that's correct.

But wait, when $ \theta = \pi/2 $, we had $ f(\pi/2) = \pi^2 / 4 $, and when $ \theta = -\pi/2 $, we have $ 5\pi^2 / 4 $. Therefore, the maximum value of $ f(x) $ is $ 5\pi^2 / 4 $, and the minimum is $ \pi^2 / 8 $. Therefore, the range of $ f(x) $ is $ [\pi^2 / 8, 5\pi^2 / 4] $.

But wait, hold on. Let me check if this is correct. Let me verify with specific values of x.

For example, when x = 0, let's compute f(0):

$ \arccos(0) = \pi/2 $, $ \arcsin(0) = 0 $, so $ f(0) = (\pi/2)^2 + 0^2 = \pi^2 / 4 $. Which is the same as the value at $ \theta = \pi/2 $, which makes sense because when x = 0, $ \theta = \arcsin(0) = 0 $, but wait, no. Wait, earlier substitution was $ \theta = \arcsin x $, so when x = 0, $ \theta = 0 $, so then $ f(\theta) = 2(0)^2 - \pi(0) + \pi^2 / 4 = \pi^2 / 4 $. Wait, but earlier when I calculated $ f(\pi/2) $, that was when $ \theta = \pi/2 $, which would correspond to x = 1, since $ \arcsin(1) = \pi/2 $. Similarly, $ \theta = -\pi/2 $ corresponds to x = -1. So when x = 1, $ f(1) = (\arccos 1)^2 + (\arcsin 1)^2 = 0^2 + (\pi/2)^2 = \pi^2 / 4 $. Similarly, when x = -1, $ f(-1) = (\arccos(-1))^2 + (\arcsin(-1))^2 = (\pi)^2 + (-\pi/2)^2 = \pi^2 + \pi^2 / 4 = 5\pi^2 / 4 $. So that checks out.

But when x = 0, we get $ \pi^2 / 4 $, which is the same as at x = 1 and x = -1? Wait, no. Wait, when x = 1, we get $ \pi^2 / 4 $, and when x = -1, we get $ 5\pi^2 / 4 $. So that's different. So the maximum is at x = -1, and the minimum is at x where theta = pi/4, which would be when arcsin x = pi/4, so x = sin(pi/4) = sqrt(2)/2. Let me check that.

So if theta = pi/4, then x = sin(theta) = sin(pi/4) = sqrt(2)/2 ≈ 0.707. Let me compute f(sqrt(2)/2):

arccos(sqrt(2)/2) = pi/4, arcsin(sqrt(2)/2) = pi/4. Therefore, f(x) = (pi/4)^2 + (pi/4)^2 = 2*(pi^2)/16 = pi^2 / 8. Which matches our previous calculation. So that's the minimum.

So, so far, the range seems to be from pi^2 / 8 to 5 pi^2 / 4. But let me check another point. For example, take x = 1/2. Then arcsin(1/2) = pi/6, arccos(1/2) = pi/3. Therefore, f(1/2) = (pi/3)^2 + (pi/6)^2 = pi^2 / 9 + pi^2 / 36 = (4 pi^2 + pi^2)/36 = 5 pi^2 / 36 ≈ 0.434 pi^2. Wait, but pi^2 / 8 is approximately 0.3927 pi^2, so 5 pi^2 / 36 is larger than that. Wait, but according to our previous conclusion, the minimum is pi^2 / 8, so this value is higher than that. So that seems okay.

Wait, but let me check if there are any other critical points. Wait, we considered the quadratic function in theta, and found that the minimum is at theta = pi/4, and the maximum at theta = -pi/2. But is there a possibility that the maximum occurs somewhere else? Wait, since the quadratic opens upwards, the maximum on the interval would be at one of the endpoints. Since the vertex is a minimum, the maximum must be at one of the endpoints. So since theta ranges from -pi/2 to pi/2, and the function is quadratic with minimum at pi/4, then the maximum is at theta = -pi/2 (which gives 5 pi^2 / 4) or theta = pi/2 (which gives pi^2 / 4). Therefore, the maximum is 5 pi^2 / 4.

But let me check another value. Let me take theta = -pi/4. Then x = sin(-pi/4) = -sqrt(2)/2. Then f(x) would be:

arccos(-sqrt(2)/2) = 3pi/4, arcsin(-sqrt(2)/2) = -pi/4. Therefore, f(x) = (3pi/4)^2 + (-pi/4)^2 = 9 pi^2 / 16 + pi^2 / 16 = 10 pi^2 / 16 = 5 pi^2 / 8 ≈ 0.625 pi^2. Which is between pi^2 / 8 ≈ 0.3927 pi^2 and 5 pi^2 / 4 ≈ 3.927 pi^2. So that seems okay.

But let me check if there's a possibility that the function could attain higher values somewhere else. For example, suppose theta approaches pi/2, but theta is limited to pi/2. So when theta is pi/2, we get pi^2 / 4. When theta is -pi/2, we get 5 pi^2 / 4. So that's the maximum.

Alternatively, maybe I can think of f(x) in terms of another substitution. Let me try to write f(x) as:

Since $ \arccos x + \arcsin x = \frac{\pi}{2} $, let me denote $ \alpha = \arccos x $, so $ \arcsin x = \frac{\pi}{2} - \alpha $. Therefore, f(x) becomes:

$ \alpha^2 + \left( \frac{\pi}{2} - \alpha \right)^2 $

Which is the same as before. Expanding:

$ \alpha^2 + \frac{\pi^2}{4} - \pi \alpha + \alpha^2 = 2\alpha^2 - \pi \alpha + \frac{\pi^2}{4} $

Which is the same quadratic as before, but now with $ \alpha = \arccos x $, which ranges from 0 to pi as x goes from 1 to -1. Wait, so if I think of alpha as ranging from 0 to pi, then the quadratic function in alpha is 2 alpha^2 - pi alpha + pi^2 / 4. So, in this case, the domain of alpha is [0, pi].

So, if I consider this quadratic function over alpha in [0, pi], then the vertex is at alpha = pi / 4, which is within the interval [0, pi]. Therefore, the minimum is at alpha = pi / 4, and the maximum occurs at one of the endpoints. Let me check the endpoints:

At alpha = 0: f(0) = 0 - 0 + pi^2 / 4 = pi^2 / 4.

At alpha = pi: f(pi) = 2 pi^2 - pi^2 + pi^2 / 4 = pi^2 + pi^2 / 4 = 5 pi^2 / 4.

So same result. Therefore, the range is [pi^2 / 8, 5 pi^2 / 4].

But let me check if there's another way to approach this problem. Maybe using calculus? Let me try taking the derivative of f(x) with respect to x and find critical points.

First, f(x) = (arccos x)^2 + (arcsin x)^2. Let me compute f'(x):

The derivative of (arccos x)^2 is 2 arccos x * (-1 / sqrt(1 - x^2)).

The derivative of (arcsin x)^2 is 2 arcsin x * (1 / sqrt(1 - x^2)).

Therefore, f'(x) = -2 arccos x / sqrt(1 - x^2) + 2 arcsin x / sqrt(1 - x^2) = [2 (arcsin x - arccos x)] / sqrt(1 - x^2).

But since arccos x + arcsin x = pi / 2, we can write arcsin x = pi / 2 - arccos x. Therefore, substituting:

arcsin x - arccos x = (pi / 2 - arccos x) - arccos x = pi / 2 - 2 arccos x.

Therefore, f'(x) = [2 (pi / 2 - 2 arccos x)] / sqrt(1 - x^2) = [pi - 4 arccos x] / sqrt(1 - x^2).

Set derivative equal to zero to find critical points:

pi - 4 arccos x = 0 => arccos x = pi / 4 => x = cos(pi / 4) = sqrt(2)/2 ≈ 0.707.

Therefore, the critical point is at x = sqrt(2)/2. Let me check the value of f(x) there, which we already did earlier: pi^2 / 8.

Now, to confirm whether this is a minimum or maximum, we can check the second derivative or analyze the sign of the first derivative around that point. Alternatively, since we already have the quadratic expression and know that it's a minimum, we can be confident that this critical point is a minimum.

Therefore, the function f(x) has a minimum at x = sqrt(2)/2 with value pi^2 / 8, and maximum at the endpoints x = -1 and x = 1. Wait, but earlier when we checked x = 1, f(1) = pi^2 / 4, and x = -1 gives f(-1) = 5 pi^2 / 4. Therefore, the maximum is at x = -1.

But why is the maximum at x = -1? Let me think. Because when x approaches -1, arccos x approaches pi, and arcsin x approaches -pi/2. Therefore, their squares are pi^2 and (pi/2)^2, so adding them gives pi^2 + pi^2 / 4 = 5 pi^2 / 4. Whereas at x = 1, arccos x is 0 and arcsin x is pi/2, so their squares add up to pi^2 / 4. Therefore, the maximum occurs at x = -1.

Therefore, the range of f(x) is from pi^2 / 8 to 5 pi^2 / 4.

But let me check if there are any other critical points. The derivative f'(x) is [pi - 4 arccos x] / sqrt(1 - x^2). The denominator sqrt(1 - x^2) is always positive for x in (-1, 1). Therefore, the sign of f'(x) depends on the numerator pi - 4 arccos x.

So, when is pi - 4 arccos x positive? When arccos x < pi / 4. Since arccos x decreases as x increases, so arccos x < pi / 4 when x > cos(pi / 4) = sqrt(2)/2. Similarly, when arccos x > pi / 4, which is when x < sqrt(2)/2, then pi - 4 arccos x is negative. Therefore, the derivative is positive when x > sqrt(2)/2, negative when x < sqrt(2)/2. Therefore, the function f(x) is decreasing on [-1, sqrt(2)/2] and increasing on [sqrt(2)/2, 1]. Therefore, the critical point at x = sqrt(2)/2 is indeed a minimum.

Therefore, the minimum value is pi^2 / 8, and the maximum occurs at the left endpoint x = -1, which is 5 pi^2 / 4.

Wait, but let me check the value at x = -1. As I did before, arccos(-1) = pi, arcsin(-1) = -pi/2. Therefore, f(-1) = pi^2 + (pi/2)^2 = pi^2 + pi^2 / 4 = 5 pi^2 / 4.

But what about the value at x = 1? It's pi^2 / 4, which is less than 5 pi^2 / 4. Therefore, the maximum is indeed at x = -1.

Therefore, the range of f(x) is [pi^2 / 8, 5 pi^2 / 4].

But let me check if there's a possibility that somewhere else in the domain, the function could attain a higher value. For example, suppose we take x approaching -1 from the right. Let me take x approaching -1, so arccos x approaches pi, arcsin x approaches -pi/2. Therefore, f(x) approaches pi^2 + (pi/2)^2 = 5 pi^2 / 4. Similarly, as x approaches 1, f(x) approaches 0 + (pi/2)^2 = pi^2 / 4. Therefore, the maximum is indeed achieved at x = -1, and the minimum at x = sqrt(2)/2.

Therefore, the range is from pi^2 / 8 to 5 pi^2 / 4.

But let me check if there's a mistake in the calculation of the quadratic function. Let me re-express f(x) in terms of theta again.

Earlier, I had f(theta) = 2 theta^2 - pi theta + pi^2 / 4. Then, since theta ranges from -pi/2 to pi/2, and the quadratic has its minimum at theta = pi/4, which is within the interval. Therefore, the minimum value is pi^2 / 8, and the maximum is at theta = -pi/2, which gives 5 pi^2 / 4.

Alternatively, if I think of theta as ranging from -pi/2 to pi/2, then the maximum of the quadratic function would be at the endpoint with the larger value. Since the quadratic opens upwards, the maximum occurs at the endpoint farthest from the vertex. The vertex is at theta = pi/4. The distance from pi/4 to -pi/2 is pi/4 + pi/2 = 3 pi / 4, whereas the distance from pi/4 to pi/2 is pi/4. Therefore, the endpoint theta = -pi/2 is farther from the vertex, so the function attains a higher value there.

Therefore, all checks confirm that the range is [pi^2 / 8, 5 pi^2 / 4].

