For what value of c does the conclusion of th | vedclass

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(A) The Mean Value Theorem states that for a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$,there exists at least one point $c

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$2 \log_3 e$

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(A) The Mean Value Theorem states that for a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$,there exists at least one point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = \log_e x$,$a = 1$,and $b = 3$.The derivative is $f'(x) = \frac{1}{x}$.Applying the formula:$f'(c) = \frac{f(3) - f(1)}{3 - 1}$$\frac{1}{c} = \frac{\log_e 3 - \log_e 1}{2}$Since $\log_e 1 = 0$,we have:$\frac{1}{c} = \frac{\log_e 3}{2}$$c = \frac{2}{\log_e 3} = 2 \log_3 e$.However,checking the options provided,the expression for $f'(c)$ itself is $\frac{1}{2} \log_e 3$. If the question asks for the value of $f'(c)$,the answer is $\frac{1}{2} \log_e 3$. If it asks for $c$,$c = \frac{2}{\log_e 3}$.

For what value of $c$ does the conclusion of the Mean Value Theorem hold for the function $f(x) = \log_e x$ on the interval $[1, 3]$?

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