Question

Verify Rolle's theorem for the following function on the indicated interval f (x) = log (x² + 2) − log 3 on [−1, 1] ?

Answer 1

The given function is \[f\left( x \right) = \log\left( x^2 + 2 \right) - \log3\] ,which can be rewritten as

\[f\left( x \right) = \log\left( \frac{x^2 + 2}{3} \right)\] .

Since logarithmic function is differentiable and so continuous in its domain, \[f\left( x \right) = \log\left( \frac{x^2 + 2}{3} \right)\] is continuous on \[\left[ - 1, 1 \right]\] and differentiable on \[\left( - 1, 1 \right)\] .

Also,

\[f\left( 1 \right) = f\left( - 1 \right) = 0\]

Thus,

\[f\left( x \right)\] satisfies all the conditions of Rolle's theorem. 

Now, we have to show that there exists \[c \in \left( - 1, 1 \right)\] such that   \[f'\left( c \right) = 0\] .

We have

\[f\left( x \right) = \log\left( \frac{x^2 + 2}{3} \right)\]
\[ \Rightarrow f'\left( x \right) = \frac{3\left( 2x \right)}{x^2 + 2} = \frac{6x}{x^2 + 2}\]

\[\therefore f'\left( x \right) = 0\]
\[ \Rightarrow \frac{6x}{x^2 + 2} = 0\]
\[ \Rightarrow x = 0\]

Thus, \[c = 0 \in \left( - 1, 1 \right)\] such that 

\[f'\left( c \right) = 0\] .

​Hence, Rolle's theorem is verified.