Question

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`?

Answer 1

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)}`.