Explanation:
Okay, so I need to find the area of an isosceles right triangle where the altitude to the hypotenuse is 4√2. Hmm, let me think. First, let me recall some properties of isosceles right triangles.

An isosceles right triangle has two legs that are equal in length and a hypotenuse. The angles are 45°, 45°, and 90°. The area of a triangle is generally (base × height)/2. But here, they mention the altitude to the hypotenuse. So maybe I need to relate the altitude to the hypotenuse with the area?

Wait, let me remember. In any triangle, the area can be calculated using any side as the base, with the corresponding height. So if I take the hypotenuse as the base, then the altitude to the hypotenuse would be the height. Therefore, the area is (hypotenuse × altitude)/2. But I also know that the area can be calculated using the legs. Since it's an isosceles right triangle, the legs are equal. Let me denote the legs as 'a'. Then the hypotenuse would be a√2, right? Because in a right triangle, hypotenuse squared is equal to the sum of the squares of the legs. So, hypotenuse = √(a² + a²) = √(2a²) = a√2.

So, the area can be written in two ways: one using the legs and one using the hypotenuse and the altitude. Let me write that down.

First, area using legs: (a × a)/2 = a²/2.

Second, area using hypotenuse and altitude: (hypotenuse × altitude)/2 = (a√2 × 4√2)/2. Let me compute that.

Multiplying a√2 by 4√2: a × 4 × √2 × √2 = 4a × (√2 × √2) = 4a × 2 = 8a. Then divide by 2: 8a / 2 = 4a. So the area is 4a.

But we also have the area as a²/2. Therefore, setting them equal: a²/2 = 4a. Let me solve for a. Multiply both sides by 2: a² = 8a. Then subtract 8a: a² - 8a = 0. Factor: a(a - 8) = 0. So, a = 0 or a = 8. Since a is a length, it can't be 0, so a = 8.

Therefore, the legs are 8 each. Then the area is a²/2 = 8² / 2 = 64 / 2 = 32. So the area is 32? Let me check if that makes sense with the altitude.

Alternatively, using the hypotenuse and altitude: hypotenuse is a√2 = 8√2. Then area is (8√2 × 4√2)/2. Let me compute that again. 8√2 × 4√2 = 32 × (√2 × √2) = 32 × 2 = 64. Then divide by 2: 64 / 2 = 32. Yep, same result. So that seems consistent.

Wait, but let me think again. Is there another way to find the altitude? Maybe using properties of isosceles right triangles? Let me recall. In an isosceles right triangle, the altitude to the hypotenuse is also the median and the angle bisector. Wait, but in this case, since it's an isosceles right triangle, the altitude to the hypotenuse divides the triangle into two smaller congruent isosceles right triangles? Or maybe not? Let me visualize.

If I have an isosceles right triangle ABC with right angle at C, and legs AC and BC equal. Then the hypotenuse is AB. The altitude from C to AB would be CD, where D is the foot of the altitude on AB. In this case, since the triangle is isosceles, D is the midpoint of AB. Therefore, AD = DB = AB/2. Also, CD is the altitude.

But in an isosceles right triangle, the altitude to the hypotenuse can be calculated in terms of the legs. Let me try to find CD in terms of a.

Since area is (a²)/2, and also area is (AB × CD)/2. AB is a√2, so (a√2 × CD)/2 = (a²)/2. Therefore, solving for CD: (a√2 × CD)/2 = a²/2 ⇒ a√2 × CD = a² ⇒ CD = a / √2.

But they told us that CD is 4√2. Therefore, a / √2 = 4√2 ⇒ a = 4√2 × √2 = 4 × 2 = 8. Which is the same result as before. Therefore, legs are 8, area is 8² / 2 = 32. So that's consistent.

Alternatively, maybe I can use Pythagoras theorem on the smaller triangles. Since CD is the altitude, then triangle ADC is a right triangle with legs CD and AD, and hypotenuse AC. Wait, AC is a leg of the original triangle, which is 8.

So, AD is half of AB, which is (a√2)/2. Since a is 8, AB is 8√2, so AD is 4√2. Then CD is given as 4√2. Therefore, triangle ADC has legs of 4√2 and 4√2, so hypotenuse is sqrt( (4√2)^2 + (4√2)^2 ) = sqrt( 32 + 32 ) = sqrt(64) = 8. Which matches AC = 8. Perfect, that checks out.

So all the steps are consistent. Therefore, the area is 32.

But let me just make sure I didn't make any miscalculations. Let me go through once again.

Given altitude to hypotenuse is 4√2. In an isosceles right triangle, altitude to hypotenuse is equal to (leg)/√2. Wait, earlier we had CD = a / √2. So if CD = 4√2, then a = CD × √2 = 4√2 × √2 = 4 × 2 = 8. Then area is (a²)/2 = 64 / 2 = 32. Yep.

Alternatively, if I didn't remember that formula, I could derive it. Let me try that.

