How do you determine all values of c that satisfy the mean value theorem on the interval [1,3] for # f(x)= ln x^2#?

1 Answer
Bill K.
May 6, 2015

First take the derivative of #f# to get #f'(x)=\frac{1}{x^{2}}\cdot 2x=\frac{2}{x}# (this can also be done without the Chain Rule by using a property of logarithms: #f(x)=2\ln(x)# so #f'(x)=2/x#).

Next, calculate the difference quotient #\frac{f(3)-f(1)}{3-1}=\frac{ln(9)-ln(1)}{2}=\frac{\ln(9)}{2}#.

Now set these equal to each other and solve: #f'(c)=\frac{f(3)-f(1)}{3-1}\rightarrow \frac{2}{c}=\frac{ln(9)}{2}\rightarrow c=\frac{4}{ln(9)}\approx 1.82048#, which is in the interval #[1,3}#.

Also note that #ln(9)=ln(3^{2})=2ln(3)# so this answer can also be written as #c=\frac{2}{ln(3)}#.

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