For the function


For the function $\mathrm{f}(\mathrm{x})=x+\frac{1}{x}$

(a) $x=1$ is a point of maximum

(b) $x=-1$ is a point of minimum

(c) maximum value $>$ minimum value

(d) maximum value $<$ minimum value


(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


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

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

$\Rightarrow x^{2}=1$

$\Rightarrow x=\pm 1$


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

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

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


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

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

The local minimum value is given by


The local maximum value is given by


$\therefore$ Maximum value $<$ Minimum value

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