But just to be thorough, let me check another point. Let me take x = -sqrt(2)/2. Then, arcsin(-sqrt(2)/2) = -pi/4, arccos(-sqrt(2)/2) = 3pi/4. Therefore, f(x) = (3pi/4)^2 + (-pi/4)^2 = 9 pi^2 / 16 + pi^2 / 16 = 10 pi^2 / 16 = 5 pi^2 / 8 ≈ 1.9635 pi^2. Wait, pi^2 is approximately 9.8696, so 5 pi^2 / 8 ≈ 6.172, and 5 pi^2 / 4 ≈ 12.337. So 5 pi^2 / 8 is between pi^2 / 8 ≈ 1.234 and 5 pi^2 / 4. So that's okay.

Another check: take x = 0. Then f(0) = (pi/2)^2 + 0 = pi^2 / 4 ≈ 2.467, which is between pi^2 / 8 ≈ 1.234 and 5 pi^2 / 4 ≈ 12.337.

So, all these points lie within the interval [pi^2 / 8, 5 pi^2 / 4].

Therefore, I think I can be confident that the range of f(x) is from pi squared over 8 to 5 pi squared over 4.

But let me just check if there's a possibility that the function could attain a higher value somewhere else. For example, suppose I take theta = -pi/2 + epsilon, where epsilon is a small positive number. Then, theta approaches -pi/2 from the right. Then, f(theta) = 2 theta^2 - pi theta + pi^2 / 4. Let me compute the limit as theta approaches -pi/2:

As theta approaches -pi/2, theta^2 approaches (pi/2)^2, -pi theta approaches -pi*(-pi/2) = pi^2 / 2, so total is 2*(pi^2 / 4) + pi^2 / 2 + pi^2 / 4 = pi^2 / 2 + pi^2 / 2 + pi^2 / 4 = pi^2 + pi^2 / 4 = 5 pi^2 / 4. So that's consistent.

Alternatively, if I take theta approaching pi/2, then f(theta) approaches pi^2 / 4.

Therefore, there are no higher values than 5 pi^2 / 4.

Therefore, the range is [pi²/8, 5pi²/4].

But let me check if the problem says "range of f(x)", so it's the set of all possible outputs. Since we have found that the minimum is pi²/8 and maximum is 5 pi²/4, and the function is continuous on [-1, 1], by the Intermediate Value Theorem, the range is the closed interval between these two values.

Therefore, the answer should be \boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.

**Final Answer**
The range of $ f(x) $ is \boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.

To determine the **range** of the function
$$
f(x) = (\arccos x)^2 + (\arcsin x)^2,
$$
we begin by analyzing the relationship between the inverse trigonometric functions involved.

---

### Step 1: Use the Identity

We recall the identity:
$$
\arccos x + \arcsin x = \frac{\pi}{2}, \quad \text{for all } x \in [-1, 1].
$$
Let us define:
$$
\theta = \arcsin x \Rightarrow \arccos x = \frac{\pi}{2} - \theta.
$$
Substituting into the function:
$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2.
$$

---

### Step 2: Expand and Simplify

Expanding the expression:
$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2 = \frac{\pi^2}{4} - \pi \theta + \theta^2 + \theta^2 = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}.
$$

So, we now have:
$$
f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}, \quad \text{where } \theta = \arcsin x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].
$$

---

### Step 3: Analyze the Quadratic Function

This is a quadratic in $\theta$, opening **upwards** (since the coefficient of $\theta^2$ is positive). Therefore, the **minimum** occurs at the vertex, and the **maximum** occurs at one of the endpoints of the interval.

#### Vertex of the Quadratic

The vertex of a quadratic $ a\theta^2 + b\theta + c $ is at:
$$
\theta = -\frac{b}{2a} = \frac{\pi}{4}.
$$

Substitute $\theta = \frac{\pi}{4}$ into the function:
$$
f\left( \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} \right)^2 - \pi \left( \frac{\pi}{4} \right) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}.
$$

So, the **minimum** value of $ f(x) $ is:
$$
\boxed{\frac{\pi^2}{8}}.
$$

#### Evaluate at Endpoints

- At $\theta = \frac{\pi}{2}$ (i.e., $x = 1$):
$$
f\left( \frac{\pi}{2} \right) = 2\left( \frac{\pi}{2} \right)^2 - \pi \left( \frac{\pi}{2} \right) + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}.
$$

- At $\theta = -\frac{\pi}{2}$ (i.e., $x = -1$):
$$
f\left( -\frac{\pi}{2} \right) = 2\left( -\frac{\pi}{2} \right)^2 - \pi \left( -\frac{\pi}{2} \right) + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}.
$$

So, the **maximum** value of $ f(x) $ is:
$$
\boxed{\frac{5\pi^2}{4}}.
$$

---

### Step 4: Conclude the Range

Since $ f(x) $ is continuous on the interval $[-1, 1]$, and we have found the minimum and maximum values, the **range** of $ f(x) $ is the closed interval:
$$
\boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.
$$

[Assistant]
Explanation:
Okay, so I need to find the range of the function $ f(x) = (\arccos x)^2 + (\arcsin x)^2 $. Hmm, let me start by recalling some basic properties of inverse trigonometric functions.

First, I know that the domain of both $ \arccos x $ and $ \arcsin x $ is $ [-1, 1] $. So, the domain of $ f(x) $ is also $ [-1, 1] $. That's straightforward. Now, the question is about the range, so I need to figure out all the possible values that $ f(x) $ can take as $ x $ varies from -1 to 1.

Let me think about the relationship between $ \arccos x $ and $ \arcsin x $. I remember there's an identity that connects them. Specifically, $ \arccos x + \arcsin x = \frac{\pi}{2} $ for all $ x $ in $ [-1, 1] $. Is that right? Let me verify with a quick example. If $ x = 0 $, then $ \arccos 0 = \frac{\pi}{2} $ and $ \arcsin 0 = 0 $, so their sum is $ \frac{\pi}{2} $. If $ x = 1 $, $ \arccos 1 = 0 $ and $ \arcsin 1 = \frac{\pi}{2} $, so again the sum is $ \frac{\pi}{2} $. Similarly for $ x = -1 $, $ \arccos(-1) = \pi $ and $ \arcsin(-1) = -\frac{\pi}{2} $, so their sum is $ \pi - \frac{\pi}{2} = \frac{\pi}{2} $. Okay, so that identity seems to hold.

So, if I let $ \theta = \arcsin x $, then $ \arccos x = \frac{\pi}{2} - \theta $. Therefore, substituting into $ f(x) $, we get:

$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2
$$

Let me expand that:

$$
\left( \frac{\pi}{2} - \theta \right)^2 + \theta^2 = \left( \frac{\pi^2}{4} - \pi \theta + \theta^2 \right) + \theta^2 = \frac{\pi^2}{4} - \pi \theta + 2\theta^2
$$

So, $ f(x) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. But since $ \theta = \arcsin x $, and $ x \in [-1, 1] $, $ \theta $ ranges from $ -\frac{\pi}{2} $ to $ \frac{\pi}{2} $. Wait, but $ \arcsin x $ actually gives values in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, right? So $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

But maybe I can express $ f(x) $ in terms of $ \theta $, and then find its range over $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. Alternatively, since $ \theta $ is $ \arcsin x $, maybe I can also think of $ f(x) $ as a function of $ \theta $, and then find its minimum and maximum values.

Alternatively, maybe I can consider $ f(x) $ as a quadratic function in terms of $ \theta $, and then find its minimum or maximum. Let me try that.

So, $ f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. This is a quadratic function in $ \theta $, opening upwards because the coefficient of $ \theta^2 $ is positive (2). Therefore, its minimum occurs at the vertex, and the maximum will occur at one of the endpoints of the interval $ \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.

First, let me find the vertex of this quadratic. The vertex occurs at $ \theta = \frac{\pi}{4} $, since for a quadratic $ ax^2 + bx + c $, the vertex is at $ x = -\frac{b}{2a} $. Here, $ a = 2 $, $ b = -\pi $, so:

$$
\theta = \frac{\pi}{2 \times 2} = \frac{\pi}{4}
$$

Wait, no. Wait, the formula is $ -b/(2a) $. So here, $ b = -\pi $, so:

$$
\theta = -(-\pi)/(2 \times 2) = \pi / 4
$$

Yes, that's correct. So the vertex is at $ \theta = \pi/4 $. Since the parabola opens upwards, this is the minimum point. Therefore, the minimum value of $ f(\theta) $ is at $ \theta = \pi/4 $, and the maximum occurs at one of the endpoints $ \theta = -\pi/2 $ or $ \theta = \pi/2 $.

So let me compute the value of $ f(\theta) $ at these points.

First, at $ \theta = \pi/4 $:

$$
f(\pi/4) = 2\left( \frac{\pi}{4} \right)^2 - \pi \left( \frac{\pi}{4} \right) + \frac{\pi^2}{4}
$$

Let me compute each term:

First term: $ 2 \times \frac{\pi^2}{16} = \frac{\pi^2}{8} $

Second term: $ -\pi \times \frac{\pi}{4} = -\frac{\pi^2}{4} $

Third term: $ \frac{\pi^2}{4} $

Adding them up:

$ \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8} $

So the minimum value of $ f(x) $ is $ \frac{\pi^2}{8} $.

Now, let's check the endpoints. First, $ \theta = \pi/2 $:

$$
f(\pi/2) = 2\left( \frac{\pi}{2} \right)^2 - \pi \left( \frac{\pi}{2} \right) + \frac{\pi^2}{4}
$$

Calculating each term:

First term: $ 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $

Second term: $ -\pi \times \frac{\pi}{2} = -\frac{\pi^2}{2} $

Third term: $ \frac{\pi^2}{4} $

Adding them up:

$ \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4} $

Similarly, for $ \theta = -\pi/2 $:

$$
f(-\pi/2) = 2\left( -\frac{\pi}{2} \right)^2 - \pi \left( -\frac{\pi}{2} \right) + \frac{\pi^2}{4}
$$

Calculating each term:

First term: $ 2 \times \frac{\pi^2}{4} = \frac{\pi^2}{2} $

Second term: $ -\pi \times (-\frac{\pi}{2}) = \frac{\pi^2}{2} $

Third term: $ \frac{\pi^2}{4} $

Adding them up:

$ \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \pi^2 + \frac{\pi^2}{4} = \frac{5\pi^2}{4} $

Wait, hold on. Wait, that seems like a big jump. Let me check again.

Wait, $ f(-\pi/2) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4} $. So substituting $ \theta = -\pi/2 $:

First term: $ 2 \times (-\pi/2)^2 = 2 \times \pi^2 / 4 = \pi^2 / 2 $

Second term: $ -\pi \times (-\pi/2) = \pi^2 / 2 $

Third term: $ \pi^2 / 4 $

So total is $ \pi^2 / 2 + \pi^2 / 2 + \pi^2 / 4 = (\pi^2 + \pi^2 / 4) = (5\pi^2)/4 $. Yes, that's correct.

But wait, when $ \theta = \pi/2 $, we had $ f(\pi/2) = \pi^2 / 4 $, and when $ \theta = -\pi/2 $, we have $ 5\pi^2 / 4 $. Therefore, the maximum value of $ f(x) $ is $ 5\pi^2 / 4 $, and the minimum is $ \pi^2 / 8 $. Therefore, the range of $ f(x) $ is $ [\pi^2 / 8, 5\pi^2 / 4] $.

But wait, hold on. Let me check if this is correct. Let me verify with specific values of x.

For example, when x = 0, let's compute f(0):

$ \arccos(0) = \pi/2 $, $ \arcsin(0) = 0 $, so $ f(0) = (\pi/2)^2 + 0^2 = \pi^2 / 4 $. Which is the same as the value at $ \theta = \pi/2 $, which makes sense because when x = 0, $ \theta = \arcsin(0) = 0 $, but wait, no. Wait, earlier substitution was $ \theta = \arcsin x $, so when x = 0, $ \theta = 0 $, so then $ f(\theta) = 2(0)^2 - \pi(0) + \pi^2 / 4 = \pi^2 / 4 $. Wait, but earlier when I calculated $ f(\pi/2) $, that was when $ \theta = \pi/2 $, which would correspond to x = 1, since $ \arcsin(1) = \pi/2 $. Similarly, $ \theta = -\pi/2 $ corresponds to x = -1. So when x = 1, $ f(1) = (\arccos 1)^2 + (\arcsin 1)^2 = 0^2 + (\pi/2)^2 = \pi^2 / 4 $. Similarly, when x = -1, $ f(-1) = (\arccos(-1))^2 + (\arcsin(-1))^2 = (\pi)^2 + (-\pi/2)^2 = \pi^2 + \pi^2 / 4 = 5\pi^2 / 4 $. So that checks out.

