Let h(x)=log_(2)(x). Can we use the mean value theorem to say the equation h^(')(x)=1 has a solution where 1

Question

Let  h(x)=log_(2)(x). Can we use the mean value theorem to say the equation  h^(')(x)=1 has a solution where  1 < x < 2 ? Choose 1 answer: (A) No, since the function is not differentiable on that interval. (B) No, since the average rate of change of  h over the interval  1 <= x <= 2 isn't equal to 1 . (C) Yes, both conditions for using the mean value theorem have been met.

Bytelearn · Accepted Answer

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) = (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 (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.Therefore, by the mean value theorem, there exists at least one c in the interval (1, 2) such that h'(c) = 1.

Let $h(x)=\log _{2}(x)$ . $\newline$ Can we use the mean value theorem to say the equation $h^{\prime}(x)=1$ has a solution where \( 1

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Let $h(x)=\log _{2}(x)$ . $\newline$ Can we use the mean value theorem to say the equation $h^{\prime}(x)=1$ has a solution where \( 1