Full solution

Q. Let

h(x)=\log _{2}(x)

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Can we use the mean value theorem to say the equation

h^{\prime}(x)=1

has a solution where

1<x<2

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Choose

1

answer:

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(A) No, since the function is not differentiable on that interval.

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(B) No, since the average rate of change of

h

over the interval

1 \leq x \leq 2

isn't equal to

1

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Recall Mean Value Theorem: First, let's recall the mean value theorem (MVT). The MVT states that if a function $f$ is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , then there exists at least one $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$ .
Check Function Conditions: Now, let's check if the function $h(x) = \log_2(x)$ meets the conditions of the MVT on the interval $[1, 2]$ . The function $h(x)$ is continuous and differentiable on the interval $[1, 2]$ because the logarithmic function is continuous and differentiable for all positive $x$ values.
Calculate Average Rate of Change: Next, we calculate the average rate of change of $h$ over the interval $[1, 2]$ . This is given by $\frac{h(2) - h(1)}{2 - 1}$ .
Find $h(2)$ and $h(1)$ : We calculate $h(2)$ and $h(1)$ . Since $h(x) = \log_2(x)$ , we have $h(2) = \log_2(2) = 1$ and $h(1) = \log_2(1) = 0$ .
Calculate Average Rate of Change: Now, we find the average rate of change: $(h(2) - h(1)) / (2 - 1) = (1 - 0) / (2 - 1) = 1 / 1 = 1$ .
Apply Mean Value Theorem: Since the average rate of change of $h$ over the interval $[1, 2]$ is equal to $1$ , and the function $h(x)$ is continuous and differentiable on that interval, both conditions for using the mean value theorem have been met.
Apply Mean Value Theorem: Since the average rate of change of $h$ over the interval $[1, 2]$ is equal to $1$ , and the function $h(x)$ is continuous and differentiable on that interval, both conditions for using the mean value theorem have been met.Therefore, by the mean value theorem, there exists at least one $c$ in the interval $(1, 2)$ such that $h'(c) = 1$ .