But when x = 0, we get $ \pi^2 / 4 $, which is the same as at x = 1 and x = -1? Wait, no. Wait, when x = 1, we get $ \pi^2 / 4 $, and when x = -1, we get $ 5\pi^2 / 4 $. So that's different. So the maximum is at x = -1, and the minimum is at x where theta = pi/4, which would be when arcsin x = pi/4, so x = sin(pi/4) = sqrt(2)/2. Let me check that.

So if theta = pi/4, then x = sin(theta) = sin(pi/4) = sqrt(2)/2 ≈ 0.707. Let me compute f(sqrt(2)/2):

arccos(sqrt(2)/2) = pi/4, arcsin(sqrt(2)/2) = pi/4. Therefore, f(x) = (pi/4)^2 + (pi/4)^2 = 2*(pi^2)/16 = pi^2 / 8. Which matches our previous calculation. So that's the minimum.

So, so far, the range seems to be from pi^2 / 8 to 5 pi^2 / 4. But let me check another point. For example, take x = 1/2. Then arcsin(1/2) = pi/6, arccos(1/2) = pi/3. Therefore, f(1/2) = (pi/3)^2 + (pi/6)^2 = pi^2 / 9 + pi^2 / 36 = (4 pi^2 + pi^2)/36 = 5 pi^2 / 36 ≈ 0.434 pi^2. Wait, but pi^2 / 8 is approximately 0.3927 pi^2, so 5 pi^2 / 36 is larger than that. Wait, but according to our previous conclusion, the minimum is pi^2 / 8, so this value is higher than that. So that seems okay.

Wait, but let me check if there are any other critical points. Wait, we considered the quadratic function in theta, and found that the minimum is at theta = pi/4, and the maximum at theta = -pi/2. But is there a possibility that the maximum occurs somewhere else? Wait, since the quadratic opens upwards, the maximum on the interval would be at one of the endpoints. Since the vertex is a minimum, the maximum must be at one of the endpoints. So since theta ranges from -pi/2 to pi/2, and the function is quadratic with minimum at pi/4, then the maximum is at theta = -pi/2 (which gives 5 pi^2 / 4) or theta = pi/2 (which gives pi^2 / 4). Therefore, the maximum is 5 pi^2 / 4.

But let me check another value. Let me take theta = -pi/4. Then x = sin(-pi/4) = -sqrt(2)/2. Then f(x) would be:

arccos(-sqrt(2)/2) = 3pi/4, arcsin(-sqrt(2)/2) = -pi/4. Therefore, f(x) = (3pi/4)^2 + (-pi/4)^2 = 9 pi^2 / 16 + pi^2 / 16 = 10 pi^2 / 16 = 5 pi^2 / 8 ≈ 0.625 pi^2. Which is between pi^2 / 8 ≈ 0.3927 pi^2 and 5 pi^2 / 4 ≈ 3.927 pi^2. So that seems okay.

But let me check if there's a possibility that the function could attain higher values somewhere else. For example, suppose theta approaches pi/2, but theta is limited to pi/2. So when theta is pi/2, we get pi^2 / 4. When theta is -pi/2, we get 5 pi^2 / 4. So that's the maximum.

Alternatively, maybe I can think of f(x) in terms of another substitution. Let me try to write f(x) as:

Since $ \arccos x + \arcsin x = \frac{\pi}{2} $, let me denote $ \alpha = \arccos x $, so $ \arcsin x = \frac{\pi}{2} - \alpha $. Therefore, f(x) becomes:

$ \alpha^2 + \left( \frac{\pi}{2} - \alpha \right)^2 $

Which is the same as before. Expanding:

$ \alpha^2 + \frac{\pi^2}{4} - \pi \alpha + \alpha^2 = 2\alpha^2 - \pi \alpha + \frac{\pi^2}{4} $

Which is the same quadratic as before, but now with $ \alpha = \arccos x $, which ranges from 0 to pi as x goes from 1 to -1. Wait, so if I think of alpha as ranging from 0 to pi, then the quadratic function in alpha is 2 alpha^2 - pi alpha + pi^2 / 4. So, in this case, the domain of alpha is [0, pi].

So, if I consider this quadratic function over alpha in [0, pi], then the vertex is at alpha = pi / 4, which is within the interval [0, pi]. Therefore, the minimum is at alpha = pi / 4, and the maximum occurs at one of the endpoints. Let me check the endpoints:

At alpha = 0: f(0) = 0 - 0 + pi^2 / 4 = pi^2 / 4.

At alpha = pi: f(pi) = 2 pi^2 - pi^2 + pi^2 / 4 = pi^2 + pi^2 / 4 = 5 pi^2 / 4.

So same result. Therefore, the range is [pi^2 / 8, 5 pi^2 / 4].

But let me check if there's another way to approach this problem. Maybe using calculus? Let me try taking the derivative of f(x) with respect to x and find critical points.

First, f(x) = (arccos x)^2 + (arcsin x)^2. Let me compute f'(x):

The derivative of (arccos x)^2 is 2 arccos x * (-1 / sqrt(1 - x^2)).

The derivative of (arcsin x)^2 is 2 arcsin x * (1 / sqrt(1 - x^2)).

Therefore, f'(x) = -2 arccos x / sqrt(1 - x^2) + 2 arcsin x / sqrt(1 - x^2) = [2 (arcsin x - arccos x)] / sqrt(1 - x^2).

But since arccos x + arcsin x = pi / 2, we can write arcsin x = pi / 2 - arccos x. Therefore, substituting:

arcsin x - arccos x = (pi / 2 - arccos x) - arccos x = pi / 2 - 2 arccos x.

Therefore, f'(x) = [2 (pi / 2 - 2 arccos x)] / sqrt(1 - x^2) = [pi - 4 arccos x] / sqrt(1 - x^2).

Set derivative equal to zero to find critical points:

pi - 4 arccos x = 0 => arccos x = pi / 4 => x = cos(pi / 4) = sqrt(2)/2 ≈ 0.707.

Therefore, the critical point is at x = sqrt(2)/2. Let me check the value of f(x) there, which we already did earlier: pi^2 / 8.

Now, to confirm whether this is a minimum or maximum, we can check the second derivative or analyze the sign of the first derivative around that point. Alternatively, since we already have the quadratic expression and know that it's a minimum, we can be confident that this critical point is a minimum.

Therefore, the function f(x) has a minimum at x = sqrt(2)/2 with value pi^2 / 8, and maximum at the endpoints x = -1 and x = 1. Wait, but earlier when we checked x = 1, f(1) = pi^2 / 4, and x = -1 gives f(-1) = 5 pi^2 / 4. Therefore, the maximum is at x = -1.

But why is the maximum at x = -1? Let me think. Because when x approaches -1, arccos x approaches pi, and arcsin x approaches -pi/2. Therefore, their squares are pi^2 and (pi/2)^2, so adding them gives pi^2 + pi^2 / 4 = 5 pi^2 / 4. Whereas at x = 1, arccos x is 0 and arcsin x is pi/2, so their squares add up to pi^2 / 4. Therefore, the maximum occurs at x = -1.

Therefore, the range of f(x) is from pi^2 / 8 to 5 pi^2 / 4.

But let me check if there are any other critical points. The derivative f'(x) is [pi - 4 arccos x] / sqrt(1 - x^2). The denominator sqrt(1 - x^2) is always positive for x in (-1, 1). Therefore, the sign of f'(x) depends on the numerator pi - 4 arccos x.

So, when is pi - 4 arccos x positive? When arccos x < pi / 4. Since arccos x decreases as x increases, so arccos x < pi / 4 when x > cos(pi / 4) = sqrt(2)/2. Similarly, when arccos x > pi / 4, which is when x < sqrt(2)/2, then pi - 4 arccos x is negative. Therefore, the derivative is positive when x > sqrt(2)/2, negative when x < sqrt(2)/2. Therefore, the function f(x) is decreasing on [-1, sqrt(2)/2] and increasing on [sqrt(2)/2, 1]. Therefore, the critical point at x = sqrt(2)/2 is indeed a minimum.

Therefore, the minimum value is pi^2 / 8, and the maximum occurs at the left endpoint x = -1, which is 5 pi^2 / 4.

Wait, but let me check the value at x = -1. As I did before, arccos(-1) = pi, arcsin(-1) = -pi/2. Therefore, f(-1) = pi^2 + (pi/2)^2 = pi^2 + pi^2 / 4 = 5 pi^2 / 4.

But what about the value at x = 1? It's pi^2 / 4, which is less than 5 pi^2 / 4. Therefore, the maximum is indeed at x = -1.

Therefore, the range of f(x) is [pi^2 / 8, 5 pi^2 / 4].

But let me check if there's a possibility that somewhere else in the domain, the function could attain a higher value. For example, suppose we take x approaching -1 from the right. Let me take x approaching -1, so arccos x approaches pi, arcsin x approaches -pi/2. Therefore, f(x) approaches pi^2 + (pi/2)^2 = 5 pi^2 / 4. Similarly, as x approaches 1, f(x) approaches 0 + (pi/2)^2 = pi^2 / 4. Therefore, the maximum is indeed achieved at x = -1, and the minimum at x = sqrt(2)/2.

Therefore, the range is from pi^2 / 8 to 5 pi^2 / 4.

But let me check if there's a mistake in the calculation of the quadratic function. Let me re-express f(x) in terms of theta again.

Earlier, I had f(theta) = 2 theta^2 - pi theta + pi^2 / 4. Then, since theta ranges from -pi/2 to pi/2, and the quadratic has its minimum at theta = pi/4, which is within the interval. Therefore, the minimum value is pi^2 / 8, and the maximum is at theta = -pi/2, which gives 5 pi^2 / 4.

Alternatively, if I think of theta as ranging from -pi/2 to pi/2, then the maximum of the quadratic function would be at the endpoint with the larger value. Since the quadratic opens upwards, the maximum occurs at the endpoint farthest from the vertex. The vertex is at theta = pi/4. The distance from pi/4 to -pi/2 is pi/4 + pi/2 = 3 pi / 4, whereas the distance from pi/4 to pi/2 is pi/4. Therefore, the endpoint theta = -pi/2 is farther from the vertex, so the function attains a higher value there.

Therefore, all checks confirm that the range is [pi^2 / 8, 5 pi^2 / 4].

But just to be thorough, let me check another point. Let me take x = -sqrt(2)/2. Then, arcsin(-sqrt(2)/2) = -pi/4, arccos(-sqrt(2)/2) = 3pi/4. Therefore, f(x) = (3pi/4)^2 + (-pi/4)^2 = 9 pi^2 / 16 + pi^2 / 16 = 10 pi^2 / 16 = 5 pi^2 / 8 ≈ 1.9635 pi^2. Wait, pi^2 is approximately 9.8696, so 5 pi^2 / 8 ≈ 6.172, and 5 pi^2 / 4 ≈ 12.337. So 5 pi^2 / 8 is between pi^2 / 8 ≈ 1.234 and 5 pi^2 / 4. So that's okay.

Another check: take x = 0. Then f(0) = (pi/2)^2 + 0 = pi^2 / 4 ≈ 2.467, which is between pi^2 / 8 ≈ 1.234 and 5 pi^2 / 4 ≈ 12.337.

So, all these points lie within the interval [pi^2 / 8, 5 pi^2 / 4].

Therefore, I think I can be confident that the range of f(x) is from pi squared over 8 to 5 pi squared over 4.

But let me just check if there's a possibility that the function could attain a higher value somewhere else. For example, suppose I take theta = -pi/2 + epsilon, where epsilon is a small positive number. Then, theta approaches -pi/2 from the right. Then, f(theta) = 2 theta^2 - pi theta + pi^2 / 4. Let me compute the limit as theta approaches -pi/2:

As theta approaches -pi/2, theta^2 approaches (pi/2)^2, -pi theta approaches -pi*(-pi/2) = pi^2 / 2, so total is 2*(pi^2 / 4) + pi^2 / 2 + pi^2 / 4 = pi^2 / 2 + pi^2 / 2 + pi^2 / 4 = pi^2 + pi^2 / 4 = 5 pi^2 / 4. So that's consistent.

Alternatively, if I take theta approaching pi/2, then f(theta) approaches pi^2 / 4.

