Explanation:
Okay, so I need to find the largest possible value of |a| + |b| + |c| given that the absolute value of the quadratic function ax² + bx + c is less than or equal to 1 for all x between 0 and 1. Hmm, let me think about how to approach this.

First, I know that the quadratic is bounded between -1 and 1 on the interval [0, 1]. So, for every x in [0, 1], the value of ax² + bx + c is within that range. My goal is to maximize |a| + |b| + |c|.

I remember that sometimes problems like this involve using specific points in the interval to set up equations or inequalities. Since the quadratic is bounded on [0, 1], maybe I can evaluate it at certain key points, like x = 0, x = 1, and maybe the vertex of the parabola? Let me try that.

Let me first note the values of the quadratic at x = 0 and x = 1:

At x = 0: |c| ≤ 1, so c is between -1 and 1.

At x = 1: |a + b + c| ≤ 1.

Also, the vertex of the parabola is at x = -b/(2a). But since the interval is [0, 1], the vertex might lie inside this interval or outside. If it's inside, then the maximum or minimum of the quadratic could be there. However, since the quadratic is bounded between -1 and 1 on the entire interval, the maximum absolute value could occur at the endpoints or at the vertex.

But maybe I need to consider more than just the endpoints. For example, if the quadratic has its maximum or minimum inside the interval, then that point must also satisfy |ax² + bx + c| ≤ 1.

Alternatively, maybe I can use some known inequalities or optimization techniques. Since we need to maximize |a| + |b| + |c|, perhaps we can consider different cases based on the signs of a, b, c? But that might get complicated.

Alternatively, maybe I can use Lagrange multipliers? But since there are infinitely many constraints (for all x in [0,1]), that might not be straightforward.

Wait, another idea: maybe use the Chebyshev polynomials? Because Chebyshev polynomials are known for minimizing the maximum deviation from zero over an interval, which might be related. But I need to maximize |a| + |b| + |c| under the constraint that the quadratic is bounded by 1 in absolute value on [0,1].

Alternatively, maybe I can parametrize the quadratic in terms of its values at certain points. For example, if I know the values at three points, I can determine the coefficients. Since the interval is [0,1], maybe I can use the values at x=0, x=1, and x=1/2? But I need to ensure that the quadratic is bounded between -1 and 1 on the entire interval.

Alternatively, maybe consider specific quadratics that achieve the maximum. For example, if I can find a quadratic that touches the bounds 1 or -1 at several points in [0,1], then that might be the extremal case.

Let me try to think of an example. Suppose the quadratic reaches 1 at x=0 and x=1, and -1 at some point in between. Or maybe reaches 1 at one point and -1 at another.

Let me try constructing such a quadratic. Let's assume that the maximum of |ax² + bx + c| is achieved at three points in [0,1], which would be the case for the Chebyshev polynomial. But Chebyshev polynomials are usually defined on [-1,1], so maybe I need to adjust them to the interval [0,1].

The Chebyshev polynomial of degree 2 is T2(x) = 2x² - 1, which oscillates between -1 and 1 on [-1,1]. If I can map the interval [0,1] to [-1,1], maybe I can use that. Let me try that.

Let me perform a linear transformation. Let t = 2x - 1, so that when x=0, t=-1 and x=1, t=1. Then, substituting into T2(t):

T2(t) = 2t² - 1 = 2(2x - 1)² - 1 = 2(4x² -4x +1) -1 = 8x² -8x +2 -1 = 8x² -8x +1.

So, the polynomial 8x² -8x +1 oscillates between -1 and 1 on [0,1]. Let me check:

At x=0: 8(0) -8(0) +1 = 1.

At x=1: 8(1) -8(1) +1 = 1.

At x=1/2: 8*(1/4) -8*(1/2) +1 = 2 -4 +1 = -1.

So yes, it reaches 1 at x=0 and x=1, and -1 at x=1/2. Therefore, |8x² -8x +1| ≤ 1 on [0,1].

So, if we take this quadratic, then a=8, b=-8, c=1. Then |a| + |b| + |c| = 8 + 8 + 1 = 17.

Is this the maximum? Maybe. But let me check if there are other quadratics that can give a higher value.

Alternatively, maybe there's a quadratic that reaches 1 or -1 at more points or in a different configuration.

Alternatively, suppose we consider quadratics that attain maximum absolute value at the endpoints and at the midpoint. But the above example already does that.

But let me check if there's a way to get higher coefficients. For example, suppose we have a quadratic that reaches 1 at x=0, -1 at x=1/2, and 1 at x=1. Then, that's exactly the polynomial we have above.

Alternatively, suppose we have a quadratic that reaches 1 at x=0, 1 at x=1, and some other value in between. But if it's 1 at both ends, then maybe the maximum of the quadratic is 1, but then the coefficients might be smaller?

Alternatively, maybe if we have a quadratic that reaches -1 at x=0 and x=1, and 1 at the midpoint. Let me try that.

Let me suppose that at x=0, ax² + bx + c = -1, so c = -1. At x=1, a + b + c = -1. So, a + b -1 = -1 => a + b = 0 => b = -a. At x=1/2, the value is a*(1/4) + b*(1/2) + c = a/4 + (-a)/2 + (-1) = (-a/4) -1. We want this to be 1. So:

(-a/4) -1 = 1 => -a/4 = 2 => a = -8. Then b = -a = 8. So the quadratic is -8x² +8x -1. Let me check its values:

At x=0: -1, which is okay.

At x=1: -8 +8 -1 = -1, okay.

At x=1/2: -8*(1/4) +8*(1/2) -1 = -2 +4 -1 = 1, okay.

So this quadratic also oscillates between -1 and 1 on [0,1]. Then |a| + |b| + |c| = 8 + 8 + 1 = 17 again. Same as before.

So whether we have the polynomial reaching 1 at the endpoints and -1 in the middle, or -1 at the endpoints and 1 in the middle, we get the same sum.

But is 17 the maximum? Maybe there's another quadratic where |a| + |b| + |c| is larger?

Let me think. Suppose we try to maximize |a| + |b| + |c|. Since c is between -1 and 1, and similarly for the other coefficients? Wait, no. The coefficients themselves aren't bounded by 1. For example, in the previous case, a was 8, which is larger than 1. So the coefficients can be larger.

But how can we find the maximum? Maybe we need to consider different cases based on the signs of a, b, c.

Alternatively, maybe use the fact that the maximum of |ax² + bx + c| on [0,1] is 1, so we can use some optimization with constraints.

Alternatively, think of the problem as an optimization problem where we need to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x in [0,1].

But how to handle the infinite constraints? Maybe use the fact that the maximum of |ax² + bx + c| on [0,1] is equal to the maximum of |ax² + bx + c| at the endpoints and at the critical point (vertex).

So, if we can ensure that the maximum is achieved at these points, then we can set up equations.

So, suppose that the maximum of |ax² + bx + c| occurs at x=0, x=1, and at the vertex. Then, if we can set up equations so that the quadratic reaches 1 or -1 at these points, we can solve for a, b, c.

But how to know which points? For example, if the vertex is inside [0,1], then we need to check that value.

Alternatively, maybe consider that the maximum of |ax² + bx + c| on [0,1] is 1, so we can use the three points x=0, x=1, and x=1/2. But maybe not.

Alternatively, think of the quadratic as a function f(x) = ax² + bx + c. Then, since f(0) = c, f(1) = a + b + c, and the vertex is at x = -b/(2a). If the vertex is in [0,1], then f(-b/(2a)) is either a maximum or minimum.

But since we have |f(x)| ≤ 1 on [0,1], then f(0), f(1), and f(-b/(2a)) must all be between -1 and 1.

But how can we use this? Maybe consider different cases based on the position of the vertex.

Alternatively, maybe use Lagrange multipliers with constraints. But since there are infinitely many constraints, it's not straightforward.

Alternatively, think of the problem as an optimization problem where we need to maximize |a| + |b| + |c| given that |ax² + bx + c| ≤ 1 for all x in [0,1].

But maybe we can use some known results or inequalities.

Wait, another idea: since the quadratic is bounded on [0,1], we can use the concept of the maximum norm. Maybe express the coefficients in terms of the values of the function at certain points.

For example, if we know f(0), f(1), and f(1/2), we can solve for a, b, c. Let me try that.

Let me denote f(0) = c, f(1) = a + b + c, and f(1/2) = a*(1/4) + b*(1/2) + c.

So, if I have three equations:

1. c = f(0)
2. a + b + c = f(1)
3. (a/4) + (b/2) + c = f(1/2)

Then, solving for a, b, c:

From equation 1: c = f(0)

From equation 2: a + b = f(1) - c = f(1) - f(0)

From equation 3: (a/4) + (b/2) = f(1/2) - c = f(1/2) - f(0)

Let me write equation 3 as:

Multiply both sides by 4: a + 2b = 4(f(1/2) - f(0))

So, we have:

a + b = f(1) - f(0)  ...(2)

a + 2b = 4(f(1/2) - f(0)) ...(3)

Subtracting equation (2) from equation (3):

(a + 2b) - (a + b) = 4(f(1/2) - f(0)) - (f(1) - f(0))

Which simplifies to:

b = 4(f(1/2) - f(0)) - f(1) + f(0) = 4f(1/2) - 4f(0) - f(1) + f(0) = 4f(1/2) - 3f(0) - f(1)

Then, from equation (2):

a = f(1) - f(0) - b = f(1) - f(0) - [4f(1/2) - 3f(0) - f(1)] = f(1) - f(0) -4f(1/2) + 3f(0) + f(1) = 2f(1) + 2f(0) -4f(1/2)

Therefore, we have expressions for a, b, c in terms of f(0), f(1), f(1/2).

But since |f(x)| ≤ 1 on [0,1], then |f(0)| ≤ 1, |f(1)| ≤ 1, |f(1/2)| ≤ 1.

Therefore, if we want to maximize |a| + |b| + |c|, we can express this in terms of f(0), f(1), f(1/2).

Let me write down:

c = f(0)

b = 4f(1/2) - 3f(0) - f(1)

a = 2f(1) + 2f(0) -4f(1/2)

Therefore, |a| + |b| + |c| = |2f(1) + 2f(0) -4f(1/2)| + |4f(1/2) - 3f(0) - f(1)| + |f(0)|

But since f(0), f(1), f(1/2) are each between -1 and 1, maybe we can find the maximum of this expression.

But this seems complicated. However, if we assume that f(0), f(1), f(1/2) are at their extremes (i.e., ±1), then maybe we can find the maximum.

Let me try that. Let me suppose that f(0) = 1, f(1) = 1, f(1/2) = -1. Then:

a = 2*1 + 2*1 -4*(-1) = 2 + 2 +4 = 8

b = 4*(-1) -3*1 -1 = -4 -3 -1 = -8

c = 1

Which gives |a| + |b| + |c| = 8 + 8 + 1 = 17, same as before.

Alternatively, if f(0) = -1, f(1) = -1, f(1/2) = 1:

a = 2*(-1) + 2*(-1) -4*(1) = -2 -2 -4 = -8

b = 4*(1) -3*(-1) - (-1) = 4 +3 +1 = 8

c = -1

So |a| + |b| + |c| = 8 + 8 + 1 = 17 again.

Alternatively, suppose f(0) = 1, f(1) = -1, f(1/2) = 1. Let's compute:

a = 2*(-1) + 2*(1) -4*(1) = -2 + 2 -4 = -4

b = 4*(1) -3*(1) - (-1) = 4 -3 +1 = 2

c = 1

So |a| + |b| + |c| = 4 + 2 + 1 = 7, which is smaller.

Alternatively, suppose f(0) = 1, f(1) = 1, f(1/2) = 1. Then:

a = 2*1 + 2*1 -4*1 = 2 + 2 -4 = 0

b = 4*1 -3*1 -1 = 4 -3 -1 = 0

c = 1

So |a| + |b| + |c| = 0 + 0 + 1 = 1, which is small.

Alternatively, suppose f(0) = 1, f(1) = -1, f(1/2) = -1:

a = 2*(-1) + 2*(1) -4*(-1) = -2 + 2 +4 = 4

b = 4*(-1) -3*(1) - (-1) = -4 -3 +1 = -6

c = 1

So |a| + |b| + |c| = 4 + 6 + 1 = 11, which is less than 17.

Alternatively, suppose f(0) = 1, f(1) = 1, f(1/2) = 1. Wait, already did that.

