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.

Can someone please help turn my "partial" solution into a more convincing/algebraic solution.


We can show the required inequlity in an entirely algebraic manner:

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: $$4 \;=\; 2+1 +1\;\ge\; (-a+c) + 1\;\ge\; (-a+c) - a+b-c\;\ge\; -2a+b \;\ge\;2ax+b.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.