Therefore, there are no higher values than 5 pi^2 / 4.

Therefore, the range is [pi²/8, 5pi²/4].

But let me check if the problem says "range of f(x)", so it's the set of all possible outputs. Since we have found that the minimum is pi²/8 and maximum is 5 pi²/4, and the function is continuous on [-1, 1], by the Intermediate Value Theorem, the range is the closed interval between these two values.

Therefore, the answer should be \boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.

**Final Answer**
The range of $ f(x) $ is \boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.

To determine the **range** of the function
$$
f(x) = (\arccos x)^2 + (\arcsin x)^2,
$$
we begin by analyzing the relationship between the inverse trigonometric functions involved.

---

### Step 1: Use the Identity

We recall the identity:
$$
\arccos x + \arcsin x = \frac{\pi}{2}, \quad \text{for all } x \in [-1, 1].
$$
Let us define:
$$
\theta = \arcsin x \Rightarrow \arccos x = \frac{\pi}{2} - \theta.
$$
Substituting into the function:
$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2.
$$

---

### Step 2: Expand and Simplify

Expanding the expression:
$$
f(x) = \left( \frac{\pi}{2} - \theta \right)^2 + \theta^2 = \frac{\pi^2}{4} - \pi \theta + \theta^2 + \theta^2 = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}.
$$

So, we now have:
$$
f(\theta) = 2\theta^2 - \pi \theta + \frac{\pi^2}{4}, \quad \text{where } \theta = \arcsin x \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right].
$$

---

### Step 3: Analyze the Quadratic Function

This is a quadratic in $\theta$, opening **upwards** (since the coefficient of $\theta^2$ is positive). Therefore, the **minimum** occurs at the vertex, and the **maximum** occurs at one of the endpoints of the interval.

#### Vertex of the Quadratic

The vertex of a quadratic $ a\theta^2 + b\theta + c $ is at:
$$
\theta = -\frac{b}{2a} = \frac{\pi}{4}.
$$

Substitute $\theta = \frac{\pi}{4}$ into the function:
$$
f\left( \frac{\pi}{4} \right) = 2\left( \frac{\pi}{4} \right)^2 - \pi \left( \frac{\pi}{4} \right) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}.
$$

So, the **minimum** value of $ f(x) $ is:
$$
\boxed{\frac{\pi^2}{8}}.
$$

#### Evaluate at Endpoints

- At $\theta = \frac{\pi}{2}$ (i.e., $x = 1$):
$$
f\left( \frac{\pi}{2} \right) = 2\left( \frac{\pi}{2} \right)^2 - \pi \left( \frac{\pi}{2} \right) + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}.
$$

- At $\theta = -\frac{\pi}{2}$ (i.e., $x = -1$):
$$
f\left( -\frac{\pi}{2} \right) = 2\left( -\frac{\pi}{2} \right)^2 - \pi \left( -\frac{\pi}{2} \right) + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}.
$$

So, the **maximum** value of $ f(x) $ is:
$$
\boxed{\frac{5\pi^2}{4}}.
$$

---

### Step 4: Conclude the Range

Since $ f(x) $ is continuous on the interval $[-1, 1]$, and we have found the minimum and maximum values, the **range** of $ f(x) $ is the closed interval:
$$
\boxed{\left[ \dfrac{\pi^2}{8}, \dfrac{5\pi^2}{4} \right]}.
$$

Solution:
First, we claim that $\arccos x + \arcsin x = \frac{\pi}{2}$ for all $x \in [-1,1].$

Note that
\[\cos \left( \frac{\pi}{2} - \arcsin x \right) = \cos (\arccos x) = x.\]Furthermore, $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2},$ so $0 \le \frac{\pi}{2} - \arcsin x \le \pi.$  Therefore,
\[\frac{\pi}{2} - \arcsin x = \arccos x,\]so $\arccos x + \arcsin x = \frac{\pi}{2}.$

Let $\alpha = \arccos x$ and $\beta = \arcsin x,$ so $\alpha + \beta = \frac{\pi}{2}.$  Then
\begin{align*}
f(x) &= (\arccos x)^2 + (\arcsin x)^2 \\
&= \alpha^2 + \beta^2 \\
&= \left( \frac{\pi}{2} - \beta \right)^2 + \beta^2 \\
&= 2 \beta^2 - \pi \beta + \frac{\pi^2}{4} \\
&= 2 \left( \beta - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}.
\end{align*}Since $-\frac{\pi}{2} \le \beta \le \frac{\pi}{2},$ the range of $f(x)$ is $\boxed{\left[ \frac{\pi^2}{8}, \frac{5 \pi^2}{4} \right]}.$


Explanation:
Okay, so I need to find the number of dice, n, such that when you roll n fair 6-sided dice, the probability that exactly two of them show a number other than 1 is 25/216. Hmm, let me think about how to approach this.

First, let me recall some probability concepts. When dealing with probabilities involving multiple dice, especially when looking for specific numbers of successes (in this case, dice showing numbers other than 1), the binomial probability formula is usually useful. The binomial formula is:

P(k successes) = C(n, k) * p^k * (1 - p)^(n - k)

Where:
- C(n, k) is the combination of n things taken k at a time.
- p is the probability of success on a single trial.
- (1 - p) is the probability of failure.

In this problem, a "success" would be rolling a number other than 1. Let me confirm that. The problem says "exactly two of them show a number other than 1." So, each die has a probability of success (not 1) and failure (rolling a 1).

First, let me figure out the probability of success for a single die. Since each die is 6-sided, the probability of rolling a 1 is 1/6, so the probability of not rolling a 1 (i.e., rolling a 2, 3, 4, 5, or 6) is 5/6. Therefore, p = 5/6 and (1 - p) = 1/6.

But wait, the problem says the probability that exactly two of them show a number other than 1 is 25/216. So, using the binomial formula, we can set up the equation:

C(n, 2) * (5/6)^2 * (1/6)^(n - 2) = 25/216

So, I need to solve this equation for n. Let me write that down more clearly:

C(n, 2) * (25/36) * (1/6)^{n - 2} = 25/216

First, let me simplify the equation. Let me note that 25/36 is (5/6)^2, so that's correct. Let me divide both sides of the equation by 25/36 to simplify:

C(n, 2) * (1/6)^{n - 2} = (25/216) / (25/36)

Calculating the right-hand side:

(25/216) divided by (25/36) is equal to (25/216) * (36/25) = 36/216 = 1/6.

So now the equation simplifies to:

C(n, 2) * (1/6)^{n - 2} = 1/6

So, C(n, 2) * (1/6)^{n - 2} = 1/6

Let me write C(n, 2) as n(n - 1)/2. So substituting that in:

[n(n - 1)/2] * (1/6)^{n - 2} = 1/6

Let me multiply both sides by 2 to eliminate the denominator:

n(n - 1) * (1/6)^{n - 2} = 2/6 = 1/3

So:

n(n - 1) * (1/6)^{n - 2} = 1/3

Hmm, okay. Now, I need to solve for n here. Since n is an integer greater than or equal to 2 (since we are talking about exactly two dice showing numbers other than 1), let me try plugging in small integer values for n and see which one satisfies the equation.

Let me note that (1/6)^{n - 2} decreases as n increases, while n(n - 1) increases. So there might be a unique solution. Let me try n=3:

For n=3:

Left-hand side: 3*2*(1/6)^{1} = 6*(1/6) = 1. Which is 1. But the right-hand side is 1/3. So 1 ≠ 1/3. Not equal.

n=4:

Left-hand side: 4*3*(1/6)^{2} = 12*(1/36) = 12/36 = 1/3. Oh! That's equal to the right-hand side. So n=4?

Wait, so n=4 gives us 1/3, which is equal to the right-hand side. Therefore, n=4?

Wait, but let me check again. Let me go back through the steps to make sure I didn't make a mistake.

Original equation:

Probability is C(n, 2)*(5/6)^2*(1/6)^{n - 2} = 25/216.

For n=4:

C(4, 2) = 6.

(5/6)^2 = 25/36.

(1/6)^{4 - 2} = (1/6)^2 = 1/36.

Multiply all together: 6 * 25/36 * 1/36 = 6 * 25 / (36*36) = 150 / 1296.

Simplify 150/1296: divide numerator and denominator by 6: 25/216. Which is exactly the given probability. So yes, n=4 works.

But let me check n=5 just to make sure there isn't another solution.

For n=5:

C(5, 2) = 10.

(5/6)^2 = 25/36.

(1/6)^{5 - 2} = (1/6)^3 = 1/216.

Multiply all together: 10 * 25/36 * 1/216 = 250 / (36*216). Let me compute denominator: 36*216 = 7776. So 250/7776. Simplify: divide numerator and denominator by 2: 125/3888. Let me check if that's equal to 25/216. 25/216 is equal to 500/4320, and 125/3888 is approximately... Let me convert both to decimal:

25/216 ≈ 0.1157

125/3888 ≈ 0.0321. Not equal. So n=5 is not a solution.

What about n=2? Let me check n=2:

C(2, 2) = 1.

(5/6)^2 = 25/36.

(1/6)^{0} = 1.

Multiply all together: 1 * 25/36 * 1 = 25/36 ≈ 0.694, which is not 25/216 ≈ 0.1157. So no.

n=3: as before, gives 1, which is 1, which is not 25/216. So n=4 is the only solution?

Wait, but let me check n=6 just to be thorough.

n=6:

C(6, 2) = 15.

(5/6)^2 = 25/36.

(1/6)^{4} = 1/1296.

Multiply all together: 15 * 25 / 36 * 1 / 1296 = 375 / (36*1296). Let me compute denominator: 36*1296 = 46656. So 375/46656 ≈ 0.008, which is way less than 25/216. So nope.

So seems like n=4 is the only solution. Therefore, the answer is 4?

But let me check again my steps to make sure I didn't make any miscalculations.

Original equation:

C(n, 2)*(5/6)^2*(1/6)^{n - 2} = 25/216.

For n=4:

C(4,2)=6. (5/6)^2=25/36. (1/6)^{2}=1/36. Multiply them: 6*(25/36)*(1/36)= 6*25/(36*36)= 150/1296. Simplify numerator and denominator by dividing by 6: 25/216. Which matches. So yes, correct.

Therefore, n=4. So the answer is 4? Wait, but let me think again. The problem says "the probability that exactly two of them show a number other than 1 is 25/216". So, is there another possible n?

Wait, let me think again. Suppose n=3. Then, the probability of exactly two dice showing numbers other than 1. So, with n=3, the probability would be C(3,2)*(5/6)^2*(1/6)^1 = 3*(25/36)*(1/6) = 75/216 = 25/72 ≈ 0.347. Which is bigger than 25/216. For n=4, it's 25/216. For n=5, as we saw, it's 125/3888 ≈ 0.032, which is smaller. So, seems like n=4 is the only solution. Therefore, the answer is 4.

But let me check if there's another way to approach this problem, maybe through equations, to confirm.

We had earlier:

n(n - 1) * (1/6)^{n - 2} = 1/3

Let me write that equation again:

n(n - 1) * (1/6)^{n - 2} = 1/3

Let me denote k = n - 2. Then n = k + 2. Substitute into the equation:

(k + 2)(k + 1) * (1/6)^k = 1/3

So:

(k + 2)(k + 1) / 6^k = 1/3

Multiply both sides by 6^k:

(k + 2)(k + 1) = (6^k)/3 = 2*6^{k - 1}

So:

(k + 2)(k + 1) = 2*6^{k - 1}

Now, let me try small integer values for k. Since n must be at least 2, k = n - 2 ≥ 0. So k ≥ 0.

Let me try k=0:

Left side: (0 + 2)(0 + 1) = 2*1 = 2

Right side: 2*6^{-1} = 2*(1/6) = 1/3. Not equal.

k=1:

Left: (1 + 2)(1 + 1) = 3*2 = 6

Right: 2*6^{0} = 2*1 = 2. Not equal.

k=2:

Left: (2 + 2)(2 + 1) = 4*3 = 12

Right: 2*6^{1} = 12. Equal! So k=2 is a solution. Therefore, k=2 implies n = k + 2 = 4. Which matches our previous result.

k=3:

Left: (3 + 2)(3 + 1) = 5*4 = 20

Right: 2*6^{2} = 2*36 = 72. Not equal.

k=4:

Left: 6*5=30

Right: 2*6^3=2*216=432. Not equal.

