a. determine whether the mean value theorem applies to the function f(x)=6x^(1/9) on the interval -512,512…

a. determine whether the mean value theorem applies to the function f(x)=6x^(1/9) on the interval -512,512. b. if so, find or approximate the point(s) that are guaranteed to exist by the mean value theorem. a. choose the correct answer below. a. f(x) is continuous on(-512,512) and is differentiable on -512,512. therefore, the mean value theorem applies to the given function. b. the mean value theorem does not apply to the given function because f(x) is not differentiable on (-512,512). c. the mean value theorem does not apply to the given function because f(x) is not continuous on -512,512. d. f(x) is continuous on -512,512 and is differentiable on (-512,512). therefore, the mean value theorem applies to the given function.

a. determine whether the mean value theorem applies to the function f(x)=6x^(1/9) on the interval -512,512. b. if so, find or approximate the point(s) that are guaranteed to exist by the mean value theorem. a. choose the correct answer below. a. f(x) is continuous on(-512,512) and is differentiable on -512,512. therefore, the mean value theorem applies to the given function. b. the mean value theorem does not apply to the given function because f(x) is not differentiable on (-512,512). c. the mean value theorem does not apply to the given function because f(x) is not continuous on -512,512. d. f(x) is continuous on -512,512 and is differentiable on (-512,512). therefore, the mean value theorem applies to the given function.

Answer

Explanation:

Step1: Recall Mean - Value Theorem conditions

The Mean - Value Theorem states that if (y = f(x)) 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^{\prime}(c)=\frac{f(b)-f(a)}{b - a}).

Step2: Analyze the continuity of (f(x)=6x^{\frac{1}{9}})

The function (y = f(x)=6x^{\frac{1}{9}}) is a power - function. The domain of (y = x^{\frac{1}{9}}=\sqrt[9]{x}) is all real numbers, so (f(x)=6x^{\frac{1}{9}}) is continuous on the closed interval ([-512,512]).

Step3: Analyze the differentiability of (f(x)=6x^{\frac{1}{9}})

Find the derivative using the power rule (y = ax^{n}\Rightarrow y^{\prime}=anx^{n - 1}). So (f^{\prime}(x)=6\times\frac{1}{9}x^{\frac{1}{9}-1}=\frac{2}{3}x^{-\frac{8}{9}}=\frac{2}{3x^{\frac{8}{9}}}). The function (f^{\prime}(x)) is not defined at (x = 0), but (0\in(-512,512)). So (f(x)) is not differentiable on ((-512,512)).

Answer:

B. The Mean Value Theorem does not apply to the given function because (f(x)) is not differentiable on ((-512,512)).