Let f(x)=2^(x)-sin(pi x). Below is Rafael's attempt to write a formal justification for the fact that the equation f^(')(x)=(1)/(4) has a solution where -2

Question

Let  f(x)=2^(x)-sin(pi x). Below is Rafael's attempt to write a formal justification for the fact that the equation  f^(')(x)=(1)/(4) has a solution where  -2 < x < -1. Is Rafael's justification complete? If not, why? Rafael's justification: Exponential and trigonometric functions are differentiable and continuous at all points in their domain, and  -2 <= x <= -1 is within  f 's domain. So, according to the mean value theorem,  f^(')(x)=(1)/(4) must have a solution somewhere in the interval  -2 < x < -1". " Choose 1 answer: (A) Yes, Rafael's justification is complete. (B) No, Rafael didn't establish that the average rate of change of  f over  [-2,-1] is equal to  (1)/(4). (C) No, Rafael didn't establish that  f is differentiable.

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MVT Explanation: Rafael's claim is based on the Mean Value Theorem (MVT), which states that if a function is continuous on the 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) is equal to the average rate of change of the function over [a, b]. We need to check if Rafael's justification meets the conditions of the MVT.Function Continuity: First, we need to verify that the function f(x) = 2^(x) - sin(pi x) is continuous on the closed interval [-2, -1]. Since both the exponential function 2^(x) and the sine function sin(pi x) are continuous everywhere in their domain, their combination is also continuous on [-2, -1].Function Differentiability: Next, we need to check if the function f(x) is differentiable on the open interval (-2, -1). The derivative of f(x), f'(x), involves the derivative of 2^(x), which is 2^(x)ln(2), and the derivative of -sin(pi x), which is -pi cos(pi x). Both of these derivatives exist for all x, so f(x) is differentiable on (-2, -1).Average Rate of Change Calculation: Now, we need to calculate the average rate of change of f(x) over the interval [-2, -1]. The average rate of change is given by [f(b) - f(a)] / (b - a), where a = -2 and b = -1. We need to evaluate f(-2) and f(-1) to find this rate.Calculate f(-2): Calculating f(-2) gives us f(-2) = 2^(-2) - sin(pi * -2) = 1/4 - sin(-2pi) = 1/4 - 0 = 1/4, since sin(-2pi) = 0.Calculate f(-1): Calculating f(-1) gives us f(-1) = 2^(-1) - sin(pi * -1) = 1/2 - sin(-pi) = 1/2 - 0 = 1/2, since sin(-pi) = 0.Calculate Average Rate of Change: The average rate of change of f(x) over [-2, -1] is then [f(-1) - f(-2)] / (-1 - (-2)) = (1/2 - 1/4) / 1 = 1/4.MVT Application: Since the average rate of change of f(x) over [-2, -1] is 1/4, and we have established that f(x) is continuous on [-2, -1] and differentiable on (-2, -1), the Mean Value Theorem guarantees that there exists some c in (-2, -1) such that f'(c) = 1/4.Incomplete Justification: Rafael's justification is incomplete because he did not explicitly calculate the average rate of change of f(x) over the interval [-2, -1] to show that it is equal to 1/4. This calculation is a necessary condition to apply the Mean Value Theorem.

Let $f(x)=2^{x}-\sin (\pi x)$ . $\newline$ Below is Rafael's attempt to write a formal justification for the fact that the equation $f^{\prime}(x)=\frac{1}{4}$ has a solution where \( -2

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Let $f(x)=2^{x}-\sin (\pi x)$ . $\newline$ Below is Rafael's attempt to write a formal justification for the fact that the equation $f^{\prime}(x)=\frac{1}{4}$ has a solution where \( -2