If $f$ is continuous on $[-2,2]$ and thrice differentiable on $(-2,2)$ and $f(2)=-f(-2)=4$ and $f'(0)=0$ then there exist $x\in(-2,2)$ such that $f'''(x)\ge3$
I have solved that
We know that $f$ is continuous on $[-2,2]$ and thrice differentiable on $(-2,2)$, so by the Mean Value Theorem, there exists a point $c$ in $(-2,2)$ such that:
$f'(c) = (f(2) – f(-2)) / (2 – (-2))$
= $8 / 4$
= $2$
Since $f'(0) = 0$, we can conclude that there exist two points $a$ and $b$ in $(-2, 2)$ such that $f''(a) = 0$ and $f''(b) = 0$.
Now, we can apply Rolle's theorem to the function $f''(x)$ on the interval $[a, b]$. Since $f''(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, and $f''(a) = f''(b) = 0$, there exists a point $d$ in $(a, b)$ such that $f'''(d) = 0$.
We now need to show that there exists a point $x$ in $(-2, 2)$ such that $f'''(x) ≥ 3$.
Now how to approach please help…
Best Answer
Consider $g(x) = f(x) - \tfrac12x^3-f(0)(1-\tfrac14x^2)$. Then $$g(-2) = g(0) = g(2) = g'(0) = 0.$$
By Rolle's theorem, there are $a_1 \in (-2,0)$ and $b_1 \in (0,2)$ such that $g'(a_1)=g'(b_1) = 0$.
Again by Rolle's theorem, there are $a_2 \in (a_1,0)$ and $b_2 \in (0,b_1)$ such that $g''(a_2) = g''(b_2) = 0$.
One more application of Rolle's theorem shows that there is some $c \in (a_2,b_2) \subset (-2,2)$ such that $g'''(c) = 0$.
Now just note that $f'''(x) = g'''(x) + 3,$ so $f'''(c) = g'''(c) + 3 = 3$.