Let’s denote the legs as a, hypotenuse as h. Then h = a√2. The area is (a²)/2. Also, area is (h × altitude)/2. Therefore, altitude = (2 × area)/h = (2 × (a²)/2)/h = a² / h. Since h = a√2, altitude = a² / (a√2) = a / √2. So altitude is a / √2. Therefore, given altitude is 4√2, so a / √2 = 4√2 ⇒ a = 4√2 × √2 = 4 × 2 = 8. Then area is (8²)/2 = 64 / 2 = 32.

Same result. So seems like 32 is the answer.

Alternatively, maybe I can think of the altitude in terms of the hypotenuse. Let me think. In any right triangle, the altitude to the hypotenuse is equal to (product of the legs)/hypotenuse. So, altitude = (a × a)/h = a² / h. Which is same as above. Since h = a√2, altitude = a² / (a√2) = a / √2. So again, same thing.

Therefore, regardless of the approach, I get that the legs are 8, area is 32.

Wait, but let me check if there's another way someone might approach this problem and maybe get confused. For example, someone might think that the altitude is half the hypotenuse? But in an isosceles right triangle, the altitude to hypotenuse is not half the hypotenuse. Wait, in an isosceles triangle, the altitude to the base is not necessarily half the base unless it's equilateral. Wait, but in this case, the triangle is isosceles right triangle. So, the altitude to hypotenuse divides the hypotenuse into two equal parts, so AD = DB = h / 2. But the altitude itself is different.

But if someone thought that altitude is half the hypotenuse, they would be wrong. For example, if hypotenuse is 8√2, then half of that is 4√2, which is exactly the altitude given here. Wait, hold on! Wait, in this case, the altitude is 4√2, which is equal to half of the hypotenuse. Wait, hypotenuse is 8√2, half of that is 4√2. So, in this case, the altitude is equal to half the hypotenuse? Is that a general rule?

Wait, in a right triangle, the altitude to the hypotenuse is equal to (product of legs)/hypotenuse. But in an isosceles right triangle, legs are equal, so altitude is (a²)/ (a√2) = a / √2. But hypotenuse is a√2, so half of hypotenuse is (a√2)/2. So, comparing altitude and half hypotenuse:

Altitude = a / √2 = (a√2)/2, since multiplying numerator and denominator by √2: a / √2 = (a√2)/2. Therefore, altitude is equal to half the hypotenuse. So in this case, altitude is equal to half the hypotenuse. Therefore, in an isosceles right triangle, the altitude to hypotenuse is half the hypotenuse. Therefore, if altitude is 4√2, then hypotenuse is 8√2. Then area is (hypotenuse × altitude)/2 = (8√2 × 4√2)/2 = same as before, 32.

So that's another way to think about it. Since altitude is half the hypotenuse, then hypotenuse is twice the altitude. Therefore, hypotenuse = 2 × 4√2 = 8√2. Then area is (hypotenuse × altitude)/2 = (8√2 × 4√2)/2. Which again gives 32.

So that's another way to see it. Therefore, regardless of the approach, the area is 32.

But let me check once again with actual numbers. Suppose the legs are 8. Then hypotenuse is 8√2. The altitude to hypotenuse is 4√2. Let me verify that.

In a right triangle, area is (leg1 × leg2)/2 = (8 × 8)/2 = 32. Also, area is (hypotenuse × altitude)/2 = (8√2 × altitude)/2. Therefore, 32 = (8√2 × altitude)/2 ⇒ 32 × 2 = 8√2 × altitude ⇒ 64 = 8√2 × altitude ⇒ altitude = 64 / (8√2) = 8 / √2 = 4√2. Which matches the given altitude.

Therefore, all the numbers check out.

So, I think I've verified this multiple times. The area is 32. Therefore, the answer is 32.

**Final Answer**
The area of the triangle is \boxed{32}.

To find the area of the isosceles right triangle, we are given that the **altitude to the hypotenuse** is $ 4\sqrt{2} $. Let's proceed step by step using geometric and algebraic reasoning.

---

### Step 1: Understand the Triangle

An **isosceles right triangle** has:
- Two equal legs of length $ a $,
- A hypotenuse of length $ a\sqrt{2} $ (by the Pythagorean Theorem),
- Angles of $ 45^\circ, 45^\circ, 90^\circ $.

The **altitude to the hypotenuse** in such a triangle is a special property: it is **equal to half the hypotenuse**. This is because the triangle is isosceles and the altitude from the right angle to the hypotenuse divides the triangle into two smaller, congruent isosceles right triangles.

