Inequality related to Taylor polynomial Consider a function defined in $\mathbb{R}$ such that $f^{(3)}$ is contiuous in $[0,1].$ Suppose that  $f(0) = f^{'}(0) = f^{''}(0)=f^{'}(1) =f^{''}(1)=0$ and $f(1)=1.$ Prove that there exists $c \in [0,1]$ such that $f^ {(3)}(c) \geq 24.$
I tried to apply the Taylor polynomial at zero and $1$ . But I am getting anywhere. Someone could help me?
Thanks in advance
 A: Unfortunately wrong. $\exists f\in C^{3}[0,1]\ \forall c \in [0,1]\colon f(c) < 24.$
Here's such an $f$ as a counterexample:
Consider the polynomial $f(x) = 6x^5 -15x^4 + 10x^3 \in C^\infty(\mathbb R)$
$$
\begin{aligned}
f(x) &= (6x^2 -15x + 10)x^3 &f(0)&=0 &f(1)=1\\
f'(x) &= 30(x-1)^2x^2 &f'(0)&=0 &f'(1)=0\\
f''(x) &= 120(x^2-1,5+1)x &f''(0)&=0 &f''(1)=0
\end{aligned}
$$
If we assume $\exists c\in[0,1]\colon f(c)\ge24$, then there must $\exists\xi\in(0,1)$ such that $f(\xi)$ is an extremum because $f(0) < f(1) < 24$ and $f$ is continuous. Since $f$ is continuous and differentiable: $f'(\xi) = 0$ must hold. But we can simply check $$f'(x)=0 \implies x\in\{0,1\} \implies x\notin(0,1) \implies \nexists\xi\in(0,1)\colon\ f'(\xi) = 0 \implies \nexists c\in[0,1]\colon f(c)\ge24.$$
A: I'm afraid you're wrong. It is not difficult to calculate a polynomial that satisfies your conditions. Take
$$f(x)=6x^5-15x^4+10x^3$$ so you have $$f'(x)=30x^4-60x^3+30x^2\\f''(x)=120x^3-180x^2+60x$$
You can verify that $f(0) = f^{'}(0) = f^{''}(0)=f^{'}(1) =f^{''}(1)=0$ and that $f(1)=1$ All your condition are satisfied for all $c$ in $[0,1]$ you have $f(c)\lt24$ (the maximum is almost equal to $1$).
