Question

f(x) = \[\frac{1}{x^2 + 2}\] .

Answer 1

\[\text { Given }: \hspace{0.167em} f\left( x \right) = \frac{1}{x^2 + 2}\]

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

\[\text { For the local maxima or minima, we must have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow \frac{- 2x}{\left( x^2 + 2 \right)^2} = 0\]

\[ \Rightarrow x = 0\]

Now, for values close to x = 0 and to the left of 0,

\[f'\left( x \right) > 0\] .

Also, for values close to x = 0 and to the right of 0, \[f'\left( x \right) < 0\] .

Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of

\[f\left( x \right) \text { is }\frac{1}{2} .\]