Alternatively, suppose f(0) = 1, f(1) = -1, f(1/2) = something else. Maybe trying different combinations.

But maybe the maximum occurs when two of the points are at 1 or -1 and the third is at the opposite. For example, the case we had before with f(0)=1, f(1)=1, f(1/2)=-1 gives 17.

But is there a combination where |a| + |b| + |c| is larger? Let me think.

Suppose we take f(0) = 1, f(1) = 1, and f(1/2) = something else. Wait, but if we set f(1/2) to be something else, maybe we can get a larger |a| + |b| + |c|.

But how? Let me think. Let me suppose that f(0) = 1, f(1) = 1, and f(1/2) = t, where t is between -1 and 1. Then:

a = 2*1 + 2*1 -4t = 4 -4t

b = 4t -3*1 -1 = 4t -4

c = 1

Therefore, |a| + |b| + |c| = |4 -4t| + |4t -4| + 1.

Note that |4 -4t| = 4|1 - t| and |4t -4| = 4|t -1|, so they are equal. Therefore, the sum is 4|1 - t| + 4|1 - t| + 1 = 8|1 - t| + 1.

To maximize this, since t ∈ [-1, 1], |1 - t| is maximized when t is as small as possible, i.e., t = -1. Then |1 - (-1)| = 2, so 8*2 +1 = 17. Which is the same as before. So that case gives maximum 17.

Similarly, if we take other combinations. For example, suppose f(0) = 1, f(1) = -1, and f(1/2) = t. Then:

a = 2*(-1) + 2*(1) -4t = -2 + 2 -4t = -4t

b = 4t -3*(1) - (-1) = 4t -3 +1 = 4t -2

c = 1

So |a| + |b| + |c| = | -4t | + |4t -2| + 1 = 4|t| + |4t -2| + 1.

To maximize this expression over t ∈ [-1, 1]. Let me analyze this function.

Let me consider different intervals for t:

Case 1: t ≥ 0.5. Then 4t -2 ≥ 0. So expression becomes 4t + 4t -2 +1 = 8t -1. This is increasing in t, so maximum at t=1: 8*1 -1 = 7.

Case 2: 0 ≤ t < 0.5. Then 4t -2 < 0. So expression becomes 4t + |4t -2| +1 = 4t + (2 -4t) +1 = 3. So constant 3.

Case 3: t < 0. Then 4t -2 < 0. So expression becomes 4|t| + |4t -2| +1 = -4t + (2 -4t) +1 = -8t +3. This is increasing in t (since coefficient of t is negative). So maximum at t=0: -0 +3 = 3.

Therefore, maximum in this case is 7, which is less than 17.

So, this case doesn't give a higher value.

Similarly, trying other combinations. Suppose f(0) = -1, f(1) = 1, f(1/2) = t. Then:

a = 2*1 + 2*(-1) -4t = 2 -2 -4t = -4t

b = 4t -3*(-1) -1 = 4t +3 -1 = 4t +2

c = -1

So |a| + |b| + |c| = | -4t | + |4t +2| + 1 = 4|t| + |4t +2| +1.

Again, analyze over t ∈ [-1,1].

Case 1: t ≥ -0.5. Then 4t +2 ≥ 0. So expression becomes 4|t| + 4t +2 +1.

Subcases:

If t ≥ 0: 4t +4t +3 = 8t +3. Max at t=1: 8 +3=11.

If -0.5 ≤ t <0: 4*(-t) +4t +3 = -4t +4t +3 = 3.

Case 2: t < -0.5. Then 4t +2 < 0. So expression becomes 4|t| + |4t +2| +1 = -4t + (-4t -2) +1 = -8t -1.

This is increasing in t (since coefficient of t is negative). So maximum at t approaching -0.5 from the left: -8*(-0.5) -1 = 4 -1 = 3.

Therefore, maximum in this case is 11, which is still less than 17.

Another case: suppose f(0) = -1, f(1) = -1, f(1/2) = t. Then:

a = 2*(-1) + 2*(-1) -4t = -2 -2 -4t = -4 -4t

b = 4t -3*(-1) - (-1) = 4t +3 +1 = 4t +4

c = -1

So |a| + |b| + |c| = | -4 -4t | + |4t +4| +1 = 4|1 + t| + 4|t +1| +1 = 8|t +1| +1.

To maximize this over t ∈ [-1,1]. The maximum occurs when |t +1| is maximum, which is when t =1: |1 +1| = 2. So 8*2 +1 = 17.

Same as before.

So, in all these cases, the maximum seems to be 17.

But is there a case where we don't use f(0), f(1), f(1/2) but maybe other points? For example, maybe the maximum occurs when the quadratic reaches 1 or -1 at some other points?

Alternatively, maybe consider that the maximum of |a| + |b| + |c| occurs when the quadratic is extremal in some way.

Alternatively, think of the problem as an optimization problem with constraints. Let me try to set up the problem.

We need to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x ∈ [0,1].

This is a constrained optimization problem. Since the absolute values complicate things, maybe consider different cases based on the signs of a, b, c. However, since we are dealing with absolute values, maybe we can assume certain signs to maximize the sum.

For example, suppose that a, b, c are all positive. Then |a| + |b| + |c| = a + b + c. But if we want to maximize this, given that ax² + bx + c ≤ 1 for all x in [0,1]. However, since the quadratic is positive, its maximum would be at one of the endpoints or at the vertex.

But if we want to maximize a + b + c, maybe set the quadratic to be 1 at x=1, and maybe have it increasing? But if the quadratic is increasing on [0,1], then its maximum is at x=1. So, if we set ax² + bx + c = 1 at x=1, and maybe also at x=0? But if it's increasing, then at x=0, it would be less than 1.

But if we set c = 1 (so that at x=0, it's 1), and then have the quadratic increasing, then at x=1, it would be a + b + 1. To have it ≤1, we need a + b +1 ≤1 => a + b ≤0. But if we want to maximize a + b + c = a + b +1, then with a + b ≤0, the maximum is 0 +1 =1. Which is not great.

Alternatively, if we set the quadratic to be 1 at x=1 and maybe have it dip down to -1 somewhere else. But then the coefficients might be larger.

Alternatively, maybe consider that the maximum of |a| + |b| + |c| occurs when the quadratic alternates between 1 and -1 at three points in [0,1], which is the case for the Chebyshev polynomial. Which gives us 17.

But maybe there's another configuration where |a| + |b| + |c| is larger.

Alternatively, think of the problem in terms of linear algebra. Suppose we want to maximize |a| + |b| + |c| given that |ax² + bx + c| ≤ 1 for all x in [0,1].

But how to relate this? Maybe use the fact that the maximum of |ax² + bx + c| on [0,1] is equal to the maximum of |f(0)|, |f(1)|, and |f(-b/(2a))| if the vertex is in [0,1].

But if we can ensure that the maximum is achieved at these points, then we can set up equations.

Alternatively, think of the problem as follows: since the quadratic is bounded between -1 and 1 on [0,1], then we can use the Lagrange interpolation formula.

But I think I need to think differently. Let me try to use the previous result where we found that 17 is achievable. Maybe that's the maximum.

But to confirm, let me try to see if there's a quadratic that can have a larger |a| + |b| + |c|.

Suppose we take a quadratic that reaches 1 at x=0, -1 at x=1, and something else. Let me try.

Let me suppose f(0) = 1, f(1) = -1, and suppose that the maximum of |f(x)| is 1. Let me try to find such a quadratic.

Let me assume that the quadratic has its maximum at x=1/2. Let me suppose that f(1/2) = 1. Then, we have three points: (0,1), (1/2,1), (1,-1). Let me find the quadratic passing through these points.

Let me write the equations:

At x=0: c = 1.

At x=1/2: a*(1/4) + b*(1/2) + c = 1.

At x=1: a + b + c = -1.

So, c = 1.

From the third equation: a + b + 1 = -1 => a + b = -2.

From the second equation: (a/4) + (b/2) + 1 = 1 => (a/4) + (b/2) = 0 => Multiply by 4: a + 2b = 0.

So, we have:

a + b = -2

a + 2b = 0

Subtract first equation from second: b = 2. Then from a + 2 = -2 => a = -4.

Therefore, the quadratic is -4x² + 2x + 1. Let me check its values:

At x=0: 1, okay.

At x=1/2: -4*(1/4) + 2*(1/2) +1 = -1 +1 +1 = 1, okay.

At x=1: -4 + 2 +1 = -1, okay.

Now, check if this quadratic stays within [-1,1] on [0,1]. Let me check its vertex. The vertex is at x = -b/(2a) = -2/(2*(-4)) = -2 / (-8) = 1/4.

So, the vertex is at x=1/4. Let me compute f(1/4):

-4*(1/4)^2 + 2*(1/4) +1 = -4*(1/16) + 1/2 +1 = -1/4 + 1/2 +1 = (-1/4 + 2/4) + 4/4 = (1/4) + 4/4 = 5/4 = 1.25.

Oh! That's greater than 1. Which violates the condition. Therefore, this quadratic is not allowed.

So, even though we set the values at three points to be within [-1,1], the quadratic exceeds the bound in between. Therefore, this quadratic is invalid.

So, this shows that just setting three points isn't sufficient; we need to ensure that the quadratic doesn't exceed the bounds in between.

Therefore, the previous example with the Chebyshev polynomial is valid because it alternates between 1 and -1 at three points, which is the minimal maximum deviation, hence ensuring that it doesn't exceed the bounds.

So, in this case, the quadratic -8x² +8x -1 has its maximum at x=1/2, which is 1, and minimum at x=0 and x=1, which is -1. Wait, no. Wait, earlier we had 8x² -8x +1 which had maximum at x=0 and x=1 (value 1) and minimum at x=1/2 (value -1). Similarly, the other quadratic -8x² +8x -1 would have maximum at x=1/2 (value 1) and minima at x=0 and x=1 (value -1).

But in the case where I tried to set f(0)=1, f(1/2)=1, f(1)=-1, the quadratic exceeded the bound at x=1/4. Therefore, that quadratic is not valid.

Therefore, the Chebyshev polynomial case is special because it alternates between the extremes at three points, which is the case for the minimal maximum deviation.

Therefore, perhaps the maximum of |a| + |b| + |c| is indeed 17.

But let me try another approach. Let me consider the general case.

Suppose that the quadratic is bounded by 1 in absolute value on [0,1]. Then, we can use the fact that for any x in [0,1], we have:

-1 ≤ ax² + bx + c ≤ 1.

We can consider specific values of x to get inequalities. For example, at x=0: -1 ≤ c ≤ 1. At x=1: -1 ≤ a + b + c ≤ 1. Also, at x=1/2: -1 ≤ a*(1/4) + b*(1/2) + c ≤ 1.

But these are three inequalities. However, there are infinitely many more.

But maybe we can use these three inequalities to bound a, b, c.

Alternatively, think of the problem as a linear programming problem with infinitely many constraints. But since we are dealing with absolute values, it's more complex.

Alternatively, use the concept of the maximum of |a| + |b| + |c|. Let me think of |a| + |b| + |c| as a linear function, and we need to maximize it under the constraint that |ax² + bx + c| ≤ 1 for all x in [0,1].

This is a convex optimization problem, but I don't know the exact method to solve it.

Alternatively, think of the maximum of |a| + |b| + |c| as the L1 norm of the coefficients, and we need to maximize it under the L∞ norm constraint on the function.

But I don't recall the exact relation between these norms.

Alternatively, use the fact that for any x in [0,1], ax² + bx + c is between -1 and 1. Therefore, we can write:

For all x in [0,1], -1 ≤ ax² + bx + c ≤ 1.

This can be rewritten as:

For all x in [0,1], ax² + bx + c ≤ 1,

and

For all x in [0,1], -ax² - bx - c ≤ 1.

But how to use this?

Alternatively, consider that the maximum of ax² + bx + c on [0,1] is ≤1 and the minimum is ≥-1.

But how to relate this to a, b, c?

Alternatively, use the fact that the maximum of ax² + bx + c on [0,1] can be found by evaluating at critical points.

The critical point is at x = -b/(2a), provided that this is in [0,1].

So, if we can ensure that the maximum and minimum of the quadratic on [0,1] are within [-1,1], then we are okay.

But how to maximize |a| + |b| + |c|?

