If the solve the problem

Question:

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

Solution:

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

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

For the local maxima or minima, we must have

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

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

$\Rightarrow x=0$

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

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

Therefore, by first derivative test, $x=0$ is a point of local maxima and the local maximum value of $f(x)$ is $\frac{1}{2}$.

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