Write the maximum value

Solution:

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

$\Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}$

For a local maxima or a local minima, we must have

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

$\Rightarrow 1-\frac{1}{x^{2}}=0$

$\Rightarrow x^{2}=1$

$\Rightarrow x=1,-1$

But $x<0$

$\Rightarrow x=-1$

Now,

$f^{\prime \prime}(x)=\frac{1}{x^{3}}$

At $x=-1:$

$f^{\prime \prime}(-1)=\frac{2}{(-1)^{3}}=-2<0$

So, $x=-1$ is a point of local maximum.

Thus, the local maximum value is given by

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