Let g(x)=2^(x). Can we use the mean value theorem to say the equation g^(')(x)=16 has a solution where 3

Question

Let  g(x)=2^(x). Can we use the mean value theorem to say the equation  g^(')(x)=16 has a solution where  3 < x < 5 ? Choose 1 answer: (A) No, since the function is not differentiable on that interval. (B) No, since the average rate of change of  g over the interval  3 <= x <= 5 isn't equal to  1 bar(6). (C) Yes, both conditions for using the mean value theorem have been met.

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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) = (f(b) - f(a)) / (b - a). We need to check if g(x) = 2^x satisfies these conditions on the interval [3, 5].Check g(x) Conditions: The function g(x) = 2^x is a continuous and differentiable function for all real numbers, including the interval [3, 5]. Therefore, the first condition of the MVT is satisfied.Calculate Average Rate of Change: Next, we calculate the average rate of change of g(x) over the interval [3, 5]. This is given by (g(5) - g(3)) / (5 - 3) = (2^5 - 2^3) / (5 - 3) = (32 - 8) / 2 = 24 / 2 = 12.Compare Average Rate of Change: Now we compare the average rate of change, which is 12, to the value given in the problem, which is 16. Since 12 does not equal 16, the average rate of change of g over the interval [3, 5] is not equal to 16.Apply Mean Value Theorem: Since the average rate of change of g on [3, 5] is not equal to 16, we cannot guarantee that there is a point c in (3, 5) where g'(c) equals 16, based on the Mean Value Theorem. Therefore, the correct answer is (B) No, since the average rate of change of g over the interval 3 <= x <= 5 isn't equal to 16.

Let $g(x)=2^{x}$ . $\newline$ Can we use the mean value theorem to say the equation $g^{\prime}(x)=16$ has a solution where \( 3

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Let g(x)=2x g(x)=2^{x} g(x)=2x.\newlineCan we use the mean value theorem to say the equation g′(x)=16 g^{\prime}(x)=16 g′(x)=16 has a solution where \( 3

Let $g(x)=2^{x}$ . $\newline$ Can we use the mean value theorem to say the equation $g^{\prime}(x)=16$ has a solution where \( 3