Let f: R rightarrow R be a twice continuously | vedclass

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(B) Given $f(0)=f(1)=f^{\prime}(0)=0$. By Rolle's Theorem on the interval $[0, 1]$, since $f(0)=f(1)=0$, there exists at least one point $d \in (0, 1)

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$f^{\prime \prime}(c)=0$ for some $c \in R$

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(B) Given $f(0)=f(1)=f^{\prime}(0)=0$. By Rolle's Theorem on the interval $[0, 1]$, since $f(0)=f(1)=0$, there exists at least one point $d \in (0, 1)$ such that $f^{\prime}(d)=0$. Now, we have $f^{\prime}(0)=0$ and $f^{\prime}(d)=0$ where $d \in (0, 1)$. Applying Rolle's Theorem to the function $f^{\prime}(x)$ on the interval $[0, d]$, since $f^{\prime}(0)=f^{\prime}(d)=0$, there must exist at least one point $c \in (0, d)$ such that $f^{\prime \prime}(c)=0$. Thus, $f^{\prime \prime}(c)=0$ for some $c \in R$.

Let $f: R \rightarrow R$ be a twice continuously differentiable function. Let $f(0)=f(1)=f^{\prime}(0)=0$. Then,

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