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Question
f(x) = \[\frac{1}{x^2 + 2}\] .
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Solution
\[\text { Given }: \hspace{0.167em} f\left( x \right) = \frac{1}{x^2 + 2}\]
\[ \Rightarrow f'\left( x \right) = \frac{- 2x}{\left( x^2 + 2 \right)^2}\]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow \frac{- 2x}{\left( x^2 + 2 \right)^2} = 0\]
\[ \Rightarrow x = 0\]
Now, for values close to x = 0 and to the left of 0,
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