So only k=2 gives equality. Therefore, n=4 is the only solution. Therefore, the answer is 4.

Therefore, I think the answer is 4. Let me check once more with another approach.

Alternatively, think about the probability. The probability of exactly two dice not being 1, and the rest being 1. So, for exactly two dice showing numbers other than 1, the other n - 2 dice must show 1. So, the number of ways is C(n, 2), and each such outcome has probability (5/6)^2*(1/6)^{n - 2}. Which is exactly the same as before. So, the equation is correct.

Therefore, solving gives n=4. So, I think that's the answer.

**Final Answer**
The value of $ n $ is \boxed{4}.

To solve the problem, we are given that the probability of **exactly two out of $ n $** fair 6-sided dice showing a number **other than 1** is $ \frac{25}{216} $. We are to find the value of $ n $.

---

### Step 1: Understand the Probability Model

Each die has 6 faces, and the probability of rolling a number **other than 1** is:

$$
p = \frac{5}{6}
$$

The probability of rolling a **1** is:

$$
1 - p = \frac{1}{6}
$$

We are looking for the probability that **exactly two** of the $ n $ dice show a number other than 1. This is a classic **binomial probability** scenario.

The binomial probability formula is:

$$
P(k) = \binom{n}{k} p^k (1 - p)^{n - k}
$$

In our case, $ k = 2 $, $ p = \frac{5}{6} $, and $ 1 - p = \frac{1}{6} $. So the probability becomes:

$$
P(2) = \binom{n}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n - 2}
$$

We are told this equals $ \frac{25}{216} $, so we set up the equation:

$$
\binom{n}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216}
$$

---

### Step 2: Simplify the Equation

First, compute $ \left( \frac{5}{6} \right)^2 = \frac{25}{36} $. So the equation becomes:

$$
\binom{n}{2} \cdot \frac{25}{36} \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216}
$$

Divide both sides by $ \frac{25}{36} $:

$$
\binom{n}{2} \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216} \div \frac{25}{36} = \frac{36}{216} = \frac{1}{6}
$$

Now, recall that $ \binom{n}{2} = \frac{n(n - 1)}{2} $. Substituting:

$$
\frac{n(n - 1)}{2} \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{1}{6}
$$

Multiply both sides by 2:

$$
n(n - 1) \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{1}{3}
$$

---

### Step 3: Try Small Integer Values of $ n $

We now test small integer values of $ n $ to find which one satisfies the equation.

#### Try $ n = 4 $:

$$
n(n - 1) = 4 \cdot 3 = 12 \\
\left( \frac{1}{6} \right)^{n - 2} = \left( \frac{1}{6} \right)^2 = \frac{1}{36} \\
\text{LHS} = 12 \cdot \frac{1}{36} = \frac{12}{36} = \frac{1}{3}
$$

This matches the right-hand side. So $ n = 4 $ is a solution.

#### Try $ n = 3 $:

$$
n(n - 1) = 3 \cdot 2 = 6 \\
\left( \frac{1}{6} \right)^{1} = \frac{1}{6} \\
\text{LHS} = 6 \cdot \frac{1}{6} = 1 \neq \frac{1}{3}
$$

#### Try $ n = 5 $:

$$
n(n - 1) = 5 \cdot 4 = 20 \\
\left( \frac{1}{6} \right)^3 = \frac{1}{216} \\
\text{LHS} = 20 \cdot \frac{1}{216} = \frac{20}{216} = \frac{5}{54} \neq \frac{1}{3}
$$

So, only $ n = 4 $ satisfies the equation.

---

### Final Answer

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to find the number of dice, n, such that when you roll n fair 6-sided dice, the probability that exactly two of them show a number other than 1 is 25/216. Hmm, let me think about how to approach this.

First, let me recall some probability concepts. When dealing with probabilities involving multiple dice, especially when looking for specific numbers of successes (in this case, dice showing numbers other than 1), the binomial probability formula is usually useful. The binomial formula is:

P(k successes) = C(n, k) * p^k * (1 - p)^(n - k)

Where:
- C(n, k) is the combination of n things taken k at a time.
- p is the probability of success on a single trial.
- (1 - p) is the probability of failure.

In this problem, a "success" would be rolling a number other than 1. Let me confirm that. The problem says "exactly two of them show a number other than 1." So, each die has a probability of success (not 1) and failure (rolling a 1).

First, let me figure out the probability of success for a single die. Since each die is 6-sided, the probability of rolling a 1 is 1/6, so the probability of not rolling a 1 (i.e., rolling a 2, 3, 4, 5, or 6) is 5/6. Therefore, p = 5/6 and (1 - p) = 1/6.

But wait, the problem says the probability that exactly two of them show a number other than 1 is 25/216. So, using the binomial formula, we can set up the equation:

C(n, 2) * (5/6)^2 * (1/6)^(n - 2) = 25/216

So, I need to solve this equation for n. Let me write that down more clearly:

C(n, 2) * (25/36) * (1/6)^{n - 2} = 25/216

First, let me simplify the equation. Let me note that 25/36 is (5/6)^2, so that's correct. Let me divide both sides of the equation by 25/36 to simplify:

C(n, 2) * (1/6)^{n - 2} = (25/216) / (25/36)

Calculating the right-hand side:

(25/216) divided by (25/36) is equal to (25/216) * (36/25) = 36/216 = 1/6.

So now the equation simplifies to:

C(n, 2) * (1/6)^{n - 2} = 1/6

So, C(n, 2) * (1/6)^{n - 2} = 1/6

Let me write C(n, 2) as n(n - 1)/2. So substituting that in:

[n(n - 1)/2] * (1/6)^{n - 2} = 1/6

Let me multiply both sides by 2 to eliminate the denominator:

n(n - 1) * (1/6)^{n - 2} = 2/6 = 1/3

So:

n(n - 1) * (1/6)^{n - 2} = 1/3

Hmm, okay. Now, I need to solve for n here. Since n is an integer greater than or equal to 2 (since we are talking about exactly two dice showing numbers other than 1), let me try plugging in small integer values for n and see which one satisfies the equation.

Let me note that (1/6)^{n - 2} decreases as n increases, while n(n - 1) increases. So there might be a unique solution. Let me try n=3:

For n=3:

Left-hand side: 3*2*(1/6)^{1} = 6*(1/6) = 1. Which is 1. But the right-hand side is 1/3. So 1 ≠ 1/3. Not equal.

n=4:

Left-hand side: 4*3*(1/6)^{2} = 12*(1/36) = 12/36 = 1/3. Oh! That's equal to the right-hand side. So n=4?

Wait, so n=4 gives us 1/3, which is equal to the right-hand side. Therefore, n=4?

Wait, but let me check again. Let me go back through the steps to make sure I didn't make a mistake.

Original equation:

Probability is C(n, 2)*(5/6)^2*(1/6)^{n - 2} = 25/216.

For n=4:

C(4, 2) = 6.

(5/6)^2 = 25/36.

(1/6)^{4 - 2} = (1/6)^2 = 1/36.

Multiply all together: 6 * 25/36 * 1/36 = 6 * 25 / (36*36) = 150 / 1296.

Simplify 150/1296: divide numerator and denominator by 6: 25/216. Which is exactly the given probability. So yes, n=4 works.

But let me check n=5 just to make sure there isn't another solution.

For n=5:

C(5, 2) = 10.

(5/6)^2 = 25/36.

(1/6)^{5 - 2} = (1/6)^3 = 1/216.

Multiply all together: 10 * 25/36 * 1/216 = 250 / (36*216). Let me compute denominator: 36*216 = 7776. So 250/7776. Simplify: divide numerator and denominator by 2: 125/3888. Let me check if that's equal to 25/216. 25/216 is equal to 500/4320, and 125/3888 is approximately... Let me convert both to decimal:

25/216 ≈ 0.1157

125/3888 ≈ 0.0321. Not equal. So n=5 is not a solution.

What about n=2? Let me check n=2:

C(2, 2) = 1.

(5/6)^2 = 25/36.

(1/6)^{0} = 1.

Multiply all together: 1 * 25/36 * 1 = 25/36 ≈ 0.694, which is not 25/216 ≈ 0.1157. So no.

n=3: as before, gives 1, which is 1, which is not 25/216. So n=4 is the only solution?

Wait, but let me check n=6 just to be thorough.

n=6:

C(6, 2) = 15.

(5/6)^2 = 25/36.

(1/6)^{4} = 1/1296.

Multiply all together: 15 * 25 / 36 * 1 / 1296 = 375 / (36*1296). Let me compute denominator: 36*1296 = 46656. So 375/46656 ≈ 0.008, which is way less than 25/216. So nope.

So seems like n=4 is the only solution. Therefore, the answer is 4?

But let me check again my steps to make sure I didn't make any miscalculations.

Original equation:

C(n, 2)*(5/6)^2*(1/6)^{n - 2} = 25/216.

For n=4:

C(4,2)=6. (5/6)^2=25/36. (1/6)^{2}=1/36. Multiply them: 6*(25/36)*(1/36)= 6*25/(36*36)= 150/1296. Simplify numerator and denominator by dividing by 6: 25/216. Which matches. So yes, correct.

Therefore, n=4. So the answer is 4? Wait, but let me think again. The problem says "the probability that exactly two of them show a number other than 1 is 25/216". So, is there another possible n?

Wait, let me think again. Suppose n=3. Then, the probability of exactly two dice showing numbers other than 1. So, with n=3, the probability would be C(3,2)*(5/6)^2*(1/6)^1 = 3*(25/36)*(1/6) = 75/216 = 25/72 ≈ 0.347. Which is bigger than 25/216. For n=4, it's 25/216. For n=5, as we saw, it's 125/3888 ≈ 0.032, which is smaller. So, seems like n=4 is the only solution. Therefore, the answer is 4.

But let me check if there's another way to approach this problem, maybe through equations, to confirm.

We had earlier:

n(n - 1) * (1/6)^{n - 2} = 1/3

Let me write that equation again:

n(n - 1) * (1/6)^{n - 2} = 1/3

Let me denote k = n - 2. Then n = k + 2. Substitute into the equation:

(k + 2)(k + 1) * (1/6)^k = 1/3

So:

(k + 2)(k + 1) / 6^k = 1/3

Multiply both sides by 6^k:

(k + 2)(k + 1) = (6^k)/3 = 2*6^{k - 1}

So:

(k + 2)(k + 1) = 2*6^{k - 1}

Now, let me try small integer values for k. Since n must be at least 2, k = n - 2 ≥ 0. So k ≥ 0.

Let me try k=0:

Left side: (0 + 2)(0 + 1) = 2*1 = 2

Right side: 2*6^{-1} = 2*(1/6) = 1/3. Not equal.

k=1:

Left: (1 + 2)(1 + 1) = 3*2 = 6

Right: 2*6^{0} = 2*1 = 2. Not equal.

k=2:

Left: (2 + 2)(2 + 1) = 4*3 = 12

Right: 2*6^{1} = 12. Equal! So k=2 is a solution. Therefore, k=2 implies n = k + 2 = 4. Which matches our previous result.

k=3:

Left: (3 + 2)(3 + 1) = 5*4 = 20

Right: 2*6^{2} = 2*36 = 72. Not equal.

k=4:

Left: 6*5=30

Right: 2*6^3=2*216=432. Not equal.

So only k=2 gives equality. Therefore, n=4 is the only solution. Therefore, the answer is 4.

Therefore, I think the answer is 4. Let me check once more with another approach.

Alternatively, think about the probability. The probability of exactly two dice not being 1, and the rest being 1. So, for exactly two dice showing numbers other than 1, the other n - 2 dice must show 1. So, the number of ways is C(n, 2), and each such outcome has probability (5/6)^2*(1/6)^{n - 2}. Which is exactly the same as before. So, the equation is correct.

Therefore, solving gives n=4. So, I think that's the answer.

**Final Answer**
The value of $ n $ is \boxed{4}.

To solve the problem, we are given that the probability of **exactly two out of $ n $** fair 6-sided dice showing a number **other than 1** is $ \frac{25}{216} $. We are to find the value of $ n $.

---

### Step 1: Understand the Probability Model

Each die has 6 faces, and the probability of rolling a number **other than 1** is:

$$
p = \frac{5}{6}
$$

The probability of rolling a **1** is:

$$
1 - p = \frac{1}{6}
$$

We are looking for the probability that **exactly two** of the $ n $ dice show a number other than 1. This is a classic **binomial probability** scenario.

