For the function

Solution:

(d) maximum value $<$ minimum value

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

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

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

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

$\Rightarrow x^{2}=1$

$\Rightarrow x=\pm 1$

Now,

$f^{\prime \prime}(x)=\frac{2}{x^{3}}$

$\Rightarrow f^{\prime \prime}(1)=\frac{2}{1}=2>0$

So, $x=1$ is a local minima.

Also,

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

So, $x=-1$ is a local maxima.

The local minimum value is given by

$f(1)=2$

The local maximum value is given by

$f(-1)=-2$

$\therefore$ Maximum value $<$ Minimum value