So, if the altitude is $ 4\sqrt{2} $, then the hypotenuse is:
$$
\text{Hypotenuse} = 2 \times \text{Altitude} = 2 \times 4\sqrt{2} = 8\sqrt{2}
$$

---

### Step 2: Use the Area Formula

We can compute the area of the triangle in two equivalent ways:

#### Method 1: Using the legs
$$
\text{Area} = \frac{1}{2} \times a \times a = \frac{a^2}{2}
$$

#### Method 2: Using the hypotenuse and the altitude
$$
\text{Area} = \frac{1}{2} \times \text{Hypotenuse} \times \text{Altitude} = \frac{1}{2} \times 8\sqrt{2} \times 4\sqrt{2}
$$

Compute the product:
$$
8\sqrt{2} \times 4\sqrt{2} = 32 \times 2 = 64
$$
$$
\text{Area} = \frac{64}{2} = 32
$$

---

### Step 3: Verify Consistency

We can also find the value of $ a $ using the relationship between the altitude and the legs. In an isosceles right triangle, the altitude to the hypotenuse is:
$$
\text{Altitude} = \frac{a}{\sqrt{2}}
$$

Given that the altitude is $ 4\sqrt{2} $, we solve:
$$
\frac{a}{\sqrt{2}} = 4\sqrt{2} \Rightarrow a = 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8
$$

Now compute the area using the legs:
$$
\text{Area} = \frac{1}{2} \times 8 \times 8 = \frac{64}{2} = 32
$$

---

### Final Answer

All methods confirm that the area of the triangle is:

$$
\boxed{32}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of an isosceles right triangle where the altitude to the hypotenuse is 4√2. Hmm, let me think. First, let me recall some properties of isosceles right triangles.

An isosceles right triangle has two legs that are equal in length and a hypotenuse. The angles are 45°, 45°, and 90°. The area of a triangle is generally (base × height)/2. But here, they mention the altitude to the hypotenuse. So maybe I need to relate the altitude to the hypotenuse with the area?

Wait, let me remember. In any triangle, the area can be calculated using any side as the base, with the corresponding height. So if I take the hypotenuse as the base, then the altitude to the hypotenuse would be the height. Therefore, the area is (hypotenuse × altitude)/2. But I also know that the area can be calculated using the legs. Since it's an isosceles right triangle, the legs are equal. Let me denote the legs as 'a'. Then the hypotenuse would be a√2, right? Because in a right triangle, hypotenuse squared is equal to the sum of the squares of the legs. So, hypotenuse = √(a² + a²) = √(2a²) = a√2.

So, the area can be written in two ways: one using the legs and one using the hypotenuse and the altitude. Let me write that down.

First, area using legs: (a × a)/2 = a²/2.

Second, area using hypotenuse and altitude: (hypotenuse × altitude)/2 = (a√2 × 4√2)/2. Let me compute that.

Multiplying a√2 by 4√2: a × 4 × √2 × √2 = 4a × (√2 × √2) = 4a × 2 = 8a. Then divide by 2: 8a / 2 = 4a. So the area is 4a.

But we also have the area as a²/2. Therefore, setting them equal: a²/2 = 4a. Let me solve for a. Multiply both sides by 2: a² = 8a. Then subtract 8a: a² - 8a = 0. Factor: a(a - 8) = 0. So, a = 0 or a = 8. Since a is a length, it can't be 0, so a = 8.

Therefore, the legs are 8 each. Then the area is a²/2 = 8² / 2 = 64 / 2 = 32. So the area is 32? Let me check if that makes sense with the altitude.

Alternatively, using the hypotenuse and altitude: hypotenuse is a√2 = 8√2. Then area is (8√2 × 4√2)/2. Let me compute that again. 8√2 × 4√2 = 32 × (√2 × √2) = 32 × 2 = 64. Then divide by 2: 64 / 2 = 32. Yep, same result. So that seems consistent.

Wait, but let me think again. Is there another way to find the altitude? Maybe using properties of isosceles right triangles? Let me recall. In an isosceles right triangle, the altitude to the hypotenuse is also the median and the angle bisector. Wait, but in this case, since it's an isosceles right triangle, the altitude to the hypotenuse divides the triangle into two smaller congruent isosceles right triangles? Or maybe not? Let me visualize.

If I have an isosceles right triangle ABC with right angle at C, and legs AC and BC equal. Then the hypotenuse is AB. The altitude from C to AB would be CD, where D is the foot of the altitude on AB. In this case, since the triangle is isosceles, D is the midpoint of AB. Therefore, AD = DB = AB/2. Also, CD is the altitude.

But in an isosceles right triangle, the altitude to the hypotenuse can be calculated in terms of the legs. Let me try to find CD in terms of a.

Since area is (a²)/2, and also area is (AB × CD)/2. AB is a√2, so (a√2 × CD)/2 = (a²)/2. Therefore, solving for CD: (a√2 × CD)/2 = a²/2 ⇒ a√2 × CD = a² ⇒ CD = a / √2.

But they told us that CD is 4√2. Therefore, a / √2 = 4√2 ⇒ a = 4√2 × √2 = 4 × 2 = 8. Which is the same result as before. Therefore, legs are 8, area is 8² / 2 = 32. So that's consistent.

Alternatively, maybe I can use Pythagoras theorem on the smaller triangles. Since CD is the altitude, then triangle ADC is a right triangle with legs CD and AD, and hypotenuse AC. Wait, AC is a leg of the original triangle, which is 8.

