Prove there's $M>0$ such that: $f(x)\le Mx^2$

Question

Let $f:[-1,1]\to\mathbb{R}$, three-times differentiable function and $f(0)=0$, $f(x)\ge 0$ for all $x\in[-1,1]$. Prove there's $M>0$ such that $f(x)\le Mx^2$. Hint: use Taylor formula.

So following the hint and the fact that $f$ differentiable three times:
$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + R_2(x)$$

$f(0) = 0$. Simplifying:
$$f(x) = f'(0)x + \frac{f''(0)}{2}x^2 + R_2(x)$$

We also know that $f(x)\ge 0$ for all $x\in [-1,1]$. Therefore:

$$f(x) = f'(0)x + \frac{f''(0)}{2}x^2 + R_2(x) \ge 0$$

What should I do next?

Answer 1

Hint: $[-1,1]$ is closed and bounded (compact), and $f',f''$ is bounded on $[-1,1]$.