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Let $f:[a, b] \rightarrow \mathbb{R}$ be continuous in $[a, b]$, differentiable in $(a, b)$ and $f(a)=0=f(b)$. Then
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1177 Upvotes
Verified Answer
The correct answer is:
there exists at least one point $c \in(a, b)$ for which $f^{\prime}(c)=f(c)$
$:$ Let $g(x)=e^{-x} f(x)$
$$
\mathrm{g}(\mathrm{a})=\mathrm{g}(\mathrm{b})=0
$$
By Rolle's theorem, for atleast one $c \in(a, b)$ where $\mathrm{g}^{\prime}(\mathrm{c})=0$
or $e^{-c} f^{\prime}(c)-f(c) e^{-c}=0$
or $e^{-c}\left(f^{\prime}(c)-f(c)\right)=0$
or $f^{\prime}(c)=f(c) \quad$ for atleast one $c \in(a, b)$
$$
\mathrm{g}(\mathrm{a})=\mathrm{g}(\mathrm{b})=0
$$
By Rolle's theorem, for atleast one $c \in(a, b)$ where $\mathrm{g}^{\prime}(\mathrm{c})=0$
or $e^{-c} f^{\prime}(c)-f(c) e^{-c}=0$
or $e^{-c}\left(f^{\prime}(c)-f(c)\right)=0$
or $f^{\prime}(c)=f(c) \quad$ for atleast one $c \in(a, b)$
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