Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$... Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$. Prove that $|f''(x)| \ge 4$ for some $x \in \left[0,\frac12\right]$.

Could someone help me with this? I'm not sure where to start.
Let $g(x)=f(x)-2x^2$. Then $g'(x)=f'(x)-4x$ and $g''(x)=f''(x) - 4$.

We have $g(0)=g(\frac 1 2)=0$, therefore there is $x_1 \in (0, \frac 1 2)$ such that $g'(x_1)=0$ (why?). Also $g'(0)=0$. Can you take it from here?
Hint: by the fundamental theorem of calculus, $f(x)=\int_0^x f'(t)\,dt$ and $f'(x)=\int_0^x f''(t)\,dt$. If $f''(x)&lt;4$ for all $x\in[0,\frac{1}{2}]$, what can you say about $f(\frac{1}{2})$?