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

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

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

$f^{\text{'}\text{'}}\left(x\right)=2a$

Given the maximum value of $f^{\text{'}\text{'}}\left(x\right)=2a=\frac{1}{2}$  $\Rightarrow a=\frac{1}{4},$

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

$\therefore f\left(x\right)=\frac{x^2}{4}+\frac{3x}{2}+\frac{13}{4}$

We know that, the vertex of the quadratic $A{x}^2+Bx+C$ is at $\left(-\frac{B}{2A}, -\frac{B^2-4AC}{4A}\right)$

Thus, the vertex of $\frac{x^2}{4}+\frac{3x}{2}+\frac{13}{4}$ 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 $f\left(-1\right)=2 \& f\left(1\right)=5$

$\Rightarrow 2\le f\left(x\right)\le 5$

$\therefore$ Least value of $\alpha$ is$5.$