So, AD is half of AB, which is (a√2)/2. Since a is 8, AB is 8√2, so AD is 4√2. Then CD is given as 4√2. Therefore, triangle ADC has legs of 4√2 and 4√2, so hypotenuse is sqrt( (4√2)^2 + (4√2)^2 ) = sqrt( 32 + 32 ) = sqrt(64) = 8. Which matches AC = 8. Perfect, that checks out.

So all the steps are consistent. Therefore, the area is 32.

But let me just make sure I didn't make any miscalculations. Let me go through once again.

Given altitude to hypotenuse is 4√2. In an isosceles right triangle, altitude to hypotenuse is equal to (leg)/√2. Wait, earlier we had CD = a / √2. So if CD = 4√2, then a = CD × √2 = 4√2 × √2 = 4 × 2 = 8. Then area is (a²)/2 = 64 / 2 = 32. Yep.

Alternatively, if I didn't remember that formula, I could derive it. Let me try that.

Let’s denote the legs as a, hypotenuse as h. Then h = a√2. The area is (a²)/2. Also, area is (h × altitude)/2. Therefore, altitude = (2 × area)/h = (2 × (a²)/2)/h = a² / h. Since h = a√2, altitude = a² / (a√2) = a / √2. So altitude is a / √2. Therefore, given altitude is 4√2, so a / √2 = 4√2 ⇒ a = 4√2 × √2 = 4 × 2 = 8. Then area is (8²)/2 = 64 / 2 = 32.

Same result. So seems like 32 is the answer.

Alternatively, maybe I can think of the altitude in terms of the hypotenuse. Let me think. In any right triangle, the altitude to the hypotenuse is equal to (product of the legs)/hypotenuse. So, altitude = (a × a)/h = a² / h. Which is same as above. Since h = a√2, altitude = a² / (a√2) = a / √2. So again, same thing.

Therefore, regardless of the approach, I get that the legs are 8, area is 32.

Wait, but let me check if there's another way someone might approach this problem and maybe get confused. For example, someone might think that the altitude is half the hypotenuse? But in an isosceles right triangle, the altitude to hypotenuse is not half the hypotenuse. Wait, in an isosceles triangle, the altitude to the base is not necessarily half the base unless it's equilateral. Wait, but in this case, the triangle is isosceles right triangle. So, the altitude to hypotenuse divides the hypotenuse into two equal parts, so AD = DB = h / 2. But the altitude itself is different.

But if someone thought that altitude is half the hypotenuse, they would be wrong. For example, if hypotenuse is 8√2, then half of that is 4√2, which is exactly the altitude given here. Wait, hold on! Wait, in this case, the altitude is 4√2, which is equal to half of the hypotenuse. Wait, hypotenuse is 8√2, half of that is 4√2. So, in this case, the altitude is equal to half the hypotenuse? Is that a general rule?

Wait, in a right triangle, the altitude to the hypotenuse is equal to (product of legs)/hypotenuse. But in an isosceles right triangle, legs are equal, so altitude is (a²)/ (a√2) = a / √2. But hypotenuse is a√2, so half of hypotenuse is (a√2)/2. So, comparing altitude and half hypotenuse:

Altitude = a / √2 = (a√2)/2, since multiplying numerator and denominator by √2: a / √2 = (a√2)/2. Therefore, altitude is equal to half the hypotenuse. So in this case, altitude is equal to half the hypotenuse. Therefore, in an isosceles right triangle, the altitude to hypotenuse is half the hypotenuse. Therefore, if altitude is 4√2, then hypotenuse is 8√2. Then area is (hypotenuse × altitude)/2 = (8√2 × 4√2)/2 = same as before, 32.

So that's another way to think about it. Since altitude is half the hypotenuse, then hypotenuse is twice the altitude. Therefore, hypotenuse = 2 × 4√2 = 8√2. Then area is (hypotenuse × altitude)/2 = (8√2 × 4√2)/2. Which again gives 32.

So that's another way to see it. Therefore, regardless of the approach, the area is 32.

But let me check once again with actual numbers. Suppose the legs are 8. Then hypotenuse is 8√2. The altitude to hypotenuse is 4√2. Let me verify that.

In a right triangle, area is (leg1 × leg2)/2 = (8 × 8)/2 = 32. Also, area is (hypotenuse × altitude)/2 = (8√2 × altitude)/2. Therefore, 32 = (8√2 × altitude)/2 ⇒ 32 × 2 = 8√2 × altitude ⇒ 64 = 8√2 × altitude ⇒ altitude = 64 / (8√2) = 8 / √2 = 4√2. Which matches the given altitude.

Therefore, all the numbers check out.

So, I think I've verified this multiple times. The area is 32. Therefore, the answer is 32.

