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 ) = 9 x + \frac { 9 } { x }, \left \frac { 1 } { 2 }, 2 \
ight \\) c = (use a comma to separate answers as needed.)

Explanation:

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

Given \(a = \frac{1}{2}\), \(b = 2\), and \(f(x)=9x+\frac{9}{x}\).
\(f(a)=f(\frac{1}{2})=9\times\frac{1}{2}+\frac{9}{\frac{1}{2}}=\frac{9}{2}+18=\frac{9 + 36}{2}=\frac{45}{2}\)
\(f(b)=f(2)=9\times2+\frac{9}{2}=18+\frac{9}{2}=\frac{36 + 9}{2}=\frac{45}{2}\)

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

\(\frac{f(b)-f(a)}{b - a}=\frac{\frac{45}{2}-\frac{45}{2}}{2-\frac{1}{2}} = 0\)

Step3: Find \(f^{\prime}(x)\)

Using the power rule \((x^n)^\prime=nx^{n - 1}\), \(f^{\prime}(x)=(9x+\frac{9}{x})^\prime=9-\frac{9}{x^{2}}\)

Step4: Solve \(f^{\prime}(c)=0\)

Set \(9-\frac{9}{c^{2}} = 0\)
\(9=\frac{9}{c^{2}}\)
Multiply both sides by \(c^{2}\): \(9c^{2}=9\)
Divide both sides by \(9\): \(c^{2}=1\)
Take square - roots: \(c=\pm1\)
Since \(c\in[\frac{1}{2},2]\), we reject \(c=- 1\)

Answer:

\(1\)