Let g(x)=sin(x). Can we use the mean value theorem to say the equation g^(')(x)=(2)/(pi) has a solution where 0

Question

Let  g(x)=sin(x). Can we use the mean value theorem to say the equation  g^(')(x)=(2)/(pi) has a solution where  0 < x < (pi)/(2) ? 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  0 <= x <= (pi)/(2) isn't equal to  (2)/(pi). (C) Yes, both conditions for using the mean value theorem have been met.

Bytelearn · Accepted Answer

Recall MVT Conditions: First, let's recall the conditions for the Mean Value Theorem (MVT). The MVT can be applied if the function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). We need to check if these conditions are met for g(x) = sin(x) on the interval [0, pi/2].Function Continuity: The function g(x) = sin(x) is continuous on the closed interval [0, pi/2] because sine is a continuous function everywhere.Function Differentiability: The function g(x) = sin(x) is differentiable on the open interval (0, pi/2) because the derivative of sine, which is cosine, exists and is continuous everywhere.Calculate Average Rate of Change: Now, let's calculate the average rate of change of g(x) = sin(x) over the interval [0, pi/2]. The average rate of change is given by (g(b) - g(a)) / (b - a), where a = 0 and b = pi/2.Substitute into Formula: Substitute a and b into the average rate of change formula to get (sin(pi/2) - sin(0)) / (pi/2 - 0) = (1 - 0) / (pi/2) = 2/pi.Verify Average Rate of Change: Since the average rate of change of g(x) over the interval [0, pi/2] is equal to 2/pi, which is the same as the value given for g'(x), the second condition for the MVT is also met.Apply Mean Value Theorem: Both conditions for the Mean Value Theorem are satisfied, so we can conclude that there exists at least one c in the open interval (0, pi/2) such that g'(c) = 2/pi.

Let $g(x)=\sin (x)$ . $\newline$ Can we use the mean value theorem to say the equation $g^{\prime}(x)=\frac{2}{\pi}$ has a solution where \( 0

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Let $g(x)=\sin (x)$ . $\newline$ Can we use the mean value theorem to say the equation $g^{\prime}(x)=\frac{2}{\pi}$ has a solution where \( 0