Alternatively, think of specific cases where the maximum of |ax² + bx + c| is achieved at three points, which would be the case for the Chebyshev polynomial.

Since we already have an example with |a| + |b| + |c| = 17, and trying other configurations gives lower values, maybe 17 is indeed the maximum.

But to be thorough, let me try another approach. Let me consider the quadratic function f(x) = ax² + bx + c.

We can express f(x) in terms of its values at three points. For example, using the basis of polynomials.

But since we have constraints on the entire interval, maybe use the concept of the maximum modulus.

Alternatively, use the fact that for any x in [0,1], we can write:

ax² + bx + c = a(x² - 1) + b(x - 0) + c(1 - 0) + something? Not sure.

Alternatively, think of the quadratic as a linear combination of basis functions. For example, the monomials 1, x, x².

But I need to relate the coefficients to the maximum of the function.

Alternatively, use the following idea: since the maximum of |f(x)| is 1 on [0,1], then the coefficients a, b, c can be related to the maximum values.

For example, consider evaluating f(x) at x=0, x=1, and x=1/2.

But earlier, we saw that if we set f(0)=1, f(1)=1, f(1/2)=-1, we get a quadratic with |a| + |b| + |c| = 17.

Alternatively, think of the following: suppose that we want to maximize |a| + |b| + |c|. Let me assume that a, b, c are positive. Then, |a| + |b| + |c| = a + b + c.

But if we want to maximize a + b + c, given that ax² + bx + c ≤ 1 for all x in [0,1].

But since the quadratic is bounded above by 1, then at x=0, c ≤1. At x=1, a + b + c ≤1. Also, the maximum of the quadratic could be at the vertex.

If the vertex is in [0,1], then the maximum value is at the vertex.

So, suppose the vertex is at x = -b/(2a). If this is in [0,1], then the maximum value is f(-b/(2a)) = c - b²/(4a).

But since we want this to be ≤1.

But this seems complicated.

Alternatively, suppose that the maximum occurs at x=1. Then, a + b + c =1. Also, since the quadratic is increasing on [0,1], which would require that the derivative at x=1 is positive. The derivative is 2ax + b. At x=1, it's 2a + b > 0.

But if we want to maximize a + b + c, with c ≤1, and a + b + c ≤1, then the maximum would be when a + b + c =1 and c is as large as possible. But if c is 1, then a + b =0. Then, the quadratic is ax² + bx +1 with a + b =0. So, b = -a. Then, the quadratic is ax² -a x +1.

To ensure that this quadratic is ≤1 on [0,1], we need to check its maximum.

The vertex is at x = -b/(2a) = a/(2a) = 1/2.

So, the value at x=1/2 is a*(1/4) -a*(1/2) +1 = -a/4 +1.

To ensure that this is ≤1, we need -a/4 +1 ≤1 => -a/4 ≤0 => a ≥0.

Also, since the quadratic is ax² -a x +1, and we want it to be ≤1 on [0,1].

At x=0: 1, okay.

At x=1: a -a +1 =1, okay.

At x=1/2: -a/4 +1 ≤1, which is okay as long as a ≥0.

But also, we need to ensure that the quadratic doesn't exceed 1 in between. However, since the maximum is at x=1/2, which is 1 -a/4. To ensure that this is ≤1, we need a ≥0.

But if we want to maximize a + b + c = a + (-a) +1 =1. So, regardless of a, the sum is 1. Therefore, this approach doesn't help us get a higher value.

Alternatively, if we consider that the maximum of the quadratic is at x=0 or x=1. For example, if the quadratic is decreasing on [0,1], then its maximum is at x=0. Then, c ≤1, and the minimum is at x=1. But then, similar to above, we can't get a higher sum.

Alternatively, suppose that the quadratic has its maximum at the vertex. Let me assume that the vertex is inside [0,1], and that the maximum value is 1. Then, we have:

f(-b/(2a)) = 1, and also f(0) ≤1, f(1) ≤1.

But how to relate this to |a| + |b| + |c|.

Alternatively, think of the quadratic as f(x) = ax² + bx + c. Let me assume that f(x) reaches 1 at x=0, x=1, and -1 at x=1/2, which is the Chebyshev case. Then, as we saw, this gives |a| + |b| + |c| = 17.

Alternatively, suppose that the quadratic reaches 1 at x=0, -1 at x=1, and 1 at x=1/2. Wait, we tried that earlier and found that the quadratic exceeded the bound.

But maybe if we adjust the coefficients so that it doesn't exceed?

Let me try again. Suppose f(0)=1, f(1)=-1, f(1/2)=1.

As before, solving:

c =1

a + b + c = -1 => a + b = -2

a*(1/4) + b*(1/2) + c =1 => a/4 + b/2 +1 =1 => a/4 + b/2 =0 => a + 2b =0

So, solving:

a + b = -2

a + 2b =0

Subtract first equation from second: b = 2. Then a = -2 -b = -4.

So, quadratic is -4x² + 2x +1.

As before, this quadratic has a maximum at x=1/4, which is -4*(1/16) + 2*(1/4) +1 = -1/4 + 1/2 +1 = 1.25, which is greater than 1. Therefore, this quadratic is invalid.

Therefore, to make sure that the quadratic doesn't exceed 1, we need to adjust the coefficients.

Suppose we want f(0)=1, f(1)=-1, and the maximum at x=1/4 is 1. Let me try to find such a quadratic.

Let me denote the quadratic as f(x) = ax² + bx + c.

We have:

f(0) = c =1.

f(1) = a + b + c = -1 => a + b = -2.

The maximum occurs at x = -b/(2a). Let me assume that this maximum is at x=1/4. Therefore:

-b/(2a) = 1/4 => -b = a/2 => b = -a/2.

From a + b = -2, substituting b:

a - a/2 = -2 => a/2 = -2 => a = -4. Then b = -(-4)/2 = 2.

So, same as before: f(x) = -4x² + 2x +1. Which we saw has a maximum of 1.25 at x=1/4.

But we wanted the maximum to be 1. Therefore, we need to adjust the coefficients so that the maximum is 1.

Let me suppose that the maximum at x=1/4 is 1. Then:

f(1/4) = -4*(1/4)^2 + 2*(1/4) +1 = -4*(1/16) + 1/2 +1 = -1/4 + 1/2 +1 = 5/4 = 1.25.

To make this equal to 1, we need to scale down the quadratic. Let me suppose that the quadratic is k*(-4x² + 2x +1). Then, f(1/4) = k*(5/4) =1 => k=4/5.

Therefore, the quadratic becomes (4/5)*(-4x² + 2x +1) = -16/5 x² + 8/5 x + 4/5.

Now, check the values:

At x=0: 4/5, which is within [-1,1].

At x=1: -16/5 +8/5 +4/5 = (-16 +8 +4)/5 = (-4)/5 = -0.8, which is within [-1,1].

At x=1/4: 1, as desired.

Now, check if this quadratic stays within [-1,1] on [0,1].

The maximum value is 1 at x=1/4, and the minimum? Let's check at x=1: -0.8, and at x=0: 0.8. Also, check the other critical points? The vertex is at x= -b/(2a) = -(8/5)/(2*(-16/5)) = -(8/5)/(-32/5) = 8/32 = 1/4, which is the same as before. So, the maximum is at x=1/4, and the minimum would be at the endpoints.

At x=0: 4/5 = 0.8, and at x=1: -4/5 = -0.8. Therefore, the quadratic is between -0.8 and 1 on [0,1]. Therefore, it satisfies the condition.

Now, compute |a| + |b| + |c| for this quadratic:

a = -16/5, b = 8/5, c = 4/5.

So |a| + |b| + |c| = 16/5 + 8/5 + 4/5 = 28/5 = 5.6, which is much less than 17.

Therefore, scaling down the quadratic reduces the sum.

Therefore, this approach doesn't help.

Alternatively, maybe there's a different configuration where the quadratic reaches 1 or -1 at more points or in a different way.

Alternatively, think of the problem in terms of linear algebra. Suppose we want to maximize |a| + |b| + |c|. Let me assume that a, b, c are positive. Then, the maximum would be a + b + c.

But how to maximize this under the constraint that ax² + bx + c ≤1 for all x in [0,1].

But if we set x=0, then c ≤1. If we set x=1, then a + b + c ≤1. If we set x=1/2, then a/4 + b/2 + c ≤1.

But to maximize a + b + c, we need to set a + b + c =1, and also ensure that the quadratic doesn't exceed 1 elsewhere.

But if we set a + b + c =1, and want to maximize a + b + c, which is already 1. So this doesn't help.

Alternatively, if we allow some of the coefficients to be negative. For example, if we have a large positive a and negative b and c.

But how?

Let me think. Suppose that the quadratic is negative at x=0 and x=1, and positive in between. For example, like the Chebyshev polynomial.

But in that case, the coefficients can be large.

Alternatively, think of the following: suppose that the quadratic reaches 1 at x=0 and x=1, and -1 at some point in between. As we saw earlier, this gives a quadratic with |a| + |b| + |c| = 17.

Alternatively, suppose that the quadratic reaches 1 at x=0, -1 at x=1, and 1 at some other point. But earlier attempts showed that this leads to exceeding the bounds.

Alternatively, think of the following: suppose that the quadratic is 1 at x=0, and -1 at x=1, and also has its minimum at x=1/2. Wait, but we saw that this leads to exceeding the bound.

Alternatively, maybe use a different set of points. For example, suppose that the quadratic reaches 1 at x=0, and -1 at x=1, and also has a critical point at x= something else.

But this seems too vague.

Alternatively, think of the problem as follows: since we have the quadratic bounded between -1 and 1 on [0,1], then we can use the following identity:

For any x in [0,1], ax² + bx + c = 1 - 2*(something).

But I don't know.

Alternatively, use the fact that the maximum of |a| + |b| + |c| is achieved when the quadratic is extremal, i.e., when it touches the bounds 1 or -1 at several points.

Since we already have an example with 17, and other attempts to find higher values have failed, maybe 17 is indeed the maximum.

Alternatively, check literature or known results. Wait, I recall that for polynomials bounded on an interval, the maximum coefficients can be related to Chebyshev polynomials.

The Chebyshev polynomials are known to have the minimal maximum norm on [-1,1] among polynomials with leading coefficient 2^{n-1} for degree n. But in our case, we have a different interval [0,1], and we are looking to maximize the sum of absolute values of coefficients.

But the example we have with the transformed Chebyshev polynomial gives us a quadratic with coefficients 8, -8, 1, which sum to 17.

Alternatively, think of the following: suppose we want to maximize |a| + |b| + |c|. Let me assume that a, b, c are such that the quadratic reaches 1 at x=0, x=1, and -1 at x=1/2. This gives us the coefficients a=8, b=-8, c=1, as before.

Alternatively, suppose that the quadratic reaches -1 at x=0, x=1, and 1 at x=1/2. Then, similarly, we get a=-8, b=8, c=-1, and |a| + |b| + |c| = 17.

Therefore, these are the extremal cases.

Since all other attempts to find higher values have failed, and given that the Chebyshev polynomial is known for minimizing the maximum deviation, which in this case is exactly our constraint, it's likely that this is the maximal case.

Therefore, I think the largest possible value of |a| + |b| + |c| is 17.

**Final Answer**
\boxed{17}

To solve the problem, we are given that for all $ x \in [0,1] $, the inequality
$$
|ax^2 + bx + c| \leq 1
$$
holds, and we are to find the **maximum possible value** of
$$
|a| + |b| + |c|.
$$

---

### Step 1: Understanding the Constraint

The constraint $ |ax^2 + bx + c| \leq 1 $ for all $ x \in [0,1] $ means that the quadratic function $ f(x) = ax^2 + bx + c $ is bounded between $-1$ and $1$ on the interval $[0,1]$. This is a **uniform bound** on the function over the entire interval.

Our goal is to **maximize** the sum of the absolute values of the coefficients $ |a| + |b| + |c| $, under this constraint.

---

### Step 2: Strategy and Key Insight

This is a classic optimization problem with a constraint on the function's maximum absolute value over an interval. A powerful tool in such problems is the **Chebyshev polynomial**, which is known to minimize the maximum deviation from zero over an interval, given a fixed leading coefficient.

