Let f: R rightarrow R be a differentiable fun | vedclass

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(B) Given that $f: R \rightarrow R$ is a differentiable function on $(a, b)$ such that $f(a) = f(b) = 0$ with $a < b$. By Rolle's Theorem,there exists

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$2$

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(B) Given that $f: R \rightarrow R$ is a differentiable function on $(a, b)$ such that $f(a) = f(b) = 0$ with $a By Rolle's Theorem,there exists at least one point $c \in (a, b)$ such that $f^{\prime}(c) = 0$. We are given $f^{\prime}(a) f^{\prime}(b) > 0$,which implies that $f^{\prime}(a)$ and $f^{\prime}(b)$ have the same sign. If $f^{\prime}(a) > 0$ and $f^{\prime}(b) > 0$,since $f(a) = f(b) = 0$,the function must increase from $a$ and eventually decrease to reach $b$. This implies there must be at least one local maximum in $(a, b)$ where $f^{\prime}(x) = 0$. However,because $f^{\prime}(b) > 0$,the function must have crossed the $x$-axis or turned back to satisfy the boundary conditions,requiring at least one more root for $f^{\prime}(x) = 0$ to account for the change in slope behavior. Thus,there must be at least $2$ roots of $f^{\prime}(x) = 0$ in the interval $(a, b)$.

Let $f: R \rightarrow R$ be a differentiable function such that $f(a)=0=f(b)$ and $f^{\prime}(a) f^{\prime}(b) > 0$ for some $a < b$. Then,the minimum number of roots of $f^{\prime}(x)=0$ in the interval $(a, b)$ is

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