**Final Answer**
The area of the triangle is \boxed{32}.

To find the area of the isosceles right triangle, we are given that the **altitude to the hypotenuse** is $ 4\sqrt{2} $. Let's proceed step by step using geometric and algebraic reasoning.

---

### Step 1: Understand the Triangle

An **isosceles right triangle** has:
- Two equal legs of length $ a $,
- A hypotenuse of length $ a\sqrt{2} $ (by the Pythagorean Theorem),
- Angles of $ 45^\circ, 45^\circ, 90^\circ $.

The **altitude to the hypotenuse** in such a triangle is a special property: it is **equal to half the hypotenuse**. This is because the triangle is isosceles and the altitude from the right angle to the hypotenuse divides the triangle into two smaller, congruent isosceles right triangles.

So, if the altitude is $ 4\sqrt{2} $, then the hypotenuse is:
$$
\text{Hypotenuse} = 2 \times \text{Altitude} = 2 \times 4\sqrt{2} = 8\sqrt{2}
$$

---

### Step 2: Use the Area Formula

We can compute the area of the triangle in two equivalent ways:

#### Method 1: Using the legs
$$
\text{Area} = \frac{1}{2} \times a \times a = \frac{a^2}{2}
$$

#### Method 2: Using the hypotenuse and the altitude
$$
\text{Area} = \frac{1}{2} \times \text{Hypotenuse} \times \text{Altitude} = \frac{1}{2} \times 8\sqrt{2} \times 4\sqrt{2}
$$

Compute the product:
$$
8\sqrt{2} \times 4\sqrt{2} = 32 \times 2 = 64
$$
$$
\text{Area} = \frac{64}{2} = 32
$$

---

### Step 3: Verify Consistency

We can also find the value of $ a $ using the relationship between the altitude and the legs. In an isosceles right triangle, the altitude to the hypotenuse is:
$$
\text{Altitude} = \frac{a}{\sqrt{2}}
$$

Given that the altitude is $ 4\sqrt{2} $, we solve:
$$
\frac{a}{\sqrt{2}} = 4\sqrt{2} \Rightarrow a = 4\sqrt{2} \times \sqrt{2} = 4 \times 2 = 8
$$

Now compute the area using the legs:
$$
\text{Area} = \frac{1}{2} \times 8 \times 8 = \frac{64}{2} = 32
$$

---

### Final Answer

All methods confirm that the area of the triangle is:

$$
\boxed{32}
$$

Solution:
In isosceles right triangle $\triangle ABC$ below, $\overline{AD}$ is the altitude to the hypotenuse.

[asy]
import olympiad;
unitsize(0.8inch);
pair A,B,C,D;
A = (0,1);
B= (1,0);
C = -B;
D = (0,0);
draw(A--B--C--A,linewidth(1));
draw(A--D,linewidth(0.8));
draw(rightanglemark(C,A,B,s=5));
draw(rightanglemark(C,D,A,s=5));
label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,S);
[/asy]

Because $\triangle ABC$ is an isosceles right triangle, $\angle ABC = 45^\circ$.  Since $\angle ADB = 90^\circ$, we know that $\angle DAB = 45^\circ$, so $\triangle ABD$ is also a 45-45-90 triangle.  Similarly, $\triangle ACD$ is a 45-45-90 triangle.  Therefore, $DB=DC = DA = 4\sqrt{2}$, so $BC = BD+DC = 8\sqrt{2}$, and  \[[ABC] = \frac{(AD)(BC)}{2} = \frac{(4\sqrt{2})(8\sqrt{2})}{2} = \boxed{32}.\]


Explanation:
Okay, so I need to find the domain of the function f(x) = (2x - 7)/√(x² - 5x + 6). Hmm, let me recall what the domain of a function is. The domain is all the real numbers x for which the function is defined. Since this function has a square root in the denominator, I need to make sure that the expression inside the square root is positive because the square root of a negative number isn't a real number, and also the denominator can't be zero because division by zero is undefined.

So, first, let me focus on the denominator: √(x² - 5x + 6). For the square root to be defined (i.e., to have a real value), the expression inside the square root, which is x² - 5x + 6, must be greater than or equal to zero. However, since this is in the denominator, it can't be zero because division by zero is undefined. Therefore, the expression inside the square root must be strictly greater than zero.

So, the first step is to solve the inequality x² - 5x + 6 > 0. Let me factor the quadratic expression. Let me see, x² - 5x + 6. To factor this, I need two numbers that multiply to 6 and add up to -5. Let me think... -2 and -3? Yes, because (-2) * (-3) = 6 and (-2) + (-3) = -5. Therefore, the quadratic factors as (x - 2)(x - 3).

So, the inequality becomes (x - 2)(x - 3) > 0. Now, I need to find the values of x for which this product is positive. To solve this inequality, I can use a sign chart or test intervals. The critical points are x = 2 and x = 3, which divide the real number line into three intervals:

1. x < 2
2. 2 < x < 3
3. x > 3

Let me pick test points in each interval to determine the sign of the product (x - 2)(x - 3).