The binomial probability formula is:

$$
P(k) = \binom{n}{k} p^k (1 - p)^{n - k}
$$

In our case, $ k = 2 $, $ p = \frac{5}{6} $, and $ 1 - p = \frac{1}{6} $. So the probability becomes:

$$
P(2) = \binom{n}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n - 2}
$$

We are told this equals $ \frac{25}{216} $, so we set up the equation:

$$
\binom{n}{2} \left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216}
$$

---

### Step 2: Simplify the Equation

First, compute $ \left( \frac{5}{6} \right)^2 = \frac{25}{36} $. So the equation becomes:

$$
\binom{n}{2} \cdot \frac{25}{36} \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216}
$$

Divide both sides by $ \frac{25}{36} $:

$$
\binom{n}{2} \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{25}{216} \div \frac{25}{36} = \frac{36}{216} = \frac{1}{6}
$$

Now, recall that $ \binom{n}{2} = \frac{n(n - 1)}{2} $. Substituting:

$$
\frac{n(n - 1)}{2} \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{1}{6}
$$

Multiply both sides by 2:

$$
n(n - 1) \cdot \left( \frac{1}{6} \right)^{n - 2} = \frac{1}{3}
$$

---

### Step 3: Try Small Integer Values of $ n $

We now test small integer values of $ n $ to find which one satisfies the equation.

#### Try $ n = 4 $:

$$
n(n - 1) = 4 \cdot 3 = 12 \\
\left( \frac{1}{6} \right)^{n - 2} = \left( \frac{1}{6} \right)^2 = \frac{1}{36} \\
\text{LHS} = 12 \cdot \frac{1}{36} = \frac{12}{36} = \frac{1}{3}
$$

This matches the right-hand side. So $ n = 4 $ is a solution.

#### Try $ n = 3 $:

$$
n(n - 1) = 3 \cdot 2 = 6 \\
\left( \frac{1}{6} \right)^{1} = \frac{1}{6} \\
\text{LHS} = 6 \cdot \frac{1}{6} = 1 \neq \frac{1}{3}
$$

#### Try $ n = 5 $:

$$
n(n - 1) = 5 \cdot 4 = 20 \\
\left( \frac{1}{6} \right)^3 = \frac{1}{216} \\
\text{LHS} = 20 \cdot \frac{1}{216} = \frac{20}{216} = \frac{5}{54} \neq \frac{1}{3}
$$

So, only $ n = 4 $ satisfies the equation.

---

### Final Answer

$$
\boxed{4}
$$

Solution: For $n$ dice, there are $\binom{n}{2}=\frac{n(n-1)}{2}$ ways to choose two of them. For each of these ways, there is a $\left( \frac{5}{6} \right)^2 \left( \frac{1}{6} \right)^{n-2}$ chance that all but the chosen two dice will roll a 1. Therefore, we need to find the value of $n$ for which $\frac{25n(n-1)}{2 \cdot 6^n}=\frac{25}{216}$, or $108n(n-1)=6^n$. Plugging in values for $n$, we can see that $n=\boxed{4}$ works and no value of $n$ less than 4 works. Now we just need to prove that no values of $n$ greater than 4 work. Note that if $n \geq 5$, then $n < 3^{n-3}$ and $n-1 < 2\cdot 2^{n-3}$. We can multiply these inequalities to get that when $n \geq 5$, we have $n(n-1) < 2\cdot 6^{n-3}$, or $108n(n-1)<6^n$.


Explanation:
Okay, so Karla drove her vehicle 12,000 miles last year. The question is asking how many gallons of gasoline she would have saved if she had driven a fuel-efficient hybrid car instead of an SUV. The hybrid averages 48 miles per gallon, and the SUV averages 15 miles per gallon. Alright, let me think about how to approach this.

First, I need to figure out how much gasoline she used with the SUV and then how much she would have used with the hybrid. Then, subtract the hybrid's gasoline usage from the SUV's to find the savings. That makes sense. Let me write that down.

So, the formula for gasoline used is total miles divided by miles per gallon. That gives gallons. So, for the SUV, gallons used would be 12,000 miles divided by 15 miles per gallon. Let me calculate that. 12,000 divided by 15. Hmm, 15 times 800 is 12,000, right? Because 15 times 100 is 1,500, so 15 times 800 is 12,000. So, that would be 800 gallons for the SUV.

Now, for the hybrid car, it's 48 miles per gallon. So, the gasoline used would be 12,000 divided by 48. Let me do that division. Hmm, 48 times 250 is 12,000? Let me check. 48 times 200 is 9,600. Then 48 times 50 is 2,400. So 9,600 + 2,400 is 12,000. Yes, so 250 gallons for the hybrid.

Therefore, the difference between the SUV and the hybrid is 800 gallons minus 250 gallons. That would be 550 gallons. So, she would have saved 550 gallons. Let me just verify my calculations again to make sure I didn't make a mistake.

First, SUV: 12,000 / 15. Let me do 12,000 divided by 15. Dividing both numerator and denominator by 5 gives 2,400 / 3, which is 800. Yep, that's correct. For the hybrid: 12,000 / 48. Let me divide numerator and denominator by 12 first. 12,000 divided by 12 is 1,000. 48 divided by 12 is 4. So now it's 1,000 / 4, which is 250. That checks out. So 800 - 250 is indeed 550.

Wait, but let me think again. Is there another way to approach this problem? Maybe by calculating the difference in fuel efficiency and then applying it somehow? But I don't think that's straightforward because the fuel efficiency is different. The total gallons saved depends on the total miles driven and the difference in fuel efficiency. But maybe I can think in terms of gallons per mile?

For the SUV, it's 1/15 gallons per mile, and for the hybrid, it's 1/48 gallons per mile. So the difference in gallons per mile is 1/15 - 1/48. Then multiply that by total miles to get total gallons saved. Let me try that method to cross-verify.

Calculating 1/15 - 1/48. To subtract these fractions, find a common denominator. The least common denominator of 15 and 48. Let's see, prime factors of 15 are 3 and 5. Prime factors of 48 are 2^4 and 3. So LCM would be 2^4 * 3 * 5 = 16 * 3 * 5 = 240. So converting both fractions:

1/15 is equal to 16/240 (since 15*16=240). 1/48 is equal to 5/240 (since 48*5=240). Therefore, 16/240 - 5/240 = 11/240 gallons per mile. Then multiply by total miles, 12,000 miles. So 12,000 * 11 / 240. Let me compute that.

First, simplify 12,000 / 240. 240 times 50 is 12,000. So 12,000 / 240 = 50. Then multiply by 11: 50 * 11 = 550. So same result, 550 gallons. Okay, that confirms the answer. So regardless of the method, I get 550 gallons saved.

Therefore, I think that's the correct answer. Let me just check once more for any possible miscalculations.

Alternatively, maybe I can think about how many gallons each car uses. For the SUV, 12,000 divided by 15. Let me do 12,000 divided by 15 step by step. 15 goes into 120 (the first three digits) 8 times (15*8=120). Then bring down the next 0, making it 0. So 15 goes into 0 zero times. Then bring down the next 0, making it 0 again. So total is 800. Yep. For the hybrid, 12,000 divided by 48. Let me do 48 into 12,000. 48 into 120 is 2 times (48*2=96), remainder 24. Bring down the 0, making it 240. 48 into 240 is 5 times (48*5=240). So that's 25. Then bring down the next 0, which is 0. So total is 250. Yep. So 800 - 250 = 550.

Therefore, I think I can be confident that the answer is 550 gallons.

Another way to think about it: if the SUV gets 15 mpg, then for every gallon, it can go 15 miles. So, 12,000 miles would require 12,000 / 15 = 800 gallons. The hybrid gets 48 mpg, so for every gallon, it can go 48 miles. Therefore, 12,000 miles would require 12,000 / 48 = 250 gallons. The difference is 800 - 250 = 550 gallons saved.

All methods lead to the same answer. Therefore, I think the answer is 550 gallons.

**Final Answer**
Karla would have saved \boxed{550} gallons of gasoline last year.

To determine how many gallons of gasoline Karla would have saved last year by driving a fuel-efficient hybrid car instead of an SUV, we need to compare the fuel consumption of both vehicles over the same distance of 12,000 miles.

---

### Step 1: Calculate Gasoline Used by the SUV

The SUV averages **15 miles per gallon**. To find out how many gallons it used for 12,000 miles:

$$
\text{Gallons used by SUV} = \frac{12,000 \text{ miles}}{15 \text{ miles per gallon}} = 800 \text{ gallons}
$$

---

### Step 2: Calculate Gasoline Used by the Hybrid

The hybrid averages **48 miles per gallon**. To find out how many gallons it would use for the same distance:

$$
\text{Gallons used by hybrid} = \frac{12,000 \text{ miles}}{48 \text{ miles per gallon}} = 250 \text{ gallons}
$$

---

### Step 3: Calculate the Difference (Gasoline Saved)

To find the amount of gasoline saved:

$$
\text{Gasoline saved} = 800 \text{ gallons} - 250 \text{ gallons} = 550 \text{ gallons}
$$

---

### Final Answer

$$
\boxed{550}
$$

[Assistant]
Explanation:
Okay, so Karla drove her vehicle 12,000 miles last year. The question is asking how many gallons of gasoline she would have saved if she had driven a fuel-efficient hybrid car instead of an SUV. The hybrid averages 48 miles per gallon, and the SUV averages 15 miles per gallon. Alright, let me think about how to approach this.

First, I need to figure out how much gasoline she used with the SUV and then how much she would have used with the hybrid. Then, subtract the hybrid's gasoline usage from the SUV's to find the savings. That makes sense. Let me write that down.

So, the formula for gasoline used is total miles divided by miles per gallon. That gives gallons. So, for the SUV, gallons used would be 12,000 miles divided by 15 miles per gallon. Let me calculate that. 12,000 divided by 15. Hmm, 15 times 800 is 12,000, right? Because 15 times 100 is 1,500, so 15 times 800 is 12,000. So, that would be 800 gallons for the SUV.

Now, for the hybrid car, it's 48 miles per gallon. So, the gasoline used would be 12,000 divided by 48. Let me do that division. Hmm, 48 times 250 is 12,000? Let me check. 48 times 200 is 9,600. Then 48 times 50 is 2,400. So 9,600 + 2,400 is 12,000. Yes, so 250 gallons for the hybrid.

Therefore, the difference between the SUV and the hybrid is 800 gallons minus 250 gallons. That would be 550 gallons. So, she would have saved 550 gallons. Let me just verify my calculations again to make sure I didn't make a mistake.

First, SUV: 12,000 / 15. Let me do 12,000 divided by 15. Dividing both numerator and denominator by 5 gives 2,400 / 3, which is 800. Yep, that's correct. For the hybrid: 12,000 / 48. Let me divide numerator and denominator by 12 first. 12,000 divided by 12 is 1,000. 48 divided by 12 is 4. So now it's 1,000 / 4, which is 250. That checks out. So 800 - 250 is indeed 550.

Wait, but let me think again. Is there another way to approach this problem? Maybe by calculating the difference in fuel efficiency and then applying it somehow? But I don't think that's straightforward because the fuel efficiency is different. The total gallons saved depends on the total miles driven and the difference in fuel efficiency. But maybe I can think in terms of gallons per mile?

For the SUV, it's 1/15 gallons per mile, and for the hybrid, it's 1/48 gallons per mile. So the difference in gallons per mile is 1/15 - 1/48. Then multiply that by total miles to get total gallons saved. Let me try that method to cross-verify.

Calculating 1/15 - 1/48. To subtract these fractions, find a common denominator. The least common denominator of 15 and 48. Let's see, prime factors of 15 are 3 and 5. Prime factors of 48 are 2^4 and 3. So LCM would be 2^4 * 3 * 5 = 16 * 3 * 5 = 240. So converting both fractions:

1/15 is equal to 16/240 (since 15*16=240). 1/48 is equal to 5/240 (since 48*5=240). Therefore, 16/240 - 5/240 = 11/240 gallons per mile. Then multiply by total miles, 12,000 miles. So 12,000 * 11 / 240. Let me compute that.

First, simplify 12,000 / 240. 240 times 50 is 12,000. So 12,000 / 240 = 50. Then multiply by 11: 50 * 11 = 550. So same result, 550 gallons. Okay, that confirms the answer. So regardless of the method, I get 550 gallons saved.

