Given function

$f(x)=ax^2+bx+c, \, x\in[-1,1]$

with conditions:

• $f(-1)=2$
• $f'(-1)=1$
• for $x\in(-1,1)$, the maximum value of $f''(x)$ is $\frac12$.

Use the second derivative condition

Since $f(x)=ax^2+bx+c$ is a quadratic,

$f'(x)=2ax+b, \qquad f''(x)=2a.$

Now $f''(x)$ is constant, so its maximum value is also $2a$. Given:

$2a=\frac12 \implies a=\frac14.$

Use the condition $f'(-1)=1$

$f'(x)=2ax+b=\frac12 x+b.$

At $x=-1$,

$f'(-1)=2a(-1)+b=-\frac12+b=1.$

Hence,

$b=\frac32.$

Use the condition $f(-1)=2$

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

Substitute $a=\frac14,\ b=\frac32$:

$\frac14-\frac32+c=2.$

$\frac14-\frac64+c=2$

$-\frac54+c=2$

$c=2+\frac54=\frac{13}{4}.$

So,

$f(x)=\frac14x^2+\frac32x+\frac{13}{4}.$

Find the least value of $\alpha$ such that $f(x)\le \alpha$ for all $x\in[-1,1]$

This is the maximum value of $f(x)$ on $[-1,1]$.

Since $a=\frac14>0$, the parabola opens upward, so the maximum on the closed interval occurs at an endpoint.

Evaluate:

$f(-1)=2$

$f(1)=\frac14+\frac32+\frac{13}{4} = \frac14+\frac64+\frac{13}{4} = \frac{20}{4}=5.$

Therefore,

$\max_{x\in[-1,1]} f(x)=5.$

Hence the least value of $\alpha$ is:

$\boxed{5}$

Comparison with stored answer

Derived answer = $5$.

Stored correct answer = $5$.

They agree.