Let f: [ - 1, 1 ] → R be defined as f ( x ) = a x ^ 2 + bx + c for...

Given $f: [ - 1, 1 ] \to R,$ $f ( x ) = a{x}^{2} + bx + c, {f}^{'} ( x ) = 2ax + b$ and {f}^{' '} ( x ) = 2a

$⇒ f ( - 1 ) = a - b + c = 2 . . . \left(1\right),$

$⇒ {f}^{'} ( - 1 ) = - 2a + b = 1 . . . \left(2\right)$ and

{f}^{' '} ( x ) = 2a

Given the maximum value of {f}^{' '} ( x ) = 2a = \dfrac{1}{2} $⇒ a = \dfrac{1}{4},$

From the equations \left(1\right)$and \left(2\right)$ we get $b = \dfrac{3}{2}$ and $c = \dfrac{13}{4} .$

$∴ f ( x ) = \dfrac{{x}^{2}}{4} + \dfrac{3x}{2} + \dfrac{13}{4}$

We know that, the vertex of the quadratic $A{x}^{2} + Bx + C$ is at \left(- \dfrac{B}{2A}, - \dfrac{{B}^{2} - 4AC}{4A}\right)$</p><p>Thus, the vertex of$\dfrac{{x}^{2}}{4} + \dfrac{3x}{2} + \dfrac{13}{4}$is at \left(- 3, 1\right),$ hence both the numbers $\pm 1$ are on the same side of the vertex and the graph of $f\left(x\right)$ is given below.

For, $x \in [ - 1, 1 ] ,$ we have $f\left(- 1\right) = 2 \text{ and } f\left(1\right) = 5$

$⇒ 2 \le f ( x ) \le 5$

$∴$ Least value of $\alpha$ is$5 .$