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

Question

Let  f(x)=(1)/(x-2). Can we use the mean value theorem to say the equation  f^(')(x)=(1)/(2) has a solution where  1 < x < 4 ? Choose 1 answer: (A) No, since the function is not differentiable on that interval. (B) No, since the average rate of change of  f over the interval  1 <= x <= 4 isn't equal to  (1)/(2). (C) Yes, both conditions for using the mean value theorem have been met.

Bytelearn · Accepted Answer

Conditions for MVT: First, let's recall the conditions for the Mean Value Theorem (MVT). The MVT can be applied to a function f on a closed interval [a, b] if: 1. f is continuous on the closed interval [a, b], and 2. f is differentiable on the open interval (a, b). We need to check if these conditions are met for f(x) = 1/(x-2) on the interval [1, 4].Check Continuity: Let's check the first condition: continuity. The function f(x) = 1/(x-2) is continuous everywhere except at x = 2, where it has a vertical asymptote. Since x = 2 is within the closed interval [1, 4], the function is not continuous on the entire interval [1, 4].Continuity Analysis: Since the function f(x) is not continuous on the entire interval [1, 4], the first condition for the Mean Value Theorem is not met. Therefore, we cannot apply the MVT to this function on the interval [1, 4].Conclusion: Given that the first condition for the MVT is not satisfied, we do not need to check the second condition (differentiability) because both conditions must be met for the MVT to apply. We can conclude that the MVT cannot be used to guarantee the existence of a solution to f'(x) = 1/2 where 1 < x < 4.

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

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