Key Concepts and Formulas

• Derivatives of Polynomials: For a function $f(x) = ax^2 + bx + c$, its first derivative is $f'(x) = 2ax + b$, and its second derivative is $f''(x) = 2a$.
• Extrema on a Closed Interval: For a continuous function $f(x)$ on a closed interval $[m, n]$, its absolute maximum and minimum values occur either at a critical point (where $f'(x) = 0$ or $f'(x)$ is undefined) within the interval, or at the endpoints of the interval.
• Completing the Square: A technique to rewrite a quadratic expression in the form $a(x-h)^2 + k$, which helps in finding the vertex of the parabola.

Step-by-Step Solution

Step 1: Determine the general derivatives of the given function. We are given the quadratic function $f(x) = ax^2 + bx + c$. We need to find its first and second derivatives to use the given conditions. $f'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b$ $f''(x) = \frac{d}{dx}(2ax + b) = 2a$ Explanation: We calculate the derivatives because the problem provides conditions involving $f'(-1)$ and the maximum value of $f''(x)$. These derivatives will allow us to use those conditions to find the unknown coefficients $a, b, c$.

Step 2: Use the information about $f''(x)$ to find the value of 'a'. The problem states: "for $x \in (-1, 1)$, the maximum value of $f''(x)$ is $\frac{1}{2}$." From Step 1, we know that $f''(x) = 2a$. Since $2a$ is a constant, its value does not change with $x$. Therefore, the maximum value of $f''(x)$ on any interval is simply $2a$. So, we can set up the equation: $2a = \frac{1}{2}$ Solving for $a$: $a = \frac{1}{4}$ Explanation: This condition directly gives us the value of 'a'. It's crucial to understand that if $f''(x)$ were a function of $x$ (e.g., $f''(x) = 2ax+d$), we would need to apply calculus techniques to find its maximum value on the interval $(-1, 1)$. However, since $f''(x)$ is a constant, its maximum value is just that constant itself.

Step 3: Use the information about $f'(-1)$ to find the value of 'b'. We are given $f'(-1) = 1$. From Step 1, $f'(x) = 2ax + b$. Substitute $x = -1$ and the value of $a = \frac{1}{4}$ (found in Step 2) into the expression for $f'(x)$: $f'(-1) = 2\left(\frac{1}{4}\right)(-1) + b = 1$ $-\frac{1}{2} + b = 1$ Solving for $b$: $b = 1 + \frac{1}{2} = \frac{3}{2}$ Explanation: With 'a' determined, we can now use the condition on the first derivative at a specific point ($x=-1$) to solve for 'b'. This is a direct substitution.

Step 4: Use the information about $f(-1)$ to find the value of 'c'. We are given $f(-1) = 2$. From the problem statement, $f(x) = ax^2 + bx + c$. Substitute $x = -1$ and the values of $a = \frac{1}{4}$ and $b = \frac{3}{2}$ (found in Steps 2 and 3) into the expression for $f(x)$: $f(-1) = \left(\frac{1}{4}\right)(-1)^2 + \left(\frac{3}{2}\right)(-1) + c = 2$ $\frac{1}{4} - \frac{3}{2} + c = 2$ To combine the fractions, find a common denominator: $\frac{1}{4} - \frac{6}{4} + c = 2$ $-\frac{5}{4} + c = 2$ Solving for $c$: $c = 2 + \frac{5}{4} = \frac{8}{4} + \frac{5}{4} = \frac{13}{4}$ Explanation: Having found 'a' and 'b', the final condition on $f(-1)$ allows us to uniquely determine 'c', thus completely defining the function $f(x)$.

Step 5: Reconstruct the function $f(x)$. Now that we have the values of $a, b, c$: $a = \frac{1}{4}$, $b = \frac{3}{2}$, $c = \frac{13}{4}$. Substitute these into the original function definition: $f(x) = \frac{1}{4}x^2 + \frac{3}{2}x + \frac{13}{4}$ We can factor out $\frac{1}{4}$ for a more compact form: $f(x) = \frac{1}{4}(x^2 + 6x + 13)$ Explanation: This step provides the explicit formula for $f(x)$, which is essential for determining its maximum value on the given interval.

Step 6: Find the maximum value of $f(x)$ on the interval $[-1, 1]$. To find the absolute maximum value of a continuous function on a closed interval, we must evaluate the function at:

1. Any critical points that lie within the interval.
2. The endpoints of the interval.

First, let's find the critical points by setting $f'(x) = 0$: We know $f'(x) = 2ax + b$. Substituting $a = \frac{1}{4}$ and $b = \frac{3}{2}$: $f'(x) = 2\left(\frac{1}{4}\right)x + \frac{3}{2} = 0$ $\frac{1}{2}x + \frac{3}{2} = 0$ $\frac{1}{2}x = -\frac{3}{2}$ $x = -3$ Since $x = -3$ is not within the interval $[-1, 1]$, it is not a candidate for the location of the maximum value.

Now, we evaluate $f(x)$ at the endpoints of the interval: $f(-1) = \frac{1}{4}((-1)^2 + 6(-1) + 13) = \frac{1}{4}(1 - 6 + 13) = \frac{1}{4}(8) = 2$ $f(1) = \frac{1}{4}((1)^2 + 6(1) + 13) = \frac{1}{4}(1 + 6 + 13) = \frac{1}{4}(20) = 5$

Since $f(1) = 5$ is greater than $f(-1) = 2$, the maximum value of $f(x)$ on the interval $[-1, 1]$ is 5.

Step 7: Find the least value of $\alpha$ such that $f(x) \le \alpha$ for $x \in [-1, 1]$. Since the maximum value of $f(x)$ on the interval $[-1, 1]$ is 5, we need to find the least value of $\alpha$ such that $f(x) \le \alpha$ for all $x$ in $[-1, 1]$. This means $\alpha$ must be greater than or equal to the maximum value of $f(x)$ on the interval. Therefore, the least value of $\alpha$ is 5.

Explanation: The value $\alpha$ represents an upper bound for the function $f(x)$ on the interval $[-1, 1]$. The least possible value for this upper bound is simply the maximum value of the function on that interval.

Common Mistakes & Tips

• Forgetting to check endpoints: When finding the maximum or minimum value of a function on a closed interval, always check the endpoints as well as critical points.
• Incorrectly calculating derivatives: Double-check your derivative calculations, especially when dealing with polynomial functions.
• Not verifying critical points: Ensure that any critical points you find are actually within the given interval before considering them as potential locations of extrema.

Summary We first found the derivatives of the given quadratic function. Then, using the given conditions on the derivatives and the function value at a point, we determined the coefficients $a, b,$ and $c$. This allowed us to define the function $f(x)$ explicitly. Finally, we found the maximum value of $f(x)$ on the interval $[-1, 1]$ by evaluating the function at its critical points and the endpoints of the interval, concluding that the maximum value is 5. Therefore, the least value of $\alpha$ such that $f(x) \le \alpha$ for all $x$ in $[-1, 1]$ is 5.

Final Answer The final answer is \boxed{5}.