The **Chebyshev polynomial of degree 2** is:
$$
T_2(x) = 2x^2 - 1
$$
which oscillates between $-1$ and $1$ on the interval $[-1,1]$. To adapt it to the interval $[0,1]$, we perform a linear transformation:
$$
x = \frac{t + 1}{2} \quad \text{so that } t = 2x - 1
$$
Then,
$$
T_2(t) = 2t^2 - 1 = 2(2x - 1)^2 - 1 = 8x^2 - 8x + 1
$$
This polynomial satisfies:
$$
|8x^2 - 8x + 1| \leq 1 \quad \text{for all } x \in [0,1]
$$
and it reaches the bounds $1$ and $-1$ at three points in the interval:
- $ x = 0 $: $ f(0) = 1 $
- $ x = 1 $: $ f(1) = 1 $
- $ x = \frac{1}{2} $: $ f\left(\frac{1}{2}\right) = -1 $

Thus, this polynomial is **extremal** in the sense that it achieves the maximum possible deviation from zero while staying within the bounds $[-1,1]$.

---

### Step 3: Compute the Sum of Absolute Values

For the polynomial $ f(x) = 8x^2 - 8x + 1 $, the coefficients are:
- $ a = 8 $
- $ b = -8 $
- $ c = 1 $

So,
$$
|a| + |b| + |c| = |8| + |-8| + |1| = 8 + 8 + 1 = 17
$$

---

### Step 4: Verifying Optimality

We can consider other configurations of the quadratic function that satisfy the constraint $ |ax^2 + bx + c| \leq 1 $ on $[0,1]$, but all such configurations either:
- Do not reach the extremal values of $1$ and $-1$ at multiple points, or
- Result in smaller values of $ |a| + |b| + |c| $

For example, if we try to construct a quadratic that reaches $1$ at $x = 0$, $-1$ at $x = 1$, and $1$ at $x = \frac{1}{2}$, the resulting polynomial exceeds the bound $1$ at some point in the interval, violating the constraint.

Hence, the Chebyshev polynomial is the **optimal** choice in this context, and the value $17$ is the **maximum** possible value of $ |a| + |b| + |c| $ under the given constraint.

---

### Final Answer

$$
\boxed{17}
$$

[Assistant]
Explanation:
Okay, so I need to find the largest possible value of |a| + |b| + |c| given that the absolute value of the quadratic function ax² + bx + c is less than or equal to 1 for all x between 0 and 1. Hmm, let me think about how to approach this.

First, I know that the quadratic is bounded between -1 and 1 on the interval [0, 1]. So, for every x in [0, 1], the value of ax² + bx + c is within that range. My goal is to maximize |a| + |b| + |c|.

I remember that sometimes problems like this involve using specific points in the interval to set up equations or inequalities. Since the quadratic is bounded on [0, 1], maybe I can evaluate it at certain key points, like x = 0, x = 1, and maybe the vertex of the parabola? Let me try that.

Let me first note the values of the quadratic at x = 0 and x = 1:

At x = 0: |c| ≤ 1, so c is between -1 and 1.

At x = 1: |a + b + c| ≤ 1.

Also, the vertex of the parabola is at x = -b/(2a). But since the interval is [0, 1], the vertex might lie inside this interval or outside. If it's inside, then the maximum or minimum of the quadratic could be there. However, since the quadratic is bounded between -1 and 1 on the entire interval, the maximum absolute value could occur at the endpoints or at the vertex.

But maybe I need to consider more than just the endpoints. For example, if the quadratic has its maximum or minimum inside the interval, then that point must also satisfy |ax² + bx + c| ≤ 1.

Alternatively, maybe I can use some known inequalities or optimization techniques. Since we need to maximize |a| + |b| + |c|, perhaps we can consider different cases based on the signs of a, b, c? But that might get complicated.

Alternatively, maybe I can use Lagrange multipliers? But since there are infinitely many constraints (for all x in [0,1]), that might not be straightforward.

Wait, another idea: maybe use the Chebyshev polynomials? Because Chebyshev polynomials are known for minimizing the maximum deviation from zero over an interval, which might be related. But I need to maximize |a| + |b| + |c| under the constraint that the quadratic is bounded by 1 in absolute value on [0,1].

Alternatively, maybe I can parametrize the quadratic in terms of its values at certain points. For example, if I know the values at three points, I can determine the coefficients. Since the interval is [0,1], maybe I can use the values at x=0, x=1, and x=1/2? But I need to ensure that the quadratic is bounded between -1 and 1 on the entire interval.

Alternatively, maybe consider specific quadratics that achieve the maximum. For example, if I can find a quadratic that touches the bounds 1 or -1 at several points in [0,1], then that might be the extremal case.

Let me try to think of an example. Suppose the quadratic reaches 1 at x=0 and x=1, and -1 at some point in between. Or maybe reaches 1 at one point and -1 at another.

Let me try constructing such a quadratic. Let's assume that the maximum of |ax² + bx + c| is achieved at three points in [0,1], which would be the case for the Chebyshev polynomial. But Chebyshev polynomials are usually defined on [-1,1], so maybe I need to adjust them to the interval [0,1].

The Chebyshev polynomial of degree 2 is T2(x) = 2x² - 1, which oscillates between -1 and 1 on [-1,1]. If I can map the interval [0,1] to [-1,1], maybe I can use that. Let me try that.

Let me perform a linear transformation. Let t = 2x - 1, so that when x=0, t=-1 and x=1, t=1. Then, substituting into T2(t):

T2(t) = 2t² - 1 = 2(2x - 1)² - 1 = 2(4x² -4x +1) -1 = 8x² -8x +2 -1 = 8x² -8x +1.

So, the polynomial 8x² -8x +1 oscillates between -1 and 1 on [0,1]. Let me check:

At x=0: 8(0) -8(0) +1 = 1.

At x=1: 8(1) -8(1) +1 = 1.

At x=1/2: 8*(1/4) -8*(1/2) +1 = 2 -4 +1 = -1.

So yes, it reaches 1 at x=0 and x=1, and -1 at x=1/2. Therefore, |8x² -8x +1| ≤ 1 on [0,1].

So, if we take this quadratic, then a=8, b=-8, c=1. Then |a| + |b| + |c| = 8 + 8 + 1 = 17.

Is this the maximum? Maybe. But let me check if there are other quadratics that can give a higher value.

Alternatively, maybe there's a quadratic that reaches 1 or -1 at more points or in a different configuration.

Alternatively, suppose we consider quadratics that attain maximum absolute value at the endpoints and at the midpoint. But the above example already does that.

But let me check if there's a way to get higher coefficients. For example, suppose we have a quadratic that reaches 1 at x=0, -1 at x=1/2, and 1 at x=1. Then, that's exactly the polynomial we have above.

Alternatively, suppose we have a quadratic that reaches 1 at x=0, 1 at x=1, and some other value in between. But if it's 1 at both ends, then maybe the maximum of the quadratic is 1, but then the coefficients might be smaller?

Alternatively, maybe if we have a quadratic that reaches -1 at x=0 and x=1, and 1 at the midpoint. Let me try that.

Let me suppose that at x=0, ax² + bx + c = -1, so c = -1. At x=1, a + b + c = -1. So, a + b -1 = -1 => a + b = 0 => b = -a. At x=1/2, the value is a*(1/4) + b*(1/2) + c = a/4 + (-a)/2 + (-1) = (-a/4) -1. We want this to be 1. So:

(-a/4) -1 = 1 => -a/4 = 2 => a = -8. Then b = -a = 8. So the quadratic is -8x² +8x -1. Let me check its values:

At x=0: -1, which is okay.

At x=1: -8 +8 -1 = -1, okay.

At x=1/2: -8*(1/4) +8*(1/2) -1 = -2 +4 -1 = 1, okay.

So this quadratic also oscillates between -1 and 1 on [0,1]. Then |a| + |b| + |c| = 8 + 8 + 1 = 17 again. Same as before.

So whether we have the polynomial reaching 1 at the endpoints and -1 in the middle, or -1 at the endpoints and 1 in the middle, we get the same sum.

But is 17 the maximum? Maybe there's another quadratic where |a| + |b| + |c| is larger?

Let me think. Suppose we try to maximize |a| + |b| + |c|. Since c is between -1 and 1, and similarly for the other coefficients? Wait, no. The coefficients themselves aren't bounded by 1. For example, in the previous case, a was 8, which is larger than 1. So the coefficients can be larger.

But how can we find the maximum? Maybe we need to consider different cases based on the signs of a, b, c.

Alternatively, maybe use the fact that the maximum of |ax² + bx + c| on [0,1] is 1, so we can use some optimization with constraints.

Alternatively, think of the problem as an optimization problem where we need to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x in [0,1].

But how to handle the infinite constraints? Maybe use the fact that the maximum of |ax² + bx + c| on [0,1] is equal to the maximum of |ax² + bx + c| at the endpoints and at the critical point (vertex).

So, if we can ensure that the maximum is achieved at these points, then we can set up equations.

So, suppose that the maximum of |ax² + bx + c| occurs at x=0, x=1, and at the vertex. Then, if we can set up equations so that the quadratic reaches 1 or -1 at these points, we can solve for a, b, c.

But how to know which points? For example, if the vertex is inside [0,1], then we need to check that value.

Alternatively, maybe consider that the maximum of |ax² + bx + c| on [0,1] is 1, so we can use the three points x=0, x=1, and x=1/2. But maybe not.

Alternatively, think of the quadratic as a function f(x) = ax² + bx + c. Then, since f(0) = c, f(1) = a + b + c, and the vertex is at x = -b/(2a). If the vertex is in [0,1], then f(-b/(2a)) is either a maximum or minimum.

But since we have |f(x)| ≤ 1 on [0,1], then f(0), f(1), and f(-b/(2a)) must all be between -1 and 1.

But how can we use this? Maybe consider different cases based on the position of the vertex.

Alternatively, maybe use Lagrange multipliers with constraints. But since there are infinitely many constraints, it's not straightforward.

Alternatively, think of the problem as an optimization problem where we need to maximize |a| + |b| + |c| given that |ax² + bx + c| ≤ 1 for all x in [0,1].

But maybe we can use some known results or inequalities.

Wait, another idea: since the quadratic is bounded on [0,1], we can use the concept of the maximum norm. Maybe express the coefficients in terms of the values of the function at certain points.

For example, if we know f(0), f(1), and f(1/2), we can solve for a, b, c. Let me try that.

Let me denote f(0) = c, f(1) = a + b + c, and f(1/2) = a*(1/4) + b*(1/2) + c.

So, if I have three equations:

1. c = f(0)
2. a + b + c = f(1)
3. (a/4) + (b/2) + c = f(1/2)

Then, solving for a, b, c:

From equation 1: c = f(0)

From equation 2: a + b = f(1) - c = f(1) - f(0)

From equation 3: (a/4) + (b/2) = f(1/2) - c = f(1/2) - f(0)

Let me write equation 3 as:

Multiply both sides by 4: a + 2b = 4(f(1/2) - f(0))

So, we have:

a + b = f(1) - f(0)  ...(2)

a + 2b = 4(f(1/2) - f(0)) ...(3)

Subtracting equation (2) from equation (3):

(a + 2b) - (a + b) = 4(f(1/2) - f(0)) - (f(1) - f(0))

Which simplifies to:

b = 4(f(1/2) - f(0)) - f(1) + f(0) = 4f(1/2) - 4f(0) - f(1) + f(0) = 4f(1/2) - 3f(0) - f(1)

Then, from equation (2):

a = f(1) - f(0) - b = f(1) - f(0) - [4f(1/2) - 3f(0) - f(1)] = f(1) - f(0) -4f(1/2) + 3f(0) + f(1) = 2f(1) + 2f(0) -4f(1/2)

Therefore, we have expressions for a, b, c in terms of f(0), f(1), f(1/2).

But since |f(x)| ≤ 1 on [0,1], then |f(0)| ≤ 1, |f(1)| ≤ 1, |f(1/2)| ≤ 1.

Therefore, if we want to maximize |a| + |b| + |c|, we can express this in terms of f(0), f(1), f(1/2).

Let me write down:

c = f(0)

b = 4f(1/2) - 3f(0) - f(1)

a = 2f(1) + 2f(0) -4f(1/2)

Therefore, |a| + |b| + |c| = |2f(1) + 2f(0) -4f(1/2)| + |4f(1/2) - 3f(0) - f(1)| + |f(0)|

But since f(0), f(1), f(1/2) are each between -1 and 1, maybe we can find the maximum of this expression.

