Here is a very old high school exam question I am trying to solve (purely for interest only):

If $a,b,c$ are real numbers such that $-1 \le ax^2+bx+c \le 1$ for $-1 \le x \le 1$ prove that $-4 \le 2ax+b \le4$ for $-1 \le x \le 1$ (Hint: Consider the functions at the end-points and at the mid-point of the interval).

I can see (graphically) that if $a>0$ then, as $2ax+b$ is the gradient function, the max and min gradients will occur when the parabola passes through the end-points $(-1,1)$ and $(1,1)$ and the mid-point $(0,-1)$. This gives 3 equations with 3 unknowns and is solved to give $a=2, b=0$ and $c=-1$. The required result follows easily from this. Due to symmetry, a<0 gives the same result.

We first note that by setting $x=0$ in our initial inequality we get: $$-1\le c\le1 \\ |c|\le1.$$
If we set $x=\pm1$ in our given inequality we will get: $$\text{(i)}\;\;-1\le a+b+c\le 1 \;\;\;\text{and}\;\;\;\text{(ii)}\;\;-1\le a-b+c\le 1.$$ Adding these two inequalities together will give us: $$-2\le 2a+2c\le2 \\ -1\le a+c\le1 \\ -2\le-c-1\le a\le-c+1\le2 \\ |a|\le2.$$ Thus, considering only $-1\le x\le1$, we see that if $a\gt 0$ then: $$2ax+b \;\le\; 2a+b \;=\; a+b+c + (a-c)\;\le\;1 + (a-c) \;\le\;1+1+2\;=\;4,$$ and $$-4 \;=\; -2-1 -1\;\le\; (-a+c) - 1\;\le\; (-a+c) - a+b-c\;\le\; -2a+b \;\le\;2ax+b.$$
Similarily, if $a\lt0$ then: $$2ax+b \;\ge\; 2a+b \;=\; a+b+c + (a-c)\;\ge\;-1 + (a-c) \;\ge\;-1-1-2\;=\;-4,$$ and similarily: $$4 \;=\; 2+1 +1\;\ge\; (-a+c) + 1\;\ge\; (-a+c) - a+b-c\;\ge\; -2a+b \;\ge\;2ax+b.$$