Question

find the value or values of c that satisfy the equation \\( \frac { f ( b ) - f ( a ) } { b - a } = f ^ { \prime } ( c ) \\) in the conclusion of the mean value theorem for the given function and interval. \\( f ( x ) = 4 x + \frac { 4 } { x } \cdot \left \frac { 1 } { 10 } \cdot 10 \
ight \\) c = (use a comma to separate answers as needed.)

Explanation:

Step1: Calculate \(f(a)\) and \(f(b)\)

Given \(a = \frac{1}{10}\), \(b = 10\), and \(f(x)=4x+\frac{4}{x}\).
\(f(a)=f(\frac{1}{10})=4\times\frac{1}{10}+\frac{4}{\frac{1}{10}}=\frac{2}{5}+40=\frac{2 + 200}{5}=\frac{202}{5}\)
\(f(b)=f(10)=4\times10+\frac{4}{10}=40+\frac{2}{5}=\frac{200 + 2}{5}=\frac{202}{5}\)

Step2: Calculate \(\frac{f(b)-f(a)}{b - a}\)

Since \(f(b)-f(a)=\frac{202}{5}-\frac{202}{5} = 0\), and \(b - a=10-\frac{1}{10}=\frac{100 - 1}{10}=\frac{99}{10}\), then \(\frac{f(b)-f(a)}{b - a}=0\)

Step3: Find \(f^{\prime}(x)\) and solve \(f^{\prime}(c)=0\)

Differentiate \(f(x)=4x+\frac{4}{x}\) using the power rule \((x^n)^\prime=nx^{n - 1}\).
\(f^{\prime}(x)=4-\frac{4}{x^{2}}\)
Set \(f^{\prime}(c)=4-\frac{4}{c^{2}} = 0\)
\(4=\frac{4}{c^{2}}\)
\(c^{2}=1\)
\(c=\pm1\)

But since the interval is \([\frac{1}{10},10]\), we discard \(c=-1\)

Answer:

\(1\)