Find the local maxima and minima of $ g(x) = \frac{1}{x^2 + 2} $.

Q: Find the local maxima and minima of $ g(x) = \frac{1}{x^2 + 2} $.

Explanation: \[ g'(x) = \frac{-2x}{(x^2 + 2)^2} \Rightarrow g'(x) = 0 \Rightarrow x = 0 \] Check sign change: - For $ x 0 $ - For $ x > 0 $, $ g'(x) < 0 $ So by first derivative test: \[ x = 0 \text{ is a local maximum} \Rightarrow g(0) = \frac{1}{2} \]

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Explanation: $$ g'(x) = \frac{-2x}{(x^2 + 2)^2} \Rightarrow g'(x) = 0 \Rightarrow x = 0 $$ Check sign change: - For $ x < 0 $, $ g'(x) > 0 $ - For $ x > 0 $, $ g'(x) < 0 $ So by first derivative test: $$ x = 0 \text{ is a local maximum} \Rightarrow g(0) = \frac{1}{2} $$

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