help complete this attempt (real analysis) Given 2 quadratic functions $f$ and $g$, such that $f(x)=ax²+bx+c$ and $g(x)=cx²+bx+a$ and
 the absolute value of $f(x)$ is less than $1$, for each $x\in[-1,1]$, prove that the absolute value of $g(x)$ is less than $2$, for each $x\in[-1,1]$.
 A: Here is a complete solution; but note that a lot of thinking went into reducing the number of cases to consider.
Multiplying through by $-1$, if necessary, we may assume $a\geq c$, and replacing $x$ by $-x$, if necessary, we may assume $b\geq0$. 
We note first that $c=f(0)\geq -1$ and $a+b+c=f(1)\leq 1$, therefore 
$$a\leq 1 -c-b\leq 2-b\ .\qquad(1)$$
Now $g(x)-f(x)=(a-c)(1-x^2)\geq0$; whence we already now that $g(x)\geq f(x)\geq-1$. If $\arg\max_x g(x)\in\{\pm 1\}$ then for all $x$ we have $g(x)\leq\max g(\pm1)=\max f(\pm1)\leq 1$.
So it remains the case that $\arg\max_x g(x)=:\tau \in\ ]{-1},1[\ $. In this case we have a local maximum at $\tau$, whence necessarily $g'(\tau)=0$ and $c<0$. It follows that $${b\over 2|c|}=\tau<1\ .\qquad (2)$$ Using (1) and (2) we now obtain
$$\max_x\ g(x)=g(\tau)= a-{b^2\over 4c}=a+{b^2\over 4|c|} \leq 2 -b +{b\over2}\leq 2\ ,$$
as stated.
The pair $f(x):=2x^2-1$, $\ g(x):=2-x^2$ shows that the proven inequality is sharp.
A: Your two functions are parabolas. The only way the extreme values are inside of $[-1,1]$ is if the vertex $x_0$ of the parabola satisfies $|x_0|<1$ . Then the maximum is reached at the vertex of the parabola if $a<0$ —parabola opens downwards —and the minimum is reached inside when $a>0$. Otherwise, if the vertex of the  parabola is not in $[-1,1]$, i.e., if the x-coordinate satisfies $|x|>1$ ,the maximum values will happen at the boundaries, i.e., at either $x=1$ or at $x=-1$. The case of the parabola opening downwards (when $a<0$)  comes down to finding the "standard form" $y-y_0=c(x- x_0)^2$, and determining the values of $x_0$  and $y_0 $. Check what happens to $f'(x)$ left- and right- of the vertex of the parabola.
Otherwise, the extreme values happen at the endpoints $\{-1,1\}$ , so that the extreme values will be $a+b+c$  (at $x=1$), and/or $a-b+c$ (at $x=-1$) unless $a<0$ and, when you express the parabola in standard form, $|x_0|<1$. 
So if $a<0$ and $|x_0|<1$ , then $y_0$ is the maximum value; otherwise, the max will be reached at either of the endpoints, and will be either $a+b+c$ , or $a-b+c$
