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Simplify. Rewrite the expression in the form 9n9^n. \newline(92)5=(9^2)^5=

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Let g(x)=2x g(x)=2^{x} .\newlineCan we use the mean value theorem to say the equation g′(x)=16 g^{\prime}(x)=16 has a solution where 3<x<5 3<x<5 ?\newlineChoose 11 answer:\newline(A) No, since the function is not differentiable on that interval.\newline(B) No, since the average rate of change of g g over the interval 3≤x≤5 3 \leq x \leq 5 isn't equal to 16‾ 1 \overline{6} .\newline(C) Yes, both conditions for using the mean value theorem have been met.
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