QUESTION IMAGE
Question
a. determine whether the mean value theorem applies to the function ( f(x)=6x^{\frac{1}{7}} ) on the interval (-128,128).
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 (-128,128) and is differentiable on ((-128,128)). 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 ((-128,128))
○ c. ( f(x) ) is continuous on ((-128,128)) and is differentiable on (-128,128). 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 (-128,128)
Step1: Check continuity
The function \(y = x^{\frac{1}{7}}=\sqrt[7]{x}\) is a root - function. Root functions \(y = x^{\frac{1}{n}}\) where \(n\) is odd (\(n = 7\) in this case) are continuous for all real \(x\). So \(f(x)=6x^{\frac{1}{7}}\) is continuous on the closed interval \([-128,128]\).
Step2: Check differentiability
Differentiate \(f(x)\) using the power rule \(y = ax^{n}\), \(y^\prime=anx^{n - 1}\). For \(f(x)=6x^{\frac{1}{7}}\), \(f^\prime(x)=6\times\frac{1}{7}x^{\frac{1}{7}-1}=\frac{6}{7}x^{-\frac{6}{7}}=\frac{6}{7x^{\frac{6}{7}}}\). The derivative \(f^\prime(x)\) exists for all \(x
eq0\) in the open interval \((-128,128)\). But \(x = 0\) is just a single point. A function is differentiable on an open interval \((a,b)\) if it is differentiable at all but a finite number of points in \((a,b)\). So \(f(x)\) is differentiable on \((-128,128)\).
Since \(f(x)\) is continuous on \([-128,128]\) and differentiable on \((-128,128)\), by the Mean Value Theorem, there exists at least one \(c\in(-128,128)\) such that \(f^\prime(c)=\frac{f(128)-f(-128)}{128 - (-128)}\).
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A. \(f(x)\) is continuous on \([-128,128]\) and is differentiable on \((-128,128)\). Therefore, the Mean Value Theorem applies to the given function.