But this seems complicated. However, if we assume that f(0), f(1), f(1/2) are at their extremes (i.e., ±1), then maybe we can find the maximum.

Let me try that. Let me suppose that f(0) = 1, f(1) = 1, f(1/2) = -1. Then:

a = 2*1 + 2*1 -4*(-1) = 2 + 2 +4 = 8

b = 4*(-1) -3*1 -1 = -4 -3 -1 = -8

c = 1

Which gives |a| + |b| + |c| = 8 + 8 + 1 = 17, same as before.

Alternatively, if f(0) = -1, f(1) = -1, f(1/2) = 1:

a = 2*(-1) + 2*(-1) -4*(1) = -2 -2 -4 = -8

b = 4*(1) -3*(-1) - (-1) = 4 +3 +1 = 8

c = -1

So |a| + |b| + |c| = 8 + 8 + 1 = 17 again.

Alternatively, suppose f(0) = 1, f(1) = -1, f(1/2) = 1. Let's compute:

a = 2*(-1) + 2*(1) -4*(1) = -2 + 2 -4 = -4

b = 4*(1) -3*(1) - (-1) = 4 -3 +1 = 2

c = 1

So |a| + |b| + |c| = 4 + 2 + 1 = 7, which is smaller.

Alternatively, suppose f(0) = 1, f(1) = 1, f(1/2) = 1. Then:

a = 2*1 + 2*1 -4*1 = 2 + 2 -4 = 0

b = 4*1 -3*1 -1 = 4 -3 -1 = 0

c = 1

So |a| + |b| + |c| = 0 + 0 + 1 = 1, which is small.

Alternatively, suppose f(0) = 1, f(1) = -1, f(1/2) = -1:

a = 2*(-1) + 2*(1) -4*(-1) = -2 + 2 +4 = 4

b = 4*(-1) -3*(1) - (-1) = -4 -3 +1 = -6

c = 1

So |a| + |b| + |c| = 4 + 6 + 1 = 11, which is less than 17.

Alternatively, suppose f(0) = 1, f(1) = 1, f(1/2) = 1. Wait, already did that.

Alternatively, suppose f(0) = 1, f(1) = -1, f(1/2) = something else. Maybe trying different combinations.

But maybe the maximum occurs when two of the points are at 1 or -1 and the third is at the opposite. For example, the case we had before with f(0)=1, f(1)=1, f(1/2)=-1 gives 17.

But is there a combination where |a| + |b| + |c| is larger? Let me think.

Suppose we take f(0) = 1, f(1) = 1, and f(1/2) = something else. Wait, but if we set f(1/2) to be something else, maybe we can get a larger |a| + |b| + |c|.

But how? Let me think. Let me suppose that f(0) = 1, f(1) = 1, and f(1/2) = t, where t is between -1 and 1. Then:

a = 2*1 + 2*1 -4t = 4 -4t

b = 4t -3*1 -1 = 4t -4

c = 1

Therefore, |a| + |b| + |c| = |4 -4t| + |4t -4| + 1.

Note that |4 -4t| = 4|1 - t| and |4t -4| = 4|t -1|, so they are equal. Therefore, the sum is 4|1 - t| + 4|1 - t| + 1 = 8|1 - t| + 1.

To maximize this, since t ∈ [-1, 1], |1 - t| is maximized when t is as small as possible, i.e., t = -1. Then |1 - (-1)| = 2, so 8*2 +1 = 17. Which is the same as before. So that case gives maximum 17.

Similarly, if we take other combinations. For example, suppose f(0) = 1, f(1) = -1, and f(1/2) = t. Then:

a = 2*(-1) + 2*(1) -4t = -2 + 2 -4t = -4t

b = 4t -3*(1) - (-1) = 4t -3 +1 = 4t -2

c = 1

So |a| + |b| + |c| = | -4t | + |4t -2| + 1 = 4|t| + |4t -2| + 1.

To maximize this expression over t ∈ [-1, 1]. Let me analyze this function.

Let me consider different intervals for t:

Case 1: t ≥ 0.5. Then 4t -2 ≥ 0. So expression becomes 4t + 4t -2 +1 = 8t -1. This is increasing in t, so maximum at t=1: 8*1 -1 = 7.

Case 2: 0 ≤ t < 0.5. Then 4t -2 < 0. So expression becomes 4t + |4t -2| +1 = 4t + (2 -4t) +1 = 3. So constant 3.

Case 3: t < 0. Then 4t -2 < 0. So expression becomes 4|t| + |4t -2| +1 = -4t + (2 -4t) +1 = -8t +3. This is increasing in t (since coefficient of t is negative). So maximum at t=0: -0 +3 = 3.

Therefore, maximum in this case is 7, which is less than 17.

So, this case doesn't give a higher value.

Similarly, trying other combinations. Suppose f(0) = -1, f(1) = 1, f(1/2) = t. Then:

a = 2*1 + 2*(-1) -4t = 2 -2 -4t = -4t

b = 4t -3*(-1) -1 = 4t +3 -1 = 4t +2

c = -1

So |a| + |b| + |c| = | -4t | + |4t +2| + 1 = 4|t| + |4t +2| +1.

Again, analyze over t ∈ [-1,1].

Case 1: t ≥ -0.5. Then 4t +2 ≥ 0. So expression becomes 4|t| + 4t +2 +1.

Subcases:

If t ≥ 0: 4t +4t +3 = 8t +3. Max at t=1: 8 +3=11.

If -0.5 ≤ t <0: 4*(-t) +4t +3 = -4t +4t +3 = 3.

Case 2: t < -0.5. Then 4t +2 < 0. So expression becomes 4|t| + |4t +2| +1 = -4t + (-4t -2) +1 = -8t -1.

This is increasing in t (since coefficient of t is negative). So maximum at t approaching -0.5 from the left: -8*(-0.5) -1 = 4 -1 = 3.

Therefore, maximum in this case is 11, which is still less than 17.

Another case: suppose f(0) = -1, f(1) = -1, f(1/2) = t. Then:

a = 2*(-1) + 2*(-1) -4t = -2 -2 -4t = -4 -4t

b = 4t -3*(-1) - (-1) = 4t +3 +1 = 4t +4

c = -1

So |a| + |b| + |c| = | -4 -4t | + |4t +4| +1 = 4|1 + t| + 4|t +1| +1 = 8|t +1| +1.

To maximize this over t ∈ [-1,1]. The maximum occurs when |t +1| is maximum, which is when t =1: |1 +1| = 2. So 8*2 +1 = 17.

Same as before.

So, in all these cases, the maximum seems to be 17.

But is there a case where we don't use f(0), f(1), f(1/2) but maybe other points? For example, maybe the maximum occurs when the quadratic reaches 1 or -1 at some other points?

Alternatively, maybe consider that the maximum of |a| + |b| + |c| occurs when the quadratic is extremal in some way.

Alternatively, think of the problem as an optimization problem with constraints. Let me try to set up the problem.

We need to maximize |a| + |b| + |c| subject to |ax² + bx + c| ≤ 1 for all x ∈ [0,1].

This is a constrained optimization problem. Since the absolute values complicate things, maybe consider different cases based on the signs of a, b, c. However, since we are dealing with absolute values, maybe we can assume certain signs to maximize the sum.

For example, suppose that a, b, c are all positive. Then |a| + |b| + |c| = a + b + c. But if we want to maximize this, given that ax² + bx + c ≤ 1 for all x in [0,1]. However, since the quadratic is positive, its maximum would be at one of the endpoints or at the vertex.

But if we want to maximize a + b + c, maybe set the quadratic to be 1 at x=1, and maybe have it increasing? But if the quadratic is increasing on [0,1], then its maximum is at x=1. So, if we set ax² + bx + c = 1 at x=1, and maybe also at x=0? But if it's increasing, then at x=0, it would be less than 1.

But if we set c = 1 (so that at x=0, it's 1), and then have the quadratic increasing, then at x=1, it would be a + b + 1. To have it ≤1, we need a + b +1 ≤1 => a + b ≤0. But if we want to maximize a + b + c = a + b +1, then with a + b ≤0, the maximum is 0 +1 =1. Which is not great.

Alternatively, if we set the quadratic to be 1 at x=1 and maybe have it dip down to -1 somewhere else. But then the coefficients might be larger.

Alternatively, maybe consider that the maximum of |a| + |b| + |c| occurs when the quadratic alternates between 1 and -1 at three points in [0,1], which is the case for the Chebyshev polynomial. Which gives us 17.

But maybe there's another configuration where |a| + |b| + |c| is larger.

Alternatively, think of the problem in terms of linear algebra. Suppose we want to maximize |a| + |b| + |c| given that |ax² + bx + c| ≤ 1 for all x in [0,1].

But how to relate this? Maybe use the fact that the maximum of |ax² + bx + c| on [0,1] is equal to the maximum of |f(0)|, |f(1)|, and |f(-b/(2a))| if the vertex is in [0,1].

But if we can ensure that the maximum is achieved at these points, then we can set up equations.

Alternatively, think of the problem as follows: since the quadratic is bounded between -1 and 1 on [0,1], then we can use the Lagrange interpolation formula.

But I think I need to think differently. Let me try to use the previous result where we found that 17 is achievable. Maybe that's the maximum.

But to confirm, let me try to see if there's a quadratic that can have a larger |a| + |b| + |c|.

Suppose we take a quadratic that reaches 1 at x=0, -1 at x=1, and something else. Let me try.

Let me suppose f(0) = 1, f(1) = -1, and suppose that the maximum of |f(x)| is 1. Let me try to find such a quadratic.

Let me assume that the quadratic has its maximum at x=1/2. Let me suppose that f(1/2) = 1. Then, we have three points: (0,1), (1/2,1), (1,-1). Let me find the quadratic passing through these points.

Let me write the equations:

At x=0: c = 1.

At x=1/2: a*(1/4) + b*(1/2) + c = 1.

At x=1: a + b + c = -1.

So, c = 1.

From the third equation: a + b + 1 = -1 => a + b = -2.

From the second equation: (a/4) + (b/2) + 1 = 1 => (a/4) + (b/2) = 0 => Multiply by 4: a + 2b = 0.

So, we have:

a + b = -2

a + 2b = 0

Subtract first equation from second: b = 2. Then from a + 2 = -2 => a = -4.

Therefore, the quadratic is -4x² + 2x + 1. Let me check its values:

At x=0: 1, okay.

At x=1/2: -4*(1/4) + 2*(1/2) +1 = -1 +1 +1 = 1, okay.

At x=1: -4 + 2 +1 = -1, okay.

Now, check if this quadratic stays within [-1,1] on [0,1]. Let me check its vertex. The vertex is at x = -b/(2a) = -2/(2*(-4)) = -2 / (-8) = 1/4.

So, the vertex is at x=1/4. Let me compute f(1/4):

-4*(1/4)^2 + 2*(1/4) +1 = -4*(1/16) + 1/2 +1 = -1/4 + 1/2 +1 = (-1/4 + 2/4) + 4/4 = (1/4) + 4/4 = 5/4 = 1.25.

Oh! That's greater than 1. Which violates the condition. Therefore, this quadratic is not allowed.

So, even though we set the values at three points to be within [-1,1], the quadratic exceeds the bound in between. Therefore, this quadratic is invalid.

So, this shows that just setting three points isn't sufficient; we need to ensure that the quadratic doesn't exceed the bounds in between.

Therefore, the previous example with the Chebyshev polynomial is valid because it alternates between 1 and -1 at three points, which is the minimal maximum deviation, hence ensuring that it doesn't exceed the bounds.

So, in this case, the quadratic -8x² +8x -1 has its maximum at x=1/2, which is 1, and minimum at x=0 and x=1, which is -1. Wait, no. Wait, earlier we had 8x² -8x +1 which had maximum at x=0 and x=1 (value 1) and minimum at x=1/2 (value -1). Similarly, the other quadratic -8x² +8x -1 would have maximum at x=1/2 (value 1) and minima at x=0 and x=1 (value -1).

But in the case where I tried to set f(0)=1, f(1/2)=1, f(1)=-1, the quadratic exceeded the bound at x=1/4. Therefore, that quadratic is not valid.

Therefore, the Chebyshev polynomial case is special because it alternates between the extremes at three points, which is the case for the minimal maximum deviation.

Therefore, perhaps the maximum of |a| + |b| + |c| is indeed 17.