Therefore, I think that's the correct answer. Let me just check once more for any possible miscalculations.

Alternatively, maybe I can think about how many gallons each car uses. For the SUV, 12,000 divided by 15. Let me do 12,000 divided by 15 step by step. 15 goes into 120 (the first three digits) 8 times (15*8=120). Then bring down the next 0, making it 0. So 15 goes into 0 zero times. Then bring down the next 0, making it 0 again. So total is 800. Yep. For the hybrid, 12,000 divided by 48. Let me do 48 into 12,000. 48 into 120 is 2 times (48*2=96), remainder 24. Bring down the 0, making it 240. 48 into 240 is 5 times (48*5=240). So that's 25. Then bring down the next 0, which is 0. So total is 250. Yep. So 800 - 250 = 550.

Therefore, I think I can be confident that the answer is 550 gallons.

Another way to think about it: if the SUV gets 15 mpg, then for every gallon, it can go 15 miles. So, 12,000 miles would require 12,000 / 15 = 800 gallons. The hybrid gets 48 mpg, so for every gallon, it can go 48 miles. Therefore, 12,000 miles would require 12,000 / 48 = 250 gallons. The difference is 800 - 250 = 550 gallons saved.

All methods lead to the same answer. Therefore, I think the answer is 550 gallons.

**Final Answer**
Karla would have saved \boxed{550} gallons of gasoline last year.

To determine how many gallons of gasoline Karla would have saved last year by driving a fuel-efficient hybrid car instead of an SUV, we need to compare the fuel consumption of both vehicles over the same distance of 12,000 miles.

---

### Step 1: Calculate Gasoline Used by the SUV

The SUV averages **15 miles per gallon**. To find out how many gallons it used for 12,000 miles:

$$
\text{Gallons used by SUV} = \frac{12,000 \text{ miles}}{15 \text{ miles per gallon}} = 800 \text{ gallons}
$$

---

### Step 2: Calculate Gasoline Used by the Hybrid

The hybrid averages **48 miles per gallon**. To find out how many gallons it would use for the same distance:

$$
\text{Gallons used by hybrid} = \frac{12,000 \text{ miles}}{48 \text{ miles per gallon}} = 250 \text{ gallons}
$$

---

### Step 3: Calculate the Difference (Gasoline Saved)

To find the amount of gasoline saved:

$$
\text{Gasoline saved} = 800 \text{ gallons} - 250 \text{ gallons} = 550 \text{ gallons}
$$

---

### Final Answer

$$
\boxed{550}
$$

Solution: Karla used $12000/15 = 800$ gallons to drive 12000 miles. Had she driven the hybrid, she would have used $12000/48 = 250$ gallons. Therefore, she would have saved $800 - 250 = \boxed{550}$ gallons.


Explanation:
Okay, so I need to find the square root of f(f(f(f(1)))), where f(x) is 2^x. Let me break this down step by step. First, let me make sure I understand the problem correctly. The function f is defined as f(x) = 2^x, which is an exponential function with base 2. The question is asking for the square root of the result of applying this function four times in succession, starting with the input 1. So, it's like f(f(f(f(1)))), and then take the square root of that.

Let me write that out more clearly. Let me denote each step:

First, compute f(1). Then, take that result and plug it into f again to get f(f(1)). Then take that result and plug it into f again to get f(f(f(1))), and then one more time to get f(f(f(f(1)))). After that, take the square root of that final value.

So, step by step:

1. Compute f(1) = 2^1 = 2.
2. Then compute f(f(1)) = f(2) = 2^2 = 4.
3. Next, compute f(f(f(1))) = f(4) = 2^4 = 16.
4. Then compute f(f(f(f(1)))) = f(16) = 2^16.
5. Finally, take the square root of 2^16.

Wait, let me check that again. Let me make sure I didn't skip any steps or make a miscalculation. Let me go through each step again.

Starting with f(1):

f(1) = 2^1 = 2. That seems straightforward.

Then f(f(1)) is f(2). Since f(2) = 2^2 = 4. Correct.

Then f(f(f(1))) is f(4). So f(4) = 2^4. Let me compute 2^4. 2*2=4, 4*2=8, 8*2=16. So 2^4 is 16. Correct.

Then f(f(f(f(1)))) is f(16). So f(16) = 2^16. Now, 2^16 is a larger number. Let me recall that 2^10 is 1024, so 2^16 is 2^10 * 2^6 = 1024 * 64. Let me compute that. 1024 * 64. Well, 1000*64 = 64,000 and 24*64 = 1,536. So adding those together: 64,000 + 1,536 = 65,536. Therefore, 2^16 is 65,536.

So then, the problem asks for the square root of that. So sqrt(65,536). Let me compute that.

But maybe there's a smarter way than just calculating 65,536 and then taking the square root. Let me think. Since 2^16 is (2^8)^2, because (2^8)^2 = 2^(8*2) = 2^16. Therefore, sqrt(2^16) is 2^(16/2) = 2^8. Which is 256.

Wait, that seems much simpler. So instead of calculating 2^16 first, which is 65,536, and then taking the square root, I can recognize that sqrt(2^16) is 2^8. Therefore, the answer is 2^8, which is 256.

But let me verify that. Let me check if sqrt(2^16) is indeed 2^8. Since sqrt(a^b) is a^(b/2), so sqrt(2^16) = 2^(16/2) = 2^8. Yes, that's correct. Therefore, the square root of 2^16 is 2^8, which is 256.

Therefore, the answer is 256.

But let me just make sure I didn't make a mistake in the earlier steps. Let me retrace:

Start with f(1) = 2^1 = 2. Then f(f(1)) = f(2) = 2^2 = 4. Then f(f(f(1))) = f(4) = 2^4 = 16. Then f(f(f(f(1)))) = f(16) = 2^16. Then sqrt(2^16) = 2^8 = 256.

Yes, that seems correct.

Alternatively, maybe I can think of the entire expression as follows. Let me denote each application of f:

Let me define:

Let’s let’s denote:

First application: f(1) = 2^1 = 2.

Second application: f(f(1)) = f(2) = 2^2 = 4.

Third application: f(f(f(1))) = f(4) = 2^4 = 16.

Fourth application: f(f(f(f(1)))) = f(16) = 2^16.

Therefore, sqrt(f(f(f(f(1))))) = sqrt(2^16) = 2^(16/2) = 2^8 = 256.

Yes, that seems consistent.

Alternatively, maybe I can think in terms of exponents. Let me think about how each application of f affects the exponent. Since f(x) = 2^x, so if I have f applied to some value, say, 2^k, then f(2^k) = 2^(2^k). Wait, no. Wait, f(x) = 2^x. So if x is 2^k, then f(x) = 2^(2^k).

But in our case, starting with 1, which is 2^0. Let me see:

First, f(1) = 2^1 = 2.

But 1 is 2^0. Then f(2^0) = 2^(2^0) = 2^1 = 2.

Then f(f(1)) = f(2) = 2^2 = 4. Which is 2^(2^1).

Then f(f(f(1))) = f(4) = 2^4 = 16, which is 2^(2^2).

Then f(f(f(f(1)))) = f(16) = 2^16 = 2^(2^4).

So, in general, applying f n times to 1 gives 2^(2^(2^(...))) with n exponents. Wait, but here, starting with 1, which is 2^0, then applying f once gives 2^1, then applying f again gives 2^(2^1), then applying f again gives 2^(2^(2^1)), and so on.

So, in this case, applying f four times would be 2^(2^(2^(2^1))). Wait, but hold on, let me check:

Wait, first application: f(1) = 2^1.

Second application: f(f(1)) = f(2^1) = 2^(2^1) = 2^2.

Third application: f(f(f(1))) = f(2^2) = 2^(2^2) = 2^4.

Fourth application: f(f(f(f(1)))) = f(2^4) = 2^(2^4) = 2^16.

Yes, so that's the same as before. Therefore, the fourth application is 2^16, so sqrt(2^16) is 2^8 = 256.

Alternatively, if I think of the tower of exponents, but in this case, since we are only applying f four times, starting from 1, it's 2^(2^(2^(2^1))) but with the exponents evaluated from the top down? Wait, no. Wait, actually, when you apply f four times, each time you're taking the previous result and exponentiating 2 to that result. So:

Start with 1.

First application: 2^1.

Second application: 2^(previous result) = 2^(2^1).

Third application: 2^(previous result) = 2^(2^(2^1)).

Fourth application: 2^(previous result) = 2^(2^(2^(2^1))).

But 2^1 is 2, so:

Fourth application: 2^(2^(2^2)) = 2^(2^4) = 2^16.

Yes, same as before. Therefore, sqrt(2^16) is 2^8.

Alternatively, maybe I can think of the entire expression as follows:

Let me denote the innermost function first. Let me write:

Let’s define:

Let’s let’s compute f(f(f(f(1)))) step by step:

Start with 1.

First, compute f(1) = 2^1 = 2.

Second, compute f(2) = 2^2 = 4.

Third, compute f(4) = 2^4 = 16.

Fourth, compute f(16) = 2^16 = 65536.

Then take the square root of 65536.

Now, sqrt(65536). Let me check what is sqrt(65536). Well, 256^2 is 65536 because 256*256. Let me verify:

256 * 256. Well, 256 is 2^8, so 2^8 * 2^8 = 2^16 = 65536. Therefore, sqrt(2^16) is 2^8 = 256.

Therefore, the answer is 256.

Alternatively, if I didn't remember that 2^16 is 65536, I could compute sqrt(65536) by recognizing that 256 squared is 65536. Let me check 256*256:

First, 200*200 = 40,000.

Then, 200*56 = 11,200.

Then, 56*200 = 11,200.

Then, 56*56 = 3,136.

Wait, but that's if I do (200 + 56)^2. Wait, but 256 is 200 + 56? Wait, no. Wait, 256 is 256. Maybe breaking it down differently. Let me do 256 * 256:

Well, 256 * 200 = 51,200.

256 * 50 = 12,800.

256 * 6 = 1,536.

Adding those together: 51,200 + 12,800 = 64,000. Then 64,000 + 1,536 = 65,536. Yes, that's correct. Therefore, sqrt(65,536) is 256.

Therefore, regardless of the method, the answer is 256.

But let me check once again if I didn't miscount the number of function applications. The problem says sqrt(f(f(f(f(1))))). So that's four f's. Let me count:

First f: f(1)

Second f: f(f(1))

Third f: f(f(f(1)))

Fourth f: f(f(f(f(1))))

Yes, four applications. So that's correct.

Alternatively, if I had miscounted and thought it was three applications, then the answer would be different, but no, the problem says four.

Alternatively, maybe I can think of the entire expression as follows:

Let me note that f(f(f(f(1)))) = 2^(2^(2^(2^1))) as per the tower of exponents. Wait, but earlier we saw that it's 2^(2^(2^2)) which is 2^(2^4) which is 2^16. Wait, but if you think of the tower, it's 2^(2^(2^(2^1))) which is 2^(2^(2^2)) which is 2^(2^4) which is 2^16. So that's the same as before.

Therefore, sqrt(2^16) is 2^8.

Alternatively, if I think of exponents, since sqrt is the same as raising to the power of 1/2, so sqrt(2^16) = (2^16)^(1/2) = 2^(16*(1/2)) = 2^8.

Yes, that's another way to see it.

Therefore, regardless of the approach, the answer is 2^8, which is 256.

Therefore, I think I can be confident that the answer is 256.

But just to make sure, let me try to compute each step numerically again:

Start with 1.

First f(1) = 2^1 = 2.

Second f(2) = 2^2 = 4.

Third f(4) = 2^4 = 16.

Fourth f(16) = 2^16.

Now, 2^10 is 1024, so 2^16 is 2^10 * 2^6 = 1024 * 64.

Calculating 1024 * 64:

Well, 1024 * 60 = 61,440.

1024 * 4 = 4,096.

Adding them together: 61,440 + 4,096 = 65,536.

So, 2^16 is 65,536.

Then sqrt(65,536). Let me check if 256 squared is 65,536.

As above, 256 * 256 = 65,536.

Alternatively, 256 is 2^8, so 2^8 * 2^8 = 2^16, which is 65,536. Therefore, sqrt(2^16) is 2^8.

Therefore, the answer is 2^8 = 256.