For the first interval, let's choose x = 1. Plugging into (1 - 2)(1 - 3) = (-1)(-2) = 2, which is positive.

For the second interval, let's choose x = 2.5. Plugging in, (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is negative.

For the third interval, let's choose x = 4. Plugging in, (4 - 2)(4 - 3) = (2)(1) = 2, which is positive.

So, the product is positive when x < 2 or x > 3. Therefore, the solution to the inequality (x - 2)(x - 3) > 0 is x ∈ (-∞, 2) ∪ (3, ∞).

But wait, the original function also has a numerator, which is 2x - 7. However, the numerator is a linear function and is defined for all real numbers. Therefore, the only restrictions on the domain come from the denominator.

Therefore, the domain of f(x) is all real numbers x such that x is in (-∞, 2) ∪ (3, ∞).

But let me double-check if there are any other restrictions. For example, sometimes when there are multiple conditions, you have to check if there's any overlap or something else. But in this case, the numerator doesn't impose any restrictions because it's a polynomial, which is defined everywhere. The denominator's square root requires the inside to be positive, which we already considered.

Wait, but just to make sure, let me check if there's any x that makes the denominator zero. The denominator is √(x² - 5x + 6). The square root is zero when x² - 5x + 6 = 0, which is when x = 2 or x = 3. Therefore, those points are excluded from the domain. So, the domain is indeed all real numbers except x = 2 and x = 3, but also considering that the expression under the square root must be positive.

But since we already solved the inequality x² - 5x + 6 > 0, which gives us x < 2 or x > 3, that already excludes x = 2 and x = 3. Therefore, combining these, the domain is x ∈ (-∞, 2) ∪ (3, ∞).

Is there anything else I need to consider? For example, sometimes functions can have other restrictions, but in this case, the function is a rational function with a square root in the denominator. The numerator is fine, so no issues there.

Let me also check if there's a possibility that the numerator could be zero, but that doesn't affect the domain. The domain is about where the function is defined, not where it equals zero. So even if the numerator is zero at some point, that point is still in the domain as long as the denominator isn't zero there.

So, solving 2x - 7 = 0 gives x = 7/2 = 3.5. Let me check if x = 3.5 is in the domain. Since 3.5 is greater than 3, it is in the interval (3, ∞), so it is included in the domain. Therefore, even though the numerator is zero there, the function is defined there.

Therefore, my conclusion is that the domain is all real numbers except those between 2 and 3, including 2 and 3. Wait, no. The domain is all real numbers except between 2 and 3, but not including 2 and 3. So, in interval notation, that's (-∞, 2) ∪ (3, ∞).

Let me just verify with a sample value. Let's take x = 1, which is in (-∞, 2). Then the denominator becomes √(1 - 5 + 6) = √(2), which is fine. The numerator is 2(1) - 7 = -5, so f(1) = -5/√2, which is defined.

Take x = 4, which is in (3, ∞). The denominator is √(16 - 20 + 6) = √(2), same as before. The numerator is 8 - 7 = 1, so f(4) = 1/√2, which is defined.

What about x = 2.5, which is between 2 and 3? The denominator would be √(6.25 - 12.5 + 6) = √(-0.25), which is not real. Therefore, x = 2.5 is not in the domain, which matches our earlier conclusion.

What about x = 2? The denominator becomes √(4 - 10 + 6) = √0 = 0, which is undefined. Similarly for x = 3. So those points are excluded.

Therefore, I think my answer is correct. The domain is all real numbers except those in the interval [2, 3]. So in interval notation, that's (-∞, 2) ∪ (3, ∞).

Wait, but let me check if there's a possibility that the quadratic could be positive in some other way. For example, if the quadratic had a different sign. But since the quadratic opens upwards (the coefficient of x² is positive), it will be positive outside the roots and negative between them. So yes, that's exactly what we found.

Therefore, I think I can confidently say that the domain is (-∞, 2) ∪ (3, ∞).

**Final Answer**
The domain of the function is \boxed{(-\infty, 2) \cup (3, \infty)}.

To determine the **domain** of the function
$$
f(x) = \frac{2x - 7}{\sqrt{x^2 - 5x + 6}},
$$
we need to identify all real values of $ x $ for which the function is defined.

---

### Step 1: Understand the Structure of the Function

The function is a **rational expression** with a **square root in the denominator**. This means:

- The **numerator** $ 2x - 7 $ is a linear function and is defined for all real numbers.
- The **denominator** $ \sqrt{x^2 - 5x + 6} $ imposes restrictions:
  - The expression inside the square root, $ x^2 - 5x + 6 $, must be **positive** (since the square root of a negative number is not real).
  - The denominator **cannot be zero**, because division by zero is undefined.