But let me try another approach. Let me consider the general case.

Suppose that the quadratic is bounded by 1 in absolute value on [0,1]. Then, we can use the fact that for any x in [0,1], we have:

-1 ≤ ax² + bx + c ≤ 1.

We can consider specific values of x to get inequalities. For example, at x=0: -1 ≤ c ≤ 1. At x=1: -1 ≤ a + b + c ≤ 1. Also, at x=1/2: -1 ≤ a*(1/4) + b*(1/2) + c ≤ 1.

But these are three inequalities. However, there are infinitely many more.

But maybe we can use these three inequalities to bound a, b, c.

Alternatively, think of the problem as a linear programming problem with infinitely many constraints. But since we are dealing with absolute values, it's more complex.

Alternatively, use the concept of the maximum of |a| + |b| + |c|. Let me think of |a| + |b| + |c| as a linear function, and we need to maximize it under the constraint that |ax² + bx + c| ≤ 1 for all x in [0,1].

This is a convex optimization problem, but I don't know the exact method to solve it.

Alternatively, think of the maximum of |a| + |b| + |c| as the L1 norm of the coefficients, and we need to maximize it under the L∞ norm constraint on the function.

But I don't recall the exact relation between these norms.

Alternatively, use the fact that for any x in [0,1], ax² + bx + c is between -1 and 1. Therefore, we can write:

For all x in [0,1], -1 ≤ ax² + bx + c ≤ 1.

This can be rewritten as:

For all x in [0,1], ax² + bx + c ≤ 1,

and

For all x in [0,1], -ax² - bx - c ≤ 1.

But how to use this?

Alternatively, consider that the maximum of ax² + bx + c on [0,1] is ≤1 and the minimum is ≥-1.

But how to relate this to a, b, c?

Alternatively, use the fact that the maximum of ax² + bx + c on [0,1] can be found by evaluating at critical points.

The critical point is at x = -b/(2a), provided that this is in [0,1].

So, if we can ensure that the maximum and minimum of the quadratic on [0,1] are within [-1,1], then we are okay.

But how to maximize |a| + |b| + |c|?

Alternatively, think of specific cases where the maximum of |ax² + bx + c| is achieved at three points, which would be the case for the Chebyshev polynomial.

Since we already have an example with |a| + |b| + |c| = 17, and trying other configurations gives lower values, maybe 17 is indeed the maximum.

But to be thorough, let me try another approach. Let me consider the quadratic function f(x) = ax² + bx + c.

We can express f(x) in terms of its values at three points. For example, using the basis of polynomials.

But since we have constraints on the entire interval, maybe use the concept of the maximum modulus.

Alternatively, use the fact that for any x in [0,1], we can write:

ax² + bx + c = a(x² - 1) + b(x - 0) + c(1 - 0) + something? Not sure.

Alternatively, think of the quadratic as a linear combination of basis functions. For example, the monomials 1, x, x².

But I need to relate the coefficients to the maximum of the function.

Alternatively, use the following idea: since the maximum of |f(x)| is 1 on [0,1], then the coefficients a, b, c can be related to the maximum values.

For example, consider evaluating f(x) at x=0, x=1, and x=1/2.

But earlier, we saw that if we set f(0)=1, f(1)=1, f(1/2)=-1, we get a quadratic with |a| + |b| + |c| = 17.

Alternatively, think of the following: suppose that we want to maximize |a| + |b| + |c|. Let me assume that a, b, c are positive. Then, |a| + |b| + |c| = a + b + c.

But if we want to maximize a + b + c, given that ax² + bx + c ≤ 1 for all x in [0,1].

But since the quadratic is bounded above by 1, then at x=0, c ≤1. At x=1, a + b + c ≤1. Also, the maximum of the quadratic could be at the vertex.

If the vertex is in [0,1], then the maximum value is at the vertex.

So, suppose the vertex is at x = -b/(2a). If this is in [0,1], then the maximum value is f(-b/(2a)) = c - b²/(4a).

But since we want this to be ≤1.

But this seems complicated.

Alternatively, suppose that the maximum occurs at x=1. Then, a + b + c =1. Also, since the quadratic is increasing on [0,1], which would require that the derivative at x=1 is positive. The derivative is 2ax + b. At x=1, it's 2a + b > 0.

But if we want to maximize a + b + c, with c ≤1, and a + b + c ≤1, then the maximum would be when a + b + c =1 and c is as large as possible. But if c is 1, then a + b =0. Then, the quadratic is ax² + bx +1 with a + b =0. So, b = -a. Then, the quadratic is ax² -a x +1.

To ensure that this quadratic is ≤1 on [0,1], we need to check its maximum.

The vertex is at x = -b/(2a) = a/(2a) = 1/2.

So, the value at x=1/2 is a*(1/4) -a*(1/2) +1 = -a/4 +1.

To ensure that this is ≤1, we need -a/4 +1 ≤1 => -a/4 ≤0 => a ≥0.

Also, since the quadratic is ax² -a x +1, and we want it to be ≤1 on [0,1].

At x=0: 1, okay.

At x=1: a -a +1 =1, okay.

At x=1/2: -a/4 +1 ≤1, which is okay as long as a ≥0.

But also, we need to ensure that the quadratic doesn't exceed 1 in between. However, since the maximum is at x=1/2, which is 1 -a/4. To ensure that this is ≤1, we need a ≥0.

But if we want to maximize a + b + c = a + (-a) +1 =1. So, regardless of a, the sum is 1. Therefore, this approach doesn't help us get a higher value.

Alternatively, if we consider that the maximum of the quadratic is at x=0 or x=1. For example, if the quadratic is decreasing on [0,1], then its maximum is at x=0. Then, c ≤1, and the minimum is at x=1. But then, similar to above, we can't get a higher sum.

Alternatively, suppose that the quadratic has its maximum at the vertex. Let me assume that the vertex is inside [0,1], and that the maximum value is 1. Then, we have:

f(-b/(2a)) = 1, and also f(0) ≤1, f(1) ≤1.

But how to relate this to |a| + |b| + |c|.

Alternatively, think of the quadratic as f(x) = ax² + bx + c. Let me assume that f(x) reaches 1 at x=0, x=1, and -1 at x=1/2, which is the Chebyshev case. Then, as we saw, this gives |a| + |b| + |c| = 17.

Alternatively, suppose that the quadratic reaches 1 at x=0, -1 at x=1, and 1 at x=1/2. Wait, we tried that earlier and found that the quadratic exceeded the bound.

But maybe if we adjust the coefficients so that it doesn't exceed?

Let me try again. Suppose f(0)=1, f(1)=-1, f(1/2)=1.

As before, solving:

c =1

a + b + c = -1 => a + b = -2

a*(1/4) + b*(1/2) + c =1 => a/4 + b/2 +1 =1 => a/4 + b/2 =0 => a + 2b =0

So, solving:

a + b = -2

a + 2b =0

Subtract first equation from second: b = 2. Then a = -2 -b = -4.

So, quadratic is -4x² + 2x +1.

As before, this quadratic has a maximum at x=1/4, which is -4*(1/16) + 2*(1/4) +1 = -1/4 + 1/2 +1 = 1.25, which is greater than 1. Therefore, this quadratic is invalid.

Therefore, to make sure that the quadratic doesn't exceed 1, we need to adjust the coefficients.

Suppose we want f(0)=1, f(1)=-1, and the maximum at x=1/4 is 1. Let me try to find such a quadratic.

Let me denote the quadratic as f(x) = ax² + bx + c.

We have:

f(0) = c =1.

f(1) = a + b + c = -1 => a + b = -2.

The maximum occurs at x = -b/(2a). Let me assume that this maximum is at x=1/4. Therefore:

-b/(2a) = 1/4 => -b = a/2 => b = -a/2.

From a + b = -2, substituting b:

a - a/2 = -2 => a/2 = -2 => a = -4. Then b = -(-4)/2 = 2.

So, same as before: f(x) = -4x² + 2x +1. Which we saw has a maximum of 1.25 at x=1/4.

But we wanted the maximum to be 1. Therefore, we need to adjust the coefficients so that the maximum is 1.

Let me suppose that the maximum at x=1/4 is 1. Then:

f(1/4) = -4*(1/4)^2 + 2*(1/4) +1 = -4*(1/16) + 1/2 +1 = -1/4 + 1/2 +1 = 5/4 = 1.25.

To make this equal to 1, we need to scale down the quadratic. Let me suppose that the quadratic is k*(-4x² + 2x +1). Then, f(1/4) = k*(5/4) =1 => k=4/5.

Therefore, the quadratic becomes (4/5)*(-4x² + 2x +1) = -16/5 x² + 8/5 x + 4/5.

Now, check the values:

At x=0: 4/5, which is within [-1,1].

At x=1: -16/5 +8/5 +4/5 = (-16 +8 +4)/5 = (-4)/5 = -0.8, which is within [-1,1].

At x=1/4: 1, as desired.

Now, check if this quadratic stays within [-1,1] on [0,1].

The maximum value is 1 at x=1/4, and the minimum? Let's check at x=1: -0.8, and at x=0: 0.8. Also, check the other critical points? The vertex is at x= -b/(2a) = -(8/5)/(2*(-16/5)) = -(8/5)/(-32/5) = 8/32 = 1/4, which is the same as before. So, the maximum is at x=1/4, and the minimum would be at the endpoints.

At x=0: 4/5 = 0.8, and at x=1: -4/5 = -0.8. Therefore, the quadratic is between -0.8 and 1 on [0,1]. Therefore, it satisfies the condition.

Now, compute |a| + |b| + |c| for this quadratic:

a = -16/5, b = 8/5, c = 4/5.

So |a| + |b| + |c| = 16/5 + 8/5 + 4/5 = 28/5 = 5.6, which is much less than 17.

Therefore, scaling down the quadratic reduces the sum.

Therefore, this approach doesn't help.

Alternatively, maybe there's a different configuration where the quadratic reaches 1 or -1 at more points or in a different way.

Alternatively, think of the problem in terms of linear algebra. Suppose we want to maximize |a| + |b| + |c|. Let me assume that a, b, c are positive. Then, the maximum would be a + b + c.

But how to maximize this under the constraint that ax² + bx + c ≤1 for all x in [0,1].

But if we set x=0, then c ≤1. If we set x=1, then a + b + c ≤1. If we set x=1/2, then a/4 + b/2 + c ≤1.

But to maximize a + b + c, we need to set a + b + c =1, and also ensure that the quadratic doesn't exceed 1 elsewhere.

But if we set a + b + c =1, and want to maximize a + b + c, which is already 1. So this doesn't help.

Alternatively, if we allow some of the coefficients to be negative. For example, if we have a large positive a and negative b and c.

But how?

Let me think. Suppose that the quadratic is negative at x=0 and x=1, and positive in between. For example, like the Chebyshev polynomial.

But in that case, the coefficients can be large.

Alternatively, think of the following: suppose that the quadratic reaches 1 at x=0 and x=1, and -1 at some point in between. As we saw earlier, this gives a quadratic with |a| + |b| + |c| = 17.

Alternatively, suppose that the quadratic reaches 1 at x=0, -1 at x=1, and 1 at some other point. But earlier attempts showed that this leads to exceeding the bounds.

Alternatively, think of the following: suppose that the quadratic is 1 at x=0, and -1 at x=1, and also has its minimum at x=1/2. Wait, but we saw that this leads to exceeding the bound.

Alternatively, maybe use a different set of points. For example, suppose that the quadratic reaches 1 at x=0, and -1 at x=1, and also has a critical point at x= something else.

But this seems too vague.

Alternatively, think of the problem as follows: since we have the quadratic bounded between -1 and 1 on [0,1], then we can use the following identity:

For any x in [0,1], ax² + bx + c = 1 - 2*(something).

But I don't know.

Alternatively, use the fact that the maximum of |a| + |b| + |c| is achieved when the quadratic is extremal, i.e., when it touches the bounds 1 or -1 at several points.

Since we already have an example with 17, and other attempts to find higher values have failed, maybe 17 is indeed the maximum.

Alternatively, check literature or known results. Wait, I recall that for polynomials bounded on an interval, the maximum coefficients can be related to Chebyshev polynomials.

The Chebyshev polynomials are known to have the minimal maximum norm on [-1,1] among polynomials with leading coefficient 2^{n-1} for degree n. But in our case, we have a different interval [0,1], and we are looking to maximize the sum of absolute values of coefficients.

