Question

a. determine whether the mean value theorem applies to the function ( f(x)=6x^{\frac{1}{3}} ) on the interval (-8,8) 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 ((-8,8)) and is differentiable on (-8,8). 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 ((-8,8)). c. ( f(x) ) is continuous on (-8,8) and is differentiable on ((-8,8)). therefore, the mean value theorem applies to the given function. d. the mean value theorem does not apply to the given function because ( f(x) ) is not continuous on (-8,8).

Explanation:

Step1: Check continuity

The function \( y = x^{\frac{1}{3}}\) is a root - function. The function \( f(x)=6x^{\frac{1}{3}}\) is a composition of a constant multiple and a root - function. Root functions \( y = x^{n}\) where \(n=\frac{1}{3}\) are continuous for all real \(x\). So \(f(x)\) is continuous on the closed interval \([-8,8]\).

Step2: Check differentiability

Differentiate \(f(x)\) using the power rule \(y = ax^{n}\), \(y^\prime=anx^{n - 1}\). For \(f(x)=6x^{\frac{1}{3}}\), \(f^\prime(x)=6\times\frac{1}{3}x^{\frac{1}{3}-1}=2x^{-\frac{2}{3}}=\frac{2}{x^{\frac{2}{3}}}\). The derivative \(f^\prime(x)\) is not defined at \(x = 0\) (since division by zero occurs when \(x = 0\)). But the Mean Value Theorem requires the function to be differentiable on the open interval \((-8,8)\). 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 \(f^\prime(c)=\frac{f(b)-f(a)}{b - a}\).

Answer:

C. \(f(x)\) is continuous on \([-8,8]\) and is differentiable on \((-8,8)\). Therefore, the Mean Value Theorem applies to the given function.