---

### Step 2: Solve the Inequality

We solve the inequality:
$$
x^2 - 5x + 6 > 0
$$

#### Factor the quadratic:
$$
x^2 - 5x + 6 = (x - 2)(x - 3)
$$

So the inequality becomes:
$$
(x - 2)(x - 3) > 0
$$

#### Determine the sign of the product:

- The critical points are $ x = 2 $ and $ x = 3 $.
- These divide the number line into three intervals:
  - $ (-\infty, 2) $
  - $ (2, 3) $
  - $ (3, \infty) $

We test the sign of the product in each interval:

- For $ x < 2 $, both $ x - 2 < 0 $ and $ x - 3 < 0 $, so the product is **positive**.
- For $ 2 < x < 3 $, $ x - 2 > 0 $ and $ x - 3 < 0 $, so the product is **negative**.
- For $ x > 3 $, both $ x - 2 > 0 $ and $ x - 3 > 0 $, so the product is **positive**.

Thus, the inequality $ (x - 2)(x - 3) > 0 $ is satisfied when:
$$
x \in (-\infty, 2) \cup (3, \infty)
$$

---

### Step 3: Exclude Points Where the Denominator is Zero

The denominator becomes zero when:
$$
x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0 \Rightarrow x = 2 \text{ or } x = 3
$$

These values must be **excluded** from the domain, even though they are not in the solution set of the inequality (since the inequality is strict: $ > 0 $, not $ \geq 0 $).

---

### Final Domain

Combining the above results, the domain of the function is all real numbers **except** those in the interval $[2, 3]$, i.e., the values of $ x $ for which the denominator is zero or negative.

So, the domain is:
$$
\boxed{(-\infty, 2) \cup (3, \infty)}
$$

[Assistant]
Explanation:
Okay, so I need to find the domain of the function f(x) = (2x - 7)/√(x² - 5x + 6). Hmm, let me recall what the domain of a function is. The domain is all the real numbers x for which the function is defined. Since this function has a square root in the denominator, I need to make sure that the expression inside the square root is positive because the square root of a negative number isn't a real number, and also the denominator can't be zero because division by zero is undefined.

So, first, let me focus on the denominator: √(x² - 5x + 6). For the square root to be defined (i.e., to have a real value), the expression inside the square root, which is x² - 5x + 6, must be greater than or equal to zero. However, since this is in the denominator, it can't be zero because division by zero is undefined. Therefore, the expression inside the square root must be strictly greater than zero.

So, the first step is to solve the inequality x² - 5x + 6 > 0. Let me factor the quadratic expression. Let me see, x² - 5x + 6. To factor this, I need two numbers that multiply to 6 and add up to -5. Let me think... -2 and -3? Yes, because (-2) * (-3) = 6 and (-2) + (-3) = -5. Therefore, the quadratic factors as (x - 2)(x - 3).

So, the inequality becomes (x - 2)(x - 3) > 0. Now, I need to find the values of x for which this product is positive. To solve this inequality, I can use a sign chart or test intervals. The critical points are x = 2 and x = 3, which divide the real number line into three intervals:

1. x < 2
2. 2 < x < 3
3. x > 3

Let me pick test points in each interval to determine the sign of the product (x - 2)(x - 3).

For the first interval, let's choose x = 1. Plugging into (1 - 2)(1 - 3) = (-1)(-2) = 2, which is positive.

For the second interval, let's choose x = 2.5. Plugging in, (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is negative.

For the third interval, let's choose x = 4. Plugging in, (4 - 2)(4 - 3) = (2)(1) = 2, which is positive.

So, the product is positive when x < 2 or x > 3. Therefore, the solution to the inequality (x - 2)(x - 3) > 0 is x ∈ (-∞, 2) ∪ (3, ∞).

But wait, the original function also has a numerator, which is 2x - 7. However, the numerator is a linear function and is defined for all real numbers. Therefore, the only restrictions on the domain come from the denominator.

Therefore, the domain of f(x) is all real numbers x such that x is in (-∞, 2) ∪ (3, ∞).

But let me double-check if there are any other restrictions. For example, sometimes when there are multiple conditions, you have to check if there's any overlap or something else. But in this case, the numerator doesn't impose any restrictions because it's a polynomial, which is defined everywhere. The denominator's square root requires the inside to be positive, which we already considered.

Wait, but just to make sure, let me check if there's any x that makes the denominator zero. The denominator is √(x² - 5x + 6). The square root is zero when x² - 5x + 6 = 0, which is when x = 2 or x = 3. Therefore, those points are excluded from the domain. So, the domain is indeed all real numbers except x = 2 and x = 3, but also considering that the expression under the square root must be positive.

But since we already solved the inequality x² - 5x + 6 > 0, which gives us x < 2 or x > 3, that already excludes x = 2 and x = 3. Therefore, combining these, the domain is x ∈ (-∞, 2) ∪ (3, ∞).

Is there anything else I need to consider? For example, sometimes functions can have other restrictions, but in this case, the function is a rational function with a square root in the denominator. The numerator is fine, so no issues there.

