Question

The Mean Value Theorem can be applied to which of the following functions on the closed interval -5,5 ? (A) f(x)=(1)/(sinx) (B) f(x)=(x-1)/(|x-1|) (C) f(x)=(x^(2))/(x^(2)-36) (D) f(x)=(x^(2))/(x^(2)-4

Answer 1

The Mean Value Theorem can be applied to option (C) since it is the only function that is continuous and differentiable on the interval [-5,5]. Options (A) and (D) have points of discontinuity, and option (B) is not differentiable within the interval.

Step-by-step explanation:

The Mean Value Theorem (MVT) can be applied to functions that are continuous on a closed interval [a, b] and differentiable on the open interval (a, b). Let's evaluate the options given:

• (A) f(x) = 1/sinx: This function is not continuous on the interval [-5,5] because sin(x) = 0 at x = 0, leading to a division by zero. Therefore, MVT cannot be applied.
• (B) f(x) = (x-1)/|x-1|: This function is not differentiable at x = 1 because of the absolute value, which causes a cusp at x = 1. Hence, MVT cannot be applied.
• (C) f(x) = x² / (x² - 36): This function has discontinuities at x = -6 and x = 6, which are outside the interval [-5,5]. Within the interval [-5,5], the function is continuous and differentiable. Thus, MVT can be applied to this function on the given interval.
• (D) f(x) = x² / (x² - 4): This function is not continuous at x = -2 and x = 2 where the denominator is zero. Therefore, MVT cannot be applied on the interval [-5,5].

Based on the analysis above, option (C) f(x) = x² / (x² - 36) is the correct choice to which the Mean Value Theorem can be applied on the interval [-5,5].

Answer 2

Based on the Mean Value Theorem's requirements, only option (C) f(x) = x² / (x² - 36) is continuous and differentiable on the closed interval [-5, 5], and thus the Mean Value Theorem can be applied.

Step-by-step explanation:

The Mean Value Theorem (MVT) states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). Let's evaluate the given functions on the interval [-5, 5] for their eligibility for the MVT application.

• (A) f(x) = 1/sinx: This function is not continuous on the given interval as sin(x) is zero at x = nπ, where n is an integer, leading to a discontinuity at x = 0 within the interval [-5, 5].
• (B) f(x) = (x - 1) / |x - 1|: The absolute value function creates a discontinuity at x = 1, meaning the function is not continuous over the whole interval.
• (C) f(x) = x² / (x² - 36): This function has discontinuities where the denominator is zero, that is, at x = ±6. Because these points are not in the interval [-5, 5], the function is continuous and differentiable on this interval, thereby satisfying the conditions for the Mean Value Theorem.
• (D) f(x) = x² / (x² - 4): Similar to option (C), this function has discontinuities at x = ±2 which are within the interval [-5, 5]; therefore, it is not eligible for the MVT application on this interval.

Based on the MVT requirements, only option (C) f(x) = x² / (x² - 36) qualifies to have the Mean Value Theorem applied on the closed interval [-5, 5].