Therefore, I think I have verified this multiple times through different methods: breaking down the exponents, calculating step by step, using properties of exponents and square roots, and even computing the numerical value. All approaches lead to the same conclusion.

Therefore, I can be confident that the answer is 256.

**Final Answer**
\boxed{256}

To solve the expression
$$
\sqrt{f(f(f(f(1))))},
$$
where $ f(x) = 2^x $, we proceed step by step, applying the function repeatedly and simplifying as we go.

---

### Step 1: Compute $ f(1) $

$$
f(1) = 2^1 = 2
$$

---

### Step 2: Compute $ f(f(1)) = f(2) $

$$
f(2) = 2^2 = 4
$$

---

### Step 3: Compute $ f(f(f(1))) = f(4) $

$$
f(4) = 2^4 = 16
$$

---

### Step 4: Compute $ f(f(f(f(1)))) = f(16) $

$$
f(16) = 2^{16}
$$

---

### Step 5: Take the square root of $ 2^{16} $

We use the property of exponents:
$$
\sqrt{2^{16}} = 2^{16/2} = 2^8
$$

Now compute $ 2^8 $:
$$
2^8 = 256
$$

---

### Final Answer

$$
\boxed{256}
$$

[Assistant]
Explanation:
Okay, so I need to find the square root of f(f(f(f(1)))), where f(x) is 2^x. Let me break this down step by step. First, let me make sure I understand the problem correctly. The function f is defined as f(x) = 2^x, which is an exponential function with base 2. The question is asking for the square root of the result of applying this function four times in succession, starting with the input 1. So, it's like f(f(f(f(1)))), and then take the square root of that.

Let me write that out more clearly. Let me denote each step:

First, compute f(1). Then, take that result and plug it into f again to get f(f(1)). Then take that result and plug it into f again to get f(f(f(1))), and then one more time to get f(f(f(f(1)))). After that, take the square root of that final value.

So, step by step:

1. Compute f(1) = 2^1 = 2.
2. Then compute f(f(1)) = f(2) = 2^2 = 4.
3. Next, compute f(f(f(1))) = f(4) = 2^4 = 16.
4. Then compute f(f(f(f(1)))) = f(16) = 2^16.
5. Finally, take the square root of 2^16.

Wait, let me check that again. Let me make sure I didn't skip any steps or make a miscalculation. Let me go through each step again.

Starting with f(1):

f(1) = 2^1 = 2. That seems straightforward.

Then f(f(1)) is f(2). Since f(2) = 2^2 = 4. Correct.

Then f(f(f(1))) is f(4). So f(4) = 2^4. Let me compute 2^4. 2*2=4, 4*2=8, 8*2=16. So 2^4 is 16. Correct.

Then f(f(f(f(1)))) is f(16). So f(16) = 2^16. Now, 2^16 is a larger number. Let me recall that 2^10 is 1024, so 2^16 is 2^10 * 2^6 = 1024 * 64. Let me compute that. 1024 * 64. Well, 1000*64 = 64,000 and 24*64 = 1,536. So adding those together: 64,000 + 1,536 = 65,536. Therefore, 2^16 is 65,536.

So then, the problem asks for the square root of that. So sqrt(65,536). Let me compute that.

But maybe there's a smarter way than just calculating 65,536 and then taking the square root. Let me think. Since 2^16 is (2^8)^2, because (2^8)^2 = 2^(8*2) = 2^16. Therefore, sqrt(2^16) is 2^(16/2) = 2^8. Which is 256.

Wait, that seems much simpler. So instead of calculating 2^16 first, which is 65,536, and then taking the square root, I can recognize that sqrt(2^16) is 2^8. Therefore, the answer is 2^8, which is 256.

But let me verify that. Let me check if sqrt(2^16) is indeed 2^8. Since sqrt(a^b) is a^(b/2), so sqrt(2^16) = 2^(16/2) = 2^8. Yes, that's correct. Therefore, the square root of 2^16 is 2^8, which is 256.

Therefore, the answer is 256.

But let me just make sure I didn't make a mistake in the earlier steps. Let me retrace:

Start with f(1) = 2^1 = 2. Then f(f(1)) = f(2) = 2^2 = 4. Then f(f(f(1))) = f(4) = 2^4 = 16. Then f(f(f(f(1)))) = f(16) = 2^16. Then sqrt(2^16) = 2^8 = 256.

Yes, that seems correct.

Alternatively, maybe I can think of the entire expression as follows. Let me denote each application of f:

Let me define:

Let’s let’s denote:

First application: f(1) = 2^1 = 2.

Second application: f(f(1)) = f(2) = 2^2 = 4.

Third application: f(f(f(1))) = f(4) = 2^4 = 16.

Fourth application: f(f(f(f(1)))) = f(16) = 2^16.

Therefore, sqrt(f(f(f(f(1))))) = sqrt(2^16) = 2^(16/2) = 2^8 = 256.

Yes, that seems consistent.

Alternatively, maybe I can think in terms of exponents. Let me think about how each application of f affects the exponent. Since f(x) = 2^x, so if I have f applied to some value, say, 2^k, then f(2^k) = 2^(2^k). Wait, no. Wait, f(x) = 2^x. So if x is 2^k, then f(x) = 2^(2^k).

But in our case, starting with 1, which is 2^0. Let me see:

First, f(1) = 2^1 = 2.

But 1 is 2^0. Then f(2^0) = 2^(2^0) = 2^1 = 2.

Then f(f(1)) = f(2) = 2^2 = 4. Which is 2^(2^1).

Then f(f(f(1))) = f(4) = 2^4 = 16, which is 2^(2^2).

Then f(f(f(f(1)))) = f(16) = 2^16 = 2^(2^4).

So, in general, applying f n times to 1 gives 2^(2^(2^(...))) with n exponents. Wait, but here, starting with 1, which is 2^0, then applying f once gives 2^1, then applying f again gives 2^(2^1), then applying f again gives 2^(2^(2^1)), and so on.

So, in this case, applying f four times would be 2^(2^(2^(2^1))). Wait, but hold on, let me check:

Wait, first application: f(1) = 2^1.

Second application: f(f(1)) = f(2^1) = 2^(2^1) = 2^2.

Third application: f(f(f(1))) = f(2^2) = 2^(2^2) = 2^4.

Fourth application: f(f(f(f(1)))) = f(2^4) = 2^(2^4) = 2^16.

Yes, so that's the same as before. Therefore, the fourth application is 2^16, so sqrt(2^16) is 2^8 = 256.

Alternatively, if I think of the tower of exponents, but in this case, since we are only applying f four times, starting from 1, it's 2^(2^(2^(2^1))) but with the exponents evaluated from the top down? Wait, no. Wait, actually, when you apply f four times, each time you're taking the previous result and exponentiating 2 to that result. So:

Start with 1.

First application: 2^1.

Second application: 2^(previous result) = 2^(2^1).

Third application: 2^(previous result) = 2^(2^(2^1)).

Fourth application: 2^(previous result) = 2^(2^(2^(2^1))).

But 2^1 is 2, so:

Fourth application: 2^(2^(2^2)) = 2^(2^4) = 2^16.

Yes, same as before. Therefore, sqrt(2^16) is 2^8.

Alternatively, maybe I can think of the entire expression as follows:

Let me denote the innermost function first. Let me write:

Let’s define:

Let’s let’s compute f(f(f(f(1)))) step by step:

Start with 1.

First, compute f(1) = 2^1 = 2.

Second, compute f(2) = 2^2 = 4.

Third, compute f(4) = 2^4 = 16.

Fourth, compute f(16) = 2^16 = 65536.

Then take the square root of 65536.

Now, sqrt(65536). Let me check what is sqrt(65536). Well, 256^2 is 65536 because 256*256. Let me verify:

256 * 256. Well, 256 is 2^8, so 2^8 * 2^8 = 2^16 = 65536. Therefore, sqrt(2^16) is 2^8 = 256.

Therefore, the answer is 256.

Alternatively, if I didn't remember that 2^16 is 65536, I could compute sqrt(65536) by recognizing that 256 squared is 65536. Let me check 256*256:

First, 200*200 = 40,000.

Then, 200*56 = 11,200.

Then, 56*200 = 11,200.

Then, 56*56 = 3,136.

Wait, but that's if I do (200 + 56)^2. Wait, but 256 is 200 + 56? Wait, no. Wait, 256 is 256. Maybe breaking it down differently. Let me do 256 * 256:

Well, 256 * 200 = 51,200.

256 * 50 = 12,800.

256 * 6 = 1,536.

Adding those together: 51,200 + 12,800 = 64,000. Then 64,000 + 1,536 = 65,536. Yes, that's correct. Therefore, sqrt(65,536) is 256.

Therefore, regardless of the method, the answer is 256.

But let me check once again if I didn't miscount the number of function applications. The problem says sqrt(f(f(f(f(1))))). So that's four f's. Let me count:

First f: f(1)

Second f: f(f(1))

Third f: f(f(f(1)))

Fourth f: f(f(f(f(1))))

Yes, four applications. So that's correct.

Alternatively, if I had miscounted and thought it was three applications, then the answer would be different, but no, the problem says four.

Alternatively, maybe I can think of the entire expression as follows:

Let me note that f(f(f(f(1)))) = 2^(2^(2^(2^1))) as per the tower of exponents. Wait, but earlier we saw that it's 2^(2^(2^2)) which is 2^(2^4) which is 2^16. Wait, but if you think of the tower, it's 2^(2^(2^(2^1))) which is 2^(2^(2^2)) which is 2^(2^4) which is 2^16. So that's the same as before.

Therefore, sqrt(2^16) is 2^8.

Alternatively, if I think of exponents, since sqrt is the same as raising to the power of 1/2, so sqrt(2^16) = (2^16)^(1/2) = 2^(16*(1/2)) = 2^8.

Yes, that's another way to see it.

Therefore, regardless of the approach, the answer is 2^8, which is 256.

Therefore, I think I can be confident that the answer is 256.

But just to make sure, let me try to compute each step numerically again:

Start with 1.

First f(1) = 2^1 = 2.

Second f(2) = 2^2 = 4.

Third f(4) = 2^4 = 16.

Fourth f(16) = 2^16.

Now, 2^10 is 1024, so 2^16 is 2^10 * 2^6 = 1024 * 64.

Calculating 1024 * 64:

Well, 1024 * 60 = 61,440.

1024 * 4 = 4,096.

Adding them together: 61,440 + 4,096 = 65,536.

So, 2^16 is 65,536.

Then sqrt(65,536). Let me check if 256 squared is 65,536.

As above, 256 * 256 = 65,536.

Alternatively, 256 is 2^8, so 2^8 * 2^8 = 2^16, which is 65,536. Therefore, sqrt(2^16) is 2^8.

Therefore, the answer is 2^8 = 256.

Therefore, I think I have verified this multiple times through different methods: breaking down the exponents, calculating step by step, using properties of exponents and square roots, and even computing the numerical value. All approaches lead to the same conclusion.

Therefore, I can be confident that the answer is 256.

**Final Answer**
\boxed{256}

To solve the expression
$$
\sqrt{f(f(f(f(1))))},
$$
where $ f(x) = 2^x $, we proceed step by step, applying the function repeatedly and simplifying as we go.

---

### Step 1: Compute $ f(1) $

$$
f(1) = 2^1 = 2
$$

---

### Step 2: Compute $ f(f(1)) = f(2) $

$$
f(2) = 2^2 = 4
$$

---

### Step 3: Compute $ f(f(f(1))) = f(4) $

$$
f(4) = 2^4 = 16
$$

---

### Step 4: Compute $ f(f(f(f(1)))) = f(16) $

$$
f(16) = 2^{16}
$$

---

### Step 5: Take the square root of $ 2^{16} $

We use the property of exponents:
$$
\sqrt{2^{16}} = 2^{16/2} = 2^8
$$

Now compute $ 2^8 $:
$$
2^8 = 256
$$

---

### Final Answer

$$
\boxed{256}
$$

Solution: We find that $f(1) = 2^1 = 2.$ Then, $f(f(1)) = f(2) = 2^2 = 4$ and $f(f(f(1))) = f(4) = 2^4 = 16.$ Therefore, $f(f(f(f(1)))) = f(16) = 2^{16}$ and so $\sqrt{f(f(f(f(1))))} = \sqrt{2^{16}} = 2^8 = \boxed{256}.$