Question

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

Answer 1

$$\text{ Given }:f\left(x\right)=\frac{1}{x^2+2}$$

$$\Rightarrow f^{\prime }\left(x\right)=\frac{-2x}{{(x^2+2)}^2}$$

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

$$f^{\prime }\left(x\right)=0$$

$$\Rightarrow \frac{-2x}{{(x^2+2)}^2}=0$$

$$\Rightarrow x=0$$

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

$$f^{\prime }\left(x\right)>0$$ .

Also, for values close to x = 0 and to the right of 0, $$f^{\prime }\left(x\right)<0$$ .

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

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