Problem source

The function $\mathrm{f}$ is defined by 
\[
\mathrm{f}(x)=ax^{2}+bx+c.
\]
Show that 
\[
\mathrm{f}'(x)=\mathrm{f}(1)\left(x+\tfrac{1}{2}\right)+\mathrm{f}(-1)\left(x-\tfrac{1}{2}\right)-2\mathrm{f}(0)x.
\]
If $a,b$ and $c$ are real and such that $\left|\mathrm{f}(x)\right|\leqslant1$ for $\left|x\right|\leqslant1$, show that $\left|\mathrm{f}'(x)\right|\leqslant4$ for $\left|x\right|\leqslant1$. 
Find particular values of $a,b$ and $c$ such that, for the corresponding function $\mathrm{f}$ of the above form $\left|\mathrm{f}(x)\right|\leqslant1$ for all $x$ with $\left|x\right|\leqslant1$ and $\mathrm{f}'(x)=4$ for some $x$ satisfying $\left|x\right|\leqslant1$.

Solution source

Let $f(x) = ax^2 + bx + c$

then 
\begin{align*}
f'(x) &= 2ax + b \\
f(0) &= c \\
f(1) &= a+b+c \\
f(-1) &= a-b+c \\
f(1)+f(-1) &= 2(a+c) \\
f(1)-f(-1) &= 2b \\
f'(x) &= x(f(1)+f(-1)) + \frac12 (f(1) - f(-1)) - 2f(0)x
\end{align*}

as required.

Since $f'(x)$ is a straight line, the maximum value is either at $1, -1$ or it's constant and either end suffices.

\begin{align*}
|f'(1)| & \leq |f(1)|\frac{3}{2} + |f(-1)| \frac12 + 2 |f(0)| \\ &\leq \frac{3}{2} + \frac12 + 2 \\
&= 4 \\
\\
|f'(-1)| & \leq |f(1)|\frac{1}{2} + |f(-1)| \frac32 + 2 |f(0)| \\ &\leq \frac{3}{2} + \frac12 + 2 \\
&= 4 \\
\end{align*}

Therefore $|f'(x)| \leq 4$.

Suppose $|f'(x)| = 4$ for some value in $x \in [-1,1]$, then it must be either $-1$ or $1$.

If $f'(1) = 4$ then $f(1) = 1, f(-1) = 1, f(0) = -1$ so $f(x) = 1+ k(x^2-1) \Rightarrow f(x) = 1+2(x^2-1) = 2x^2 -1$.

If $f'(-1) = 4$ then $f(1) = -1, f(-1) = -1, f(0) = 1 \Rightarrow f(x) = -2x^2 + 1$