For all twice differentiable functions f: mat | vedclass

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(A) Given $f(0)=f(1)=f^{\prime}(0)=0$.Apply Rolle's theorem on $f(x)$ in the interval $[0, 1]$.Since $f(0)=f(1)=0$ and $f$ is differentiable,there exi

Vedclass · Accepted Answer

$f^{\prime \prime}(x)=0,$ for some $x \in(0,1)$

Vedclass · Answer

(A) Given $f(0)=f(1)=f^{\prime}(0)=0$.Apply Rolle's theorem on $f(x)$ in the interval $[0, 1]$.Since $f(0)=f(1)=0$ and $f$ is differentiable,there exists some $\alpha \in (0, 1)$ such that $f^{\prime}(\alpha)=0$.Now,consider the function $f^{\prime}(x)$ on the interval $[0, \alpha]$.We have $f^{\prime}(0)=0$ and $f^{\prime}(\alpha)=0$.Since $f$ is twice differentiable,$f^{\prime}$ is continuous on $[0, \alpha]$ and differentiable on $(0, \alpha)$.By Rolle's theorem applied to $f^{\prime}(x)$,there exists some $\beta \in (0, \alpha)$ such that $f^{\prime \prime}(\beta)=0$.Since $\beta \in (0, \alpha) \subset (0, 1)$,it follows that $f^{\prime \prime}(x)=0$ for some $x \in (0, 1)$.

For all twice differentiable functions $f: \mathbb{R} \rightarrow \mathbb{R},$ with $f(0)=f(1)=f^{\prime}(0)=0,$ which of the following is true?

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