Let me also check if there's a possibility that the numerator could be zero, but that doesn't affect the domain. The domain is about where the function is defined, not where it equals zero. So even if the numerator is zero at some point, that point is still in the domain as long as the denominator isn't zero there.

So, solving 2x - 7 = 0 gives x = 7/2 = 3.5. Let me check if x = 3.5 is in the domain. Since 3.5 is greater than 3, it is in the interval (3, ∞), so it is included in the domain. Therefore, even though the numerator is zero there, the function is defined there.

Therefore, my conclusion is that the domain is all real numbers except those between 2 and 3, including 2 and 3. Wait, no. The domain is all real numbers except between 2 and 3, but not including 2 and 3. So, in interval notation, that's (-∞, 2) ∪ (3, ∞).

Let me just verify with a sample value. Let's take x = 1, which is in (-∞, 2). Then the denominator becomes √(1 - 5 + 6) = √(2), which is fine. The numerator is 2(1) - 7 = -5, so f(1) = -5/√2, which is defined.

Take x = 4, which is in (3, ∞). The denominator is √(16 - 20 + 6) = √(2), same as before. The numerator is 8 - 7 = 1, so f(4) = 1/√2, which is defined.

What about x = 2.5, which is between 2 and 3? The denominator would be √(6.25 - 12.5 + 6) = √(-0.25), which is not real. Therefore, x = 2.5 is not in the domain, which matches our earlier conclusion.

What about x = 2? The denominator becomes √(4 - 10 + 6) = √0 = 0, which is undefined. Similarly for x = 3. So those points are excluded.

Therefore, I think my answer is correct. The domain is all real numbers except those in the interval [2, 3]. So in interval notation, that's (-∞, 2) ∪ (3, ∞).

Wait, but let me check if there's a possibility that the quadratic could be positive in some other way. For example, if the quadratic had a different sign. But since the quadratic opens upwards (the coefficient of x² is positive), it will be positive outside the roots and negative between them. So yes, that's exactly what we found.

Therefore, I think I can confidently say that the domain is (-∞, 2) ∪ (3, ∞).

**Final Answer**
The domain of the function is \boxed{(-\infty, 2) \cup (3, \infty)}.

To determine the **domain** of the function
$$
f(x) = \frac{2x - 7}{\sqrt{x^2 - 5x + 6}},
$$
we need to identify all real values of $ x $ for which the function is defined.

---

### Step 1: Understand the Structure of the Function

The function is a **rational expression** with a **square root in the denominator**. This means:

- The **numerator** $ 2x - 7 $ is a linear function and is defined for all real numbers.
- The **denominator** $ \sqrt{x^2 - 5x + 6} $ imposes restrictions:
  - The expression inside the square root, $ x^2 - 5x + 6 $, must be **positive** (since the square root of a negative number is not real).
  - The denominator **cannot be zero**, because division by zero is undefined.

---

### Step 2: Solve the Inequality

We solve the inequality:
$$
x^2 - 5x + 6 > 0
$$

#### Factor the quadratic:
$$
x^2 - 5x + 6 = (x - 2)(x - 3)
$$

So the inequality becomes:
$$
(x - 2)(x - 3) > 0
$$

#### Determine the sign of the product:

- The critical points are $ x = 2 $ and $ x = 3 $.
- These divide the number line into three intervals:
  - $ (-\infty, 2) $
  - $ (2, 3) $
  - $ (3, \infty) $

We test the sign of the product in each interval:

- For $ x < 2 $, both $ x - 2 < 0 $ and $ x - 3 < 0 $, so the product is **positive**.
- For $ 2 < x < 3 $, $ x - 2 > 0 $ and $ x - 3 < 0 $, so the product is **negative**.
- For $ x > 3 $, both $ x - 2 > 0 $ and $ x - 3 > 0 $, so the product is **positive**.

Thus, the inequality $ (x - 2)(x - 3) > 0 $ is satisfied when:
$$
x \in (-\infty, 2) \cup (3, \infty)
$$

---

### Step 3: Exclude Points Where the Denominator is Zero

The denominator becomes zero when:
$$
x^2 - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0 \Rightarrow x = 2 \text{ or } x = 3
$$

These values must be **excluded** from the domain, even though they are not in the solution set of the inequality (since the inequality is strict: $ > 0 $, not $ \geq 0 $).

---

### Final Domain

Combining the above results, the domain of the function is all real numbers **except** those in the interval $[2, 3]$, i.e., the values of $ x $ for which the denominator is zero or negative.

So, the domain is:
$$
\boxed{(-\infty, 2) \cup (3, \infty)}
$$

Solution: The function is defined when the value inside the square root is positive, i.e. we must have $x^2-5x+6>0$.  Factoring, we get $(x-3)(x-2)>0$. So either both factors in the left hand side are negative or they are both positive. They are both negative when $x<2$. They are both positive when $x>3$. So the domain of $f(x)$ is $x<2 \text{ or } x>3$, or $x \in \boxed{(-\infty, 2) \cup (3, \infty)}$ in interval notation.