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.
 A: The condition that $|f(1)| \leq 1$ gives $-1 \leq a + b + c \leq 1$, and the condition that
$|f(-1)| \leq 1$ gives $-1 \leq a - b + c \leq 1$. Subtracting these two equations
gives $|b| \leq 1$. The condition that $|f(0)| \leq 1$ gives $|c| \leq 1$.
If $f(x)$ were monotone on $[-1,1]$, then $|f(x)|$ would be maximized at $-1$ or $1$ where $|f(x)| \leq 1$ and we would have nothing to prove. So we can assume $f(x)$ is not monotone on $[-1,1]$, which implies there is an $x_0 \in (0,1)$ where $f'(x_0) = 0$. In other words, $2ax_0 + b = 0$ or $x_0 = -{b \over 2a}$, where $f(x_0) = c - {b^2 \over 4a}$. (If $a = 0$ then the function is monotone, which is covered above.) Note that since $|x_0| < 1$ we have $|{b \over 2a}| < 1$. 
$|f(x)|$ is maximized either at an endpoint of the interval or at point where $f'(x) = 0$, which must be $x_0$. Since $|f(x)| \leq 1$ at the endpoints, it remains to show that $|f(x_0)| = |c - {b^2 \over 4a}| \leq {3 \over 2}$. But $|c - {b^2 \over 4a}| \leq |c| + |{b^2 \over 4a}| \leq |c| + {1 \over 2}|b||{b \over 2a}|$, and by the above this is less than $1 + {1 \over 2}*1*1 = {3 \over 2}$. 
A: 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.
