Let f:[a, b] rightarrow R be continuous in [a | vedclass

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(A) Define a function $g(x) = e^{-x} f(x)$.Since $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, $g(x)$ is also continuous on $[a, b]

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there exists at least one point $c \in(a, b)$ for which $f^{\prime}(c)=f(c)$

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(A) Define a function $g(x) = e^{-x} f(x)$.Since $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, $g(x)$ is also continuous on $[a, b]$ and differentiable on $(a, b)$.We have $g(a) = e^{-a} f(a) = e^{-a} \cdot 0 = 0$ and $g(b) = e^{-b} f(b) = e^{-b} \cdot 0 = 0$.Since $g(a) = g(b) = 0$, by Rolle's Theorem, there exists at least one point $c \in (a, b)$ such that $g^{\prime}(c) = 0$.Calculating the derivative: $g^{\prime}(x) = -e^{-x} f(x) + e^{-x} f^{\prime}(x) = e^{-x} (f^{\prime}(x) - f(x))$.Setting $g^{\prime}(c) = 0$, we get $e^{-c} (f^{\prime}(c) - f(c)) = 0$.Since $e^{-c} 
eq 0$ for any real $c$, we must have $f^{\prime}(c) - f(c) = 0$, which implies $f^{\prime}(c) = f(c)$ for at least one $c \in (a, b)$.

Let $f:[a, b] \rightarrow R$ be continuous in $[a, b]$, differentiable in $(a, b)$ and $f(a)=0=f(b)$. Then

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