Proof with tangent line approximation

Question

Problem: 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 $x \in \left[0,\frac12\right]$.

I know that I should use tangent line approximation somewhere in my proof, and it looks like the Mean Value theorem has some use here, but I don't have a clue of how to use these ideas to prove the above statement.

Answer 1

By second order Taylor-Lagrange formula, $$\exists c\in (0,1/2) \;:$$

$$f (1/2)=f (0 )+1/2f'(0)+1/8f''(c) $$ $$\implies 1/2=1/8f''(c) $$ $$\implies f''(c)=4$$

Answer 2

Let $g(x) = f(x) - 2x^2$. Notice that $g'(x) = f'(x)-4x$, so in particular $g'(0) = f'(0) = 0$, and $g''(x) = f''(x) - 4$. It is enough to show that $g''(x)\ge 0$ for some $x\in[0,1/2]$.

Suppose to the contrary that $g''(x)<0$ for all $x\in[0,1/2]$. Since $g'(0) = 0$, this implies that $g'(x)<0$ for all $x\in(0,1/2]$. Since $g(0) = f(0) = 0$, this implies that $g(x)<0$ for all $x\in(0,1/2]$. But $g(1/2) = f(1/2) - 2(1/2)^2 = 0$, contradiction.