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(C) Consider the function $g(x) = e^{kx} f(x)$. Since $f(x)$ is differentiable on $[a, b]$ and $e^{kx}$ is differentiable everywhere,$g(x)$ is differe

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$J(x)=0$ has at least one root in $(a, b)$

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(C) Consider the function $g(x) = e^{kx} f(x)$. Since $f(x)$ is differentiable on $[a, b]$ and $e^{kx}$ is differentiable everywhere,$g(x)$ is differentiable on $[a, b]$.
Given $f(a) = 0$ and $f(b) = 0$,we have $g(a) = e^{ka} f(a) = 0$ and $g(b) = e^{kb} f(b) = 0$.
Since $g(a) = g(b) = 0$ and $g(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$,by Rolle's Theorem,there exists at least one $c \in (a, b)$ such that $g'(c) = 0$.
Now,$g'(x) = \frac{d}{dx} (e^{kx} f(x)) = k e^{kx} f(x) + e^{kx} f'(x) = e^{kx} (f'(x) + k f(x))$.
Since $g'(c) = 0$,we have $e^{kc} (f'(c) + k f(c)) = 0$.
Because $e^{kc} 
eq 0$ for any $c$,it must be that $f'(c) + k f(c) = 0$.
Thus,$J(c) = 0$ for at least one $c \in (a, b)$.

Let $f:[a, b] \rightarrow R$ be differentiable on $[a, b]$ and $k \in R$. Let $f(a)=0=f(b)$. Also let $J(x)=f'(x)+k f(x)$. Then

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