Suppose $f$ is a real valued continuous function defined on $[-2,2]$ and is the times differentiable in $(-2,2)$. If $f(2)=-f(-2)=4$ and $f'(0)=0$ then show there exists $x\in(-2,2)$ such that $f'''(x)\ge3$.
I have tried using MVT to no avail. Tried to back calculate assuming $f'''(x)\ge3$ and then integrating. Couldn't do it.
On
Suppose, for contradiction, that $$ f'''(x)<3 $$ holds for all $x\in\left(-2,2\right)$.
By Taylor's theorem with mean-value forms of the remainder, $$ f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{6}f'''(\xi)x^3, $$ where $\xi\in\left(0,x\right)$ if $x>0$, and $\xi\in\left(x,0\right)$ if $x<0$.
Provided that $f'(0)=0$, the above equality reduces to $$ f(x)=f(0)+\frac{1}{2}f''(0)x^2+\frac{1}{6}f'''(\xi)x^3. $$ Provided that $f(2)=4$, the above equality yields $$ 4=f(2)=f(0)+\frac{1}{2}f''(0)2^2+\frac{1}{6}f'''(\xi_1)2^3=f(0)+2f''(0)+\frac{4}{3}f'''(\xi_1), $$ with $\xi_1\in\left(0,2\right)$. Note that $f'''(\xi_1)<3$ as per our assumption, this last equality implies $$ 4-f(0)-2f''(0)=\frac{4}{3}f'''(\xi)<\frac{4}{3}\cdot 3=4\iff f(0)+2f''(0)>0. $$
Similarly, thanks to $f(-2)=-4$, we have $$ -4=f(-2)=f(0)+\frac{1}{2}f''(0)(-2)^2+\frac{1}{6}f'''(\xi_2)(-2)^3=f(0)+2f''(0)-\frac{4}{3}f'''(\xi_2), $$ with $\xi_2\in\left(-2,0\right)$. Still use $f'''(\xi_2)<3$, and we obtain $$ 4+f(0)+2f''(0)=\frac{4}{3}f'''(\xi_2)<\frac{4}{3}\cdot 3=4\iff f(0)+2f''(0)<0. $$
Compare the two estimates we have for $f(0)+2f''(0)$, and it is obvious that a contradiction arises. This means that our assumption is wrong, and there must exists some $x_0\in\left(-2,2\right)$, such that $$ f'''(x_0)\ge 3. $$
Use Taylor's Theorem,
$$ f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{6}f'''(\theta)x^3,\quad x\in[-2, 2], \theta\in(-2, 2). $$
Then it follows that
$$ f(2)=-f(-2)=4 $$
which implies $$ f(0)+\frac{1}{2}f''(0)\cdot4+\frac{1}{6}f'''(\theta_1)\cdot 8=4 $$ $$ f(0)+\frac{1}{2}f''(0)\cdot4+\frac{1}{6}f'''(\theta_2)\cdot (-8)=-4 $$
Suppose $ f'''(x)<3,\forall x\in(-2, 2) $, then from the first equation we get $$ f(0)+\frac{1}{2}f''(0)\cdot4>4-\frac{4}{3}\cdot 3=0, $$
but $$ f(0)+\frac{1}{2}f''(0)\cdot4<-4+\frac{4}{3}\cdot 3=0 .$$ from the second equation. Contradicton!