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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,

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

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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 $(a, b)$.
Given $f(a)=0$ and $f(b)=0$, we have $g(a) = e^{-a} f(a) = 0$ and $g(b) = e^{-b} f(b) = 0$.
By Rolle's Theorem, there exists at least one point $c \in (a, b)$ such that $g^{\prime}(c) = 0$.
Now, $g^{\prime}(x) = \frac{d}{dx} (e^{-x} f(x)) = e^{-x} f^{\prime}(x) - e^{-x} f(x) = e^{-x} (f^{\prime}(x) - f(x))$.
Setting $g^{\prime}(c) = 0$, we get $e^{-c} (f^{\prime}(c) - f(c)) = 0$.
Since $e^{-c} 
eq 0$ for any $c$, it follows that $f^{\prime}(c) - f(c) = 0$, or $f^{\prime}(c) = f(c)$.

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

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