Prove that $|f(x)| \le \frac{3}{2}$ when $f(x)=ax^2+bx+c$ Suppose $f(x) = ax^2+bx+c$ where $x \in [-1,1]$. If $f(-1),f(0),f(1)\in [-1,1]$ show that $|f(x)| \le \frac{3}{2}$ $\forall x \in [-1,1]$.
This is how I tried: 
$f(0)=c$ 
$f(1)=a+b+c$ 
$f(-1)=a-b+c$ 
Putting $f(0)=c$ we get $f(1)-f(0)=a+b$, $f(-1)-f(0)=a-b$. Solving for $a$ and $b$ respectively we get $$a=\frac{f(1)+f(-1)-2f(0)}{2}$$ and $$b=\frac{f(1)-f(-1)}{2}$$ Then
$$|f(x)|=|(\frac{f(1)}{2})(x^2+x)+(\frac{f(-1)}{2})(x^2-x)+(f(0))(1-x^2)|$$
$$\le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|$$. Now individually the maximum of $|x^2+x|$ happens at $x=1$ , $|x^2-x|$ at $x=-1$ and $|1-x^2|$ at $x=0$.  So All I get $|f(x)| \le 3$. But I need $\frac{3}{2}$.
Thanks for the help!!
 A: Though this has been answered before, to answer the question, I am posting it here. The answer credit goes to Tom Collinge who answered my question.
The proof for 3/2 follows, though intuitively, I think that you can limit it to 5/4: it's a quadratic and if you draw an extreme graph then it should go through (-1, 1), (0, -1), (1, -1) with a minimum at (1/2, -5/4), or some reflection of this. Back to 3/2 ...
If $f$ is monotonic in [-1, 1] then trivially, $|f| \le 1$, so assume not. 
Then $f$ is differentiable and (being quadratic) can only have one extreme point at   $f'(x_0) = 0$, i.e $2ax_0 + b$ = 0, so that $ax_0 = -b/2$. The extreme value of the function at this point is therefore  $f(x_0) = x_0(-b/2) + bx_0 + c = bx_0/2 + c$.
Taking the possible values at -1, 0, and 1 gives
$|f(-1)| = |a - b + c| \le 1$; $|f(0)| = | c| \le 1$; $|f(1)| = |a + b + c| \le 1$
So $|c| \le 1 $ and it follows from these inequalites that $|b| \le 1$
Substituting in the value of $f(x_0)$, 
$|f(x_0)| = |bx_0/2 + c| \le |b|.|x_0|/2 + |c| \le 1.1/2 + 1 = 3/2$ 
Continuation of proof to show that $|f(x)| \le 5/4$
(1) If $|a| \ge 1$
At the extremum, $x_0 = -b/2a$ so by substitution  $f(x_0) = ab^2/4a^2 -b^2/2a +c$
$|f(x_0)| = |c - b^2/4a| \le |c| + |b|^2/4|a|$. We already proved $|c|, |b| \le 1$ and by assumption $|a| \ge 1$ so $|f(x_0)| \le 1 + 1/4 = 5/4$ 
(2) If $|a| \le 1$
Take the Taylor expansion of $f(x)$ about $x_0$: $f(x) = f(x_0) + (x-x_0) f'(x_0) + (x-x_0)^2 f''(z)/2$ for $z$ between $x$ and $x_0$. Note that $f'(x_0) = 0$ at the extremum, and $f''(z) = 2a $ for all $z$, so
$f(x) = f(x_0) + a(x-x_0)^2$, and $|f(x_0)| = |f(x) - a(x - x_0)^2| \le |f(x)| + |a|(x-x_0)^2$. By assumption, $|a| \le 1$, so
$|f(x_0)| \le |f(x) | + (x-x_0)^2$
For $x = -1, 0, or +1$ and $x_0$ in [-1, 1], one of these values makes $|x - x_0| \le 1/2$ (take 0 if $-1/2 \le x_0 \le +1/2 $ ; +1 if $x_0 > 1/2$; -1 if $x_0 \le -1/2$), and in all cases the corresponding function value satisfies $|f(x)| \le 1$, so at this x,
$|f(x_0)| \le 1 + (1/2)^2 = 5/4.$
A: As mentioned above, this has already been answered in 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 $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. with the better bound $5/4$. But it seems that your
proof is almost complete and slightly shorter than (or at least different from) the answers given there.
You already showed that
$$
|f(x)| \le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|
$$
and for $0 \le x \le 1$ the right-hand side is
$$
\frac{x^2+x}{2} -\frac{x^2-x}{2}+1-x^2
 = x + 1 - x^2 = \frac 54 - (x - \frac 12)^2 \le \frac 54 < \frac 32 \quad .
$$
Similarly for $-1 \le x \le 0$.
