How to prove $|f(x)|\leq \frac{3}{2}$ for all $x\in [-1, 1]$ 
Let $f(x) = ax^2 + bx + c$ where $a, b, c$ are real numbers. Suppose $f(-1),f(0), f(1) \in[-1, 1]$. Prove that $|f(x)|\leq \frac{3}{2}$ for all $x\in [-1, 1]$.

Here I need to show $|f(x)|\leq \frac{3}{2}$. It means $f(x)$ lies between $-3/2$ and $3/2$. But how can I show that. If I equate $f'(x)$ to $0$ I get $x=-b/2a$.Then stuck. Please help.

Let $I = [-1, 1]$. We know that:
$f(0) = c \in I$, $f(1) = a + b + c \in I$ and $f(-1) = a - b + c \in I$.
If we subtract the third from the second, we have that:
$f(1) - f(-1) = 2b \Rightarrow b = \frac{f(1)-f(-1)}{2} \in I$
Now, assume that $a>0$. Let $x_m$ and $x_M$ be the points where we have, respectively, the minimum and maximum of the parabola in $I$. 
The maximum point candidates are $-1$ and $1$ (borders of $I$). In both cases we will have that $f(x_M) \leq 1 < \frac{3}{2}$.
The candidates to be the minimum are $-1, 1$ (borders of $I$) and $-\frac{b}{2a}$ if this is in $I$. If the minimum is attained at $-1$ or at $1$, then we are ok since in both cases we will have that $-\frac{3}{2} < -1 \leq f(x_m)$.
If $x_m = -\frac{b}{2a} \in I$, then the minimum is $f(x_m) = c - \frac{b^2}{4a} = c + x_m \frac{b}{2}$.
We know that $b, x_m, c \in I$ and these fact yeld to $f(x_m) = c + x_m \frac{b}{2}\geq -1 +  x_m \frac{b}{2} \geq -1 + (-1)\frac{1}{2} = -\frac{3}{2}$.
This means that, in all the possible cases ($x_m = -1$ or $x_m= 1$ or $x_m = -\frac{b}{2a} \in I$), we have $f(x_m) \geq -\frac{3}{2}$. 
So we have that $|f(x)| \leq \frac{3}{2} ~ \forall x \in I \wedge a > 0$
Similar arguments holds assuming that $a < 0$.
Finally, if $a = 0$, then we have a line and minimum and maximum in $I$ are attained on the border points $-1$ and $+1$, so again we are bounded.