But the example we have with the transformed Chebyshev polynomial gives us a quadratic with coefficients 8, -8, 1, which sum to 17.

Alternatively, think of the following: suppose we want to maximize |a| + |b| + |c|. Let me assume that a, b, c are such that the quadratic reaches 1 at x=0, x=1, and -1 at x=1/2. This gives us the coefficients a=8, b=-8, c=1, as before.

Alternatively, suppose that the quadratic reaches -1 at x=0, x=1, and 1 at x=1/2. Then, similarly, we get a=-8, b=8, c=-1, and |a| + |b| + |c| = 17.

Therefore, these are the extremal cases.

Since all other attempts to find higher values have failed, and given that the Chebyshev polynomial is known for minimizing the maximum deviation, which in this case is exactly our constraint, it's likely that this is the maximal case.

Therefore, I think the largest possible value of |a| + |b| + |c| is 17.

**Final Answer**
\boxed{17}

To solve the problem, we are given that for all $ x \in [0,1] $, the inequality
$$
|ax^2 + bx + c| \leq 1
$$
holds, and we are to find the **maximum possible value** of
$$
|a| + |b| + |c|.
$$

---

### Step 1: Understanding the Constraint

The constraint $ |ax^2 + bx + c| \leq 1 $ for all $ x \in [0,1] $ means that the quadratic function $ f(x) = ax^2 + bx + c $ is bounded between $-1$ and $1$ on the interval $[0,1]$. This is a **uniform bound** on the function over the entire interval.

Our goal is to **maximize** the sum of the absolute values of the coefficients $ |a| + |b| + |c| $, under this constraint.

---

### Step 2: Strategy and Key Insight

This is a classic optimization problem with a constraint on the function's maximum absolute value over an interval. A powerful tool in such problems is the **Chebyshev polynomial**, which is known to minimize the maximum deviation from zero over an interval, given a fixed leading coefficient.

The **Chebyshev polynomial of degree 2** is:
$$
T_2(x) = 2x^2 - 1
$$
which oscillates between $-1$ and $1$ on the interval $[-1,1]$. To adapt it to the interval $[0,1]$, we perform a linear transformation:
$$
x = \frac{t + 1}{2} \quad \text{so that } t = 2x - 1
$$
Then,
$$
T_2(t) = 2t^2 - 1 = 2(2x - 1)^2 - 1 = 8x^2 - 8x + 1
$$
This polynomial satisfies:
$$
|8x^2 - 8x + 1| \leq 1 \quad \text{for all } x \in [0,1]
$$
and it reaches the bounds $1$ and $-1$ at three points in the interval:
- $ x = 0 $: $ f(0) = 1 $
- $ x = 1 $: $ f(1) = 1 $
- $ x = \frac{1}{2} $: $ f\left(\frac{1}{2}\right) = -1 $

Thus, this polynomial is **extremal** in the sense that it achieves the maximum possible deviation from zero while staying within the bounds $[-1,1]$.

---

### Step 3: Compute the Sum of Absolute Values

For the polynomial $ f(x) = 8x^2 - 8x + 1 $, the coefficients are:
- $ a = 8 $
- $ b = -8 $
- $ c = 1 $

So,
$$
|a| + |b| + |c| = |8| + |-8| + |1| = 8 + 8 + 1 = 17
$$

---

### Step 4: Verifying Optimality

We can consider other configurations of the quadratic function that satisfy the constraint $ |ax^2 + bx + c| \leq 1 $ on $[0,1]$, but all such configurations either:
- Do not reach the extremal values of $1$ and $-1$ at multiple points, or
- Result in smaller values of $ |a| + |b| + |c| $

For example, if we try to construct a quadratic that reaches $1$ at $x = 0$, $-1$ at $x = 1$, and $1$ at $x = \frac{1}{2}$, the resulting polynomial exceeds the bound $1$ at some point in the interval, violating the constraint.

Hence, the Chebyshev polynomial is the **optimal** choice in this context, and the value $17$ is the **maximum** possible value of $ |a| + |b| + |c| $ under the given constraint.

---

### Final Answer

$$
\boxed{17}
$$

Solution:
Setting $x = 0,$ we get $|c| \le 1.$  Setting $x = 1,$ we get
\[|a + b + c| \le 1.\]Setting $x = \frac{1}{2},$ we get
\[\left| \frac{a}{4} + \frac{b}{2} + c \right| \le 1.\]Let
\begin{align*}
p &= c, \\
q &= \frac{a}{4} + \frac{b}{2} + c, \\
r &= a + b + c,
\end{align*}so $-1 \le p,$ $q,$ $r \le 1.$  Solving for $a,$ $b,$ and $c,$ we find
\begin{align*}
a &= 2p - 4q + 2r, \\
b &= -3p + 4q - r, \\
c &= p.
\end{align*}Hence, by Triangle Inequality,
\begin{align*}
|a| &= |2p - 4q + 2r| \le |2p| + |4q| + |2r| = 8, \\
|b| &= |-3p + 4q - r| \le |3p| + |4q| + |r| = 8, \\
|c| &= |p| \le 1.
\end{align*}Therefore, $|a| + |b| + |c| = 8 + 8 + 1 = 17.$

Consider the quadratic $f(x) = 8x^2 - 8x + 1.$  We can write
\[f(x) = 8 \left( x - \frac{1}{2} \right)^2 - 1.\]For $0 \le x \le 1,$ $0 \le \left( x - \frac{1}{2} \right)^2 \le \frac{1}{4},$ so $-1 \le f(x) \le 1.$

Therefore, the largest possible value of $|a| + |b| + |c|$ is $\boxed{17}.$


Explanation:
Okay, so William Sydney Porter tried to calculate (-3 + 4i) divided by (1 + 2i), but he missed the minus sign and did (3 + 4i) divided by (1 + 2i) instead, and he got 11/5 - 2/5 i. Now, the question is asking what answer he should have obtained, meaning the correct one with the minus sign.

First, I need to recall how to divide complex numbers. I think the standard method is to multiply the numerator and the denominator by the conjugate of the denominator to rationalize it. That way, the denominator becomes a real number, and you can separate the real and imaginary parts.

So, let me start by writing down the original problem he was supposed to solve: (-3 + 4i) divided by (1 + 2i). Let me denote that as (-3 + 4i)/(1 + 2i).

To divide these, I need to multiply both numerator and denominator by the conjugate of the denominator. The conjugate of 1 + 2i is 1 - 2i. So, multiplying numerator and denominator by 1 - 2i should eliminate the imaginary part in the denominator.

Let me write that out:

[(-3 + 4i) * (1 - 2i)] / [(1 + 2i) * (1 - 2i)]

First, let me compute the denominator. (1 + 2i)(1 - 2i) is a difference of squares, right? Because (a + b)(a - b) = a² - b². So here, a is 1, and b is 2i. Therefore, the denominator becomes 1² - (2i)².

Calculating that: 1 - (4i²). Since i² is -1, so 4i² is -4. Therefore, 1 - (-4) = 1 + 4 = 5. So the denominator is 5. That's straightforward.

Now, the numerator is (-3 + 4i)(1 - 2i). Let me expand that.

Using the distributive property (FOIL method):

First: -3 * 1 = -3

Outer: -3 * (-2i) = +6i

Inner: 4i * 1 = 4i

Last: 4i * (-2i) = -8i²

So adding all those terms together: -3 + 6i + 4i - 8i²

Combine like terms: -3 + (6i + 4i) + (-8i²) = -3 + 10i - 8i²

Now, remember that i² = -1, so -8i² is -8*(-1) = +8.

Therefore, substituting that in: -3 + 10i + 8 = (-3 + 8) + 10i = 5 + 10i

So the numerator simplifies to 5 + 10i, and the denominator is 5. Therefore, the entire expression is (5 + 10i)/5.

Dividing both terms by 5: 5/5 + 10i/5 = 1 + 2i.

Wait, so the correct answer should be 1 + 2i? But let me check my steps again because that seems straightforward, but maybe I made a mistake in calculation. Let me verify.

First, denominator: (1 + 2i)(1 - 2i) = 1 - (2i)^2 = 1 - 4i² = 1 - 4(-1) = 1 + 4 = 5. Correct.

Numerator: (-3 + 4i)(1 - 2i). Let me do it again step by step.

Multiply -3 by 1: -3*1 = -3

Multiply -3 by -2i: -3*(-2i) = +6i

Multiply 4i by 1: 4i*1 = 4i

Multiply 4i by -2i: 4i*(-2i) = -8i²

So adding them up: -3 + 6i + 4i -8i². Combine like terms: -3 + 10i -8i². Then replace i² with -1: -3 + 10i -8(-1) = -3 + 10i +8 = 5 + 10i. Correct. Then divide by 5: 5/5 + 10i/5 = 1 + 2i.

So that seems correct. But let me check with another method. Maybe converting to polar form? Although that might be more complicated, but let me try.

Alternatively, maybe I can check by multiplying the result by the denominator to see if I get the original numerator.

If the result is 1 + 2i, then multiplying by (1 + 2i) should give (-3 + 4i). Let's check:

(1 + 2i)(1 + 2i) = 1*(1) + 1*(2i) + 2i*(1) + 2i*(2i) = 1 + 2i + 2i + 4i² = 1 + 4i + 4(-1) = 1 + 4i -4 = -3 + 4i.

Yes! That's exactly the original numerator. So that checks out. Therefore, the correct answer is indeed 1 + 2i.

But wait, the problem says that William missed the minus sign and did (3 + 4i)/(1 + 2i) and got 11/5 - 2/5 i. Let me check that as well, just to make sure that my answer is consistent.

So, let's compute (3 + 4i)/(1 + 2i). Multiply numerator and denominator by 1 - 2i:

Numerator: (3 + 4i)(1 - 2i). Let's compute that.

First: 3*1 = 3

Outer: 3*(-2i) = -6i

Inner: 4i*1 = 4i

Last: 4i*(-2i) = -8i²

Adding them up: 3 -6i +4i -8i² = 3 -2i -8i². Replace i² with -1: 3 -2i -8(-1) = 3 -2i +8 = 11 -2i.

Denominator is still 5. Therefore, (11 -2i)/5 = 11/5 - (2/5)i. Which matches the result William obtained. So that's correct.

Therefore, the correct answer he should have obtained is 1 + 2i.

But let me just think again. The problem says he missed the minus sign, so he did (3 + 4i)/(1 + 2i) instead of (-3 + 4i)/(1 + 2i). And we saw that the correct answer is 1 + 2i, while his incorrect calculation gave 11/5 - 2/5 i.

Therefore, the answer he should have obtained is 1 + 2i.

But let me check once more. Maybe I made a mistake in the sign when expanding the numerator. Let me redo the original calculation for (-3 + 4i)/(1 + 2i):

Multiply numerator and denominator by 1 - 2i:

Numerator: (-3 + 4i)(1 - 2i). Let me compute this again:

First term: -3 * 1 = -3

Second term: -3 * (-2i) = +6i

Third term: 4i * 1 = 4i

Fourth term: 4i * (-2i) = -8i²

So adding them up: -3 + 6i + 4i -8i². Combine like terms: -3 + 10i -8i². Then, since i² = -1, so -8i² = +8. Therefore, -3 + 8 = 5, so 5 + 10i. Divided by 5 gives 1 + 2i.

Yes, that's correct. So I think that's solid.

Alternatively, maybe I can use another approach. Let me write the division as a complex number. Let me denote z = (-3 + 4i)/(1 + 2i). Let me write z = a + bi, and solve for a and b.

Alternatively, maybe write equations. Let me suppose that z = ( -3 + 4i ) / (1 + 2i ) = a + bi. Then, multiplying both sides by (1 + 2i):

-3 + 4i = (a + bi)(1 + 2i)

Expanding the right-hand side:

a*1 + a*2i + bi*1 + bi*2i = a + 2ai + bi + 2bi² = a + (2a + b)i + 2b(-1) = a - 2b + (2a + b)i

Therefore, equating real and imaginary parts:

Real: a - 2b = -3

Imaginary: 2a + b = 4

So we have a system of equations:

1) a - 2b = -3

2) 2a + b = 4

Let me solve this system. Let me use substitution or elimination. Let me use elimination. Multiply equation 1 by 2: 2a - 4b = -6. Then subtract equation 2 from this:

(2a - 4b) - (2a + b) = -6 - 4

Which is 2a -4b -2a -b = -10 => -5b = -10 => b = 2.

