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Step-by-Step Solution
Step 1: Understand the Problem
We are given a function defined on the interval [-1, 1] :
f(x) = a x^2 + b x + c,
with certain conditions:
f(-1) = 2
f'(-1) = 1
\(\max f''(x) = \tfrac{1}{2}\) on \([-1,1]\)
We need to find the least upper bound \(\alpha\) such that
\[
f(x) \le \alpha \quad \text{for all } x \in [-1,1],
\]
and determine the minimum possible value of \(\alpha\).
Step 2: Use the Second Derivative Condition
The second derivative of \(f\) is
f''(x) = 2a.
Since the maximum value of \(f''(x)\) on \([-1, 1]\) is given to be \(\tfrac{1}{2}\), we have
2a = \frac{1}{2} \quad \Longrightarrow \quad a = \frac{1}{4}.
Step 3: Use the First Derivative Condition at \(x=-1\)
The first derivative of \(f\) is
f'(x) = 2ax + b.
At \(x = -1\), we have \(f'(-1) = 1\). Substituting \(a = \frac{1}{4}\) and \(x = -1\):
f'(-1) = 2\left(\frac{1}{4}\right)(-1) + b = -\frac{1}{2} + b = 1.
Solving for \(b\):
b = 1 + \frac{1}{2} = \frac{3}{2}.
Step 4: Use the Function Value Condition at \(x=-1\)
We also know \(f(-1) = 2.\) Substituting \(x = -1, a = \frac{1}{4}, b = \frac{3}{2}\) into \(f(x) = ax^2 + bx + c\):
f(-1) = \frac{1}{4} \cdot 1 - \frac{3}{2} + c = 2.
Hence,
\frac{1}{4} - \frac{3}{2} + c = 2.
Combine like terms:
\frac{1}{4} - \frac{3}{2} = \frac{1}{4} - \frac{6}{4} = -\frac{5}{4}.
So,
-\frac{5}{4} + c = 2
\quad \Longrightarrow \quad
c = 2 + \frac{5}{4} = \frac{13}{4}.
Step 5: Final Form of the Function
Thus the function becomes:
f(x) = \frac{1}{4} x^2 + \frac{3}{2} x + \frac{13}{4}.
It can be factored out as:
f(x) = \frac{1}{4} \bigl(x^2 + 6x + 13\bigr).
Step 6: Determine Maximum of \(f(x)\) on \([-1,1]\)
To find the maximum value of a quadratic on a closed interval, we check critical points (where the derivative is zero) and the endpoints.
First, set the derivative to zero:
f'(x) = 2a x + b = 2\left(\frac{1}{4}\right)x + \frac{3}{2} = \frac{1}{2}x + \frac{3}{2}.
Solve \(\frac{1}{2} x + \frac{3}{2} = 0\):
x = -3.
Since \(-3\) is not in the interval \([-1, 1]\), there are no critical points inside \([-1, 1]\). Hence, the maximum in \([-1,1]\) occurs at an endpoint.
At \(x= 1\):
f(1) = \frac{1}{4}(1^2 + 6 \cdot 1 + 13) = \frac{1}{4}(1 + 6 + 13) = \frac{20}{4} = 5.
At \(x= -1\):
f(-1) = \frac{1}{4}(1 - 6 + 13) = \frac{8}{4} = 2.
Therefore, the maximum value of \(f(x)\) on \([-1,1]\) is \(5\).
Step 7: Conclude the Least Upper Bound (\(\alpha\))
Since \(f(x)\) never exceeds \(5\) on the interval \([-1,1]\), the smallest \(\alpha\) satisfying \(f(x) \le \alpha\) for all \(x \in [-1, 1]\) is
\alpha_{\min} = 5.