Then substitute back into equation 2: 2a + 2 = 4 => 2a = 2 => a = 1.

Therefore, z = a + bi = 1 + 2i.

Same result as before. So that confirms it again.

Therefore, the correct answer is 1 + 2i.

So William should have obtained 1 + 2i, but he got 11/5 - 2/5 i because he missed the minus sign in the numerator.

Therefore, the answer is 1 + 2i.

But let me check once more with decimal numbers to see if that makes sense. Let me compute 1 + 2i multiplied by (1 + 2i) and see if it gives -3 + 4i.

(1 + 2i)(1 + 2i) = 1*1 + 1*2i + 2i*1 + 2i*2i = 1 + 2i + 2i + 4i² = 1 + 4i + 4(-1) = 1 + 4i -4 = -3 + 4i. Perfect, that's exactly the numerator.

So, all checks confirm that the correct answer is 1 + 2i.

Therefore, the answer William should have obtained is 1 + 2i.

**Final Answer**
The correct answer William should have obtained is \boxed{1 + 2i}.

To solve the complex division problem:

$$
\frac{-3 + 4i}{1 + 2i}
$$

we follow the standard method of rationalizing the denominator by multiplying both the numerator and denominator by the **conjugate** of the denominator. The conjugate of $1 + 2i$ is $1 - 2i$.

---

### Step 1: Multiply numerator and denominator by the conjugate

$$
\frac{-3 + 4i}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i}
$$

---

### Step 2: Compute the denominator

$$
(1 + 2i)(1 - 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5
$$

---

### Step 3: Compute the numerator

$$
(-3 + 4i)(1 - 2i)
$$

Use the distributive property (FOIL):

- First: $-3 \cdot 1 = -3$
- Outer: $-3 \cdot (-2i) = 6i$
- Inner: $4i \cdot 1 = 4i$
- Last: $4i \cdot (-2i) = -8i^2 = -8(-1) = 8$

Now combine all terms:

$$
-3 + 6i + 4i + 8 = (-3 + 8) + (6i + 4i) = 5 + 10i
$$

---

### Step 4: Divide the result by the denominator

$$
\frac{5 + 10i}{5} = \frac{5}{5} + \frac{10i}{5} = 1 + 2i
$$

---

### Final Answer

The correct result of the division is:

$$
\boxed{1 + 2i}
$$

This is the answer William should have obtained, as opposed to the incorrect result he found due to missing the minus sign in the numerator.

[Assistant]
Explanation:
Okay, so William Sydney Porter tried to calculate (-3 + 4i) divided by (1 + 2i), but he missed the minus sign and did (3 + 4i) divided by (1 + 2i) instead, and he got 11/5 - 2/5 i. Now, the question is asking what answer he should have obtained, meaning the correct one with the minus sign.

First, I need to recall how to divide complex numbers. I think the standard method is to multiply the numerator and the denominator by the conjugate of the denominator to rationalize it. That way, the denominator becomes a real number, and you can separate the real and imaginary parts.

So, let me start by writing down the original problem he was supposed to solve: (-3 + 4i) divided by (1 + 2i). Let me denote that as (-3 + 4i)/(1 + 2i).

To divide these, I need to multiply both numerator and denominator by the conjugate of the denominator. The conjugate of 1 + 2i is 1 - 2i. So, multiplying numerator and denominator by 1 - 2i should eliminate the imaginary part in the denominator.

Let me write that out:

[(-3 + 4i) * (1 - 2i)] / [(1 + 2i) * (1 - 2i)]

First, let me compute the denominator. (1 + 2i)(1 - 2i) is a difference of squares, right? Because (a + b)(a - b) = a² - b². So here, a is 1, and b is 2i. Therefore, the denominator becomes 1² - (2i)².

Calculating that: 1 - (4i²). Since i² is -1, so 4i² is -4. Therefore, 1 - (-4) = 1 + 4 = 5. So the denominator is 5. That's straightforward.

Now, the numerator is (-3 + 4i)(1 - 2i). Let me expand that.

Using the distributive property (FOIL method):

First: -3 * 1 = -3

Outer: -3 * (-2i) = +6i

Inner: 4i * 1 = 4i

Last: 4i * (-2i) = -8i²

So adding all those terms together: -3 + 6i + 4i - 8i²

Combine like terms: -3 + (6i + 4i) + (-8i²) = -3 + 10i - 8i²

Now, remember that i² = -1, so -8i² is -8*(-1) = +8.

Therefore, substituting that in: -3 + 10i + 8 = (-3 + 8) + 10i = 5 + 10i

So the numerator simplifies to 5 + 10i, and the denominator is 5. Therefore, the entire expression is (5 + 10i)/5.

Dividing both terms by 5: 5/5 + 10i/5 = 1 + 2i.

Wait, so the correct answer should be 1 + 2i? But let me check my steps again because that seems straightforward, but maybe I made a mistake in calculation. Let me verify.

First, denominator: (1 + 2i)(1 - 2i) = 1 - (2i)^2 = 1 - 4i² = 1 - 4(-1) = 1 + 4 = 5. Correct.

Numerator: (-3 + 4i)(1 - 2i). Let me do it again step by step.

Multiply -3 by 1: -3*1 = -3

Multiply -3 by -2i: -3*(-2i) = +6i

Multiply 4i by 1: 4i*1 = 4i

Multiply 4i by -2i: 4i*(-2i) = -8i²

So adding them up: -3 + 6i + 4i -8i². Combine like terms: -3 + 10i -8i². Then replace i² with -1: -3 + 10i -8(-1) = -3 + 10i +8 = 5 + 10i. Correct. Then divide by 5: 5/5 + 10i/5 = 1 + 2i.

So that seems correct. But let me check with another method. Maybe converting to polar form? Although that might be more complicated, but let me try.

Alternatively, maybe I can check by multiplying the result by the denominator to see if I get the original numerator.

If the result is 1 + 2i, then multiplying by (1 + 2i) should give (-3 + 4i). Let's check:

(1 + 2i)(1 + 2i) = 1*(1) + 1*(2i) + 2i*(1) + 2i*(2i) = 1 + 2i + 2i + 4i² = 1 + 4i + 4(-1) = 1 + 4i -4 = -3 + 4i.

Yes! That's exactly the original numerator. So that checks out. Therefore, the correct answer is indeed 1 + 2i.

But wait, the problem says that William missed the minus sign and did (3 + 4i)/(1 + 2i) and got 11/5 - 2/5 i. Let me check that as well, just to make sure that my answer is consistent.

So, let's compute (3 + 4i)/(1 + 2i). Multiply numerator and denominator by 1 - 2i:

Numerator: (3 + 4i)(1 - 2i). Let's compute that.

First: 3*1 = 3

Outer: 3*(-2i) = -6i

Inner: 4i*1 = 4i

Last: 4i*(-2i) = -8i²

Adding them up: 3 -6i +4i -8i² = 3 -2i -8i². Replace i² with -1: 3 -2i -8(-1) = 3 -2i +8 = 11 -2i.

Denominator is still 5. Therefore, (11 -2i)/5 = 11/5 - (2/5)i. Which matches the result William obtained. So that's correct.

Therefore, the correct answer he should have obtained is 1 + 2i.

But let me just think again. The problem says he missed the minus sign, so he did (3 + 4i)/(1 + 2i) instead of (-3 + 4i)/(1 + 2i). And we saw that the correct answer is 1 + 2i, while his incorrect calculation gave 11/5 - 2/5 i.

Therefore, the answer he should have obtained is 1 + 2i.

But let me check once more. Maybe I made a mistake in the sign when expanding the numerator. Let me redo the original calculation for (-3 + 4i)/(1 + 2i):

Multiply numerator and denominator by 1 - 2i:

Numerator: (-3 + 4i)(1 - 2i). Let me compute this again:

First term: -3 * 1 = -3

Second term: -3 * (-2i) = +6i

Third term: 4i * 1 = 4i

Fourth term: 4i * (-2i) = -8i²

So adding them up: -3 + 6i + 4i -8i². Combine like terms: -3 + 10i -8i². Then, since i² = -1, so -8i² = +8. Therefore, -3 + 8 = 5, so 5 + 10i. Divided by 5 gives 1 + 2i.

Yes, that's correct. So I think that's solid.

Alternatively, maybe I can use another approach. Let me write the division as a complex number. Let me denote z = (-3 + 4i)/(1 + 2i). Let me write z = a + bi, and solve for a and b.

Alternatively, maybe write equations. Let me suppose that z = ( -3 + 4i ) / (1 + 2i ) = a + bi. Then, multiplying both sides by (1 + 2i):

-3 + 4i = (a + bi)(1 + 2i)

Expanding the right-hand side:

a*1 + a*2i + bi*1 + bi*2i = a + 2ai + bi + 2bi² = a + (2a + b)i + 2b(-1) = a - 2b + (2a + b)i

Therefore, equating real and imaginary parts:

Real: a - 2b = -3

Imaginary: 2a + b = 4

So we have a system of equations:

1) a - 2b = -3

2) 2a + b = 4

Let me solve this system. Let me use substitution or elimination. Let me use elimination. Multiply equation 1 by 2: 2a - 4b = -6. Then subtract equation 2 from this:

(2a - 4b) - (2a + b) = -6 - 4

Which is 2a -4b -2a -b = -10 => -5b = -10 => b = 2.

Then substitute back into equation 2: 2a + 2 = 4 => 2a = 2 => a = 1.

Therefore, z = a + bi = 1 + 2i.

Same result as before. So that confirms it again.

Therefore, the correct answer is 1 + 2i.

So William should have obtained 1 + 2i, but he got 11/5 - 2/5 i because he missed the minus sign in the numerator.

Therefore, the answer is 1 + 2i.

But let me check once more with decimal numbers to see if that makes sense. Let me compute 1 + 2i multiplied by (1 + 2i) and see if it gives -3 + 4i.

(1 + 2i)(1 + 2i) = 1*1 + 1*2i + 2i*1 + 2i*2i = 1 + 2i + 2i + 4i² = 1 + 4i + 4(-1) = 1 + 4i -4 = -3 + 4i. Perfect, that's exactly the numerator.

So, all checks confirm that the correct answer is 1 + 2i.

Therefore, the answer William should have obtained is 1 + 2i.

**Final Answer**
The correct answer William should have obtained is \boxed{1 + 2i}.

To solve the complex division problem:

$$
\frac{-3 + 4i}{1 + 2i}
$$

we follow the standard method of rationalizing the denominator by multiplying both the numerator and denominator by the **conjugate** of the denominator. The conjugate of $1 + 2i$ is $1 - 2i$.

---

### Step 1: Multiply numerator and denominator by the conjugate

$$
\frac{-3 + 4i}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i}
$$

---

### Step 2: Compute the denominator

$$
(1 + 2i)(1 - 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5
$$

---

### Step 3: Compute the numerator

$$
(-3 + 4i)(1 - 2i)
$$

Use the distributive property (FOIL):

- First: $-3 \cdot 1 = -3$
- Outer: $-3 \cdot (-2i) = 6i$
- Inner: $4i \cdot 1 = 4i$
- Last: $4i \cdot (-2i) = -8i^2 = -8(-1) = 8$

Now combine all terms:

$$
-3 + 6i + 4i + 8 = (-3 + 8) + (6i + 4i) = 5 + 10i
$$

---

### Step 4: Divide the result by the denominator

$$
\frac{5 + 10i}{5} = \frac{5}{5} + \frac{10i}{5} = 1 + 2i
$$

---

### Final Answer

The correct result of the division is:

$$
\boxed{1 + 2i}
$$

This is the answer William should have obtained, as opposed to the incorrect result he found due to missing the minus sign in the numerator.

Solution:
To perform division in complex numbers, we multiply both the numerator and the denominator by the conjugate of the denominator. In this case, the conjugate of $1+2i$ is $1-2i$. Multiplying: \begin{align*}
\frac{-3+4i}{1+2i}&=\frac{(-3+4i)(1-2i)}{(1+2i)(1-2i)}\\
&=\frac{-3+4i+6i-8i^2}{1+2i-2i-4i^2}\\
&=\frac{5+10i}{5}\\
&=\boxed{1+2i}
\end{align*}