The function $f\left( x \right) = x + \dfrac{1}{x}$ has
A. A local maxima at$x = 1$ and a local minima at$x = - 1$
B. A local minima at$x = 1$ and a local maxima at$x = - 1$
C. absolute maxima at$x = 1$and absolute minima at$x = - 1$
D. absolute minima at$x = 1$and absolute maxima at$x = - 1$

Hint: First, we shall analyze the given information so that we can able to solve the given problem. Let us understand the difference between the local extrema and absolute maxima. The absolute maximum is the largest value of all values of $f\left( x \right)$ and the absolute minimum is the least value of all values of $f\left( x \right)$. The local maximum cannot be the largest of all values, but the value will be large when compared to nearby values, and similarly, the local minimum is the small value when compared to nearby values.
Formula to be used:
The power rule of differentiation is\[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]

Complete step by step answer:
We are given that$f\left( x \right) = x + \dfrac{1}{x}$.
First, we shall calculate the first derivative of the given function.
That is$\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {x + \dfrac{1}{x}} \right)$
$f'\left( x \right) = \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)$ (Here we separated the terms)

\[ = \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {{x^{ - 1}}} \right)\]
\[ = 1.{x^{1 - 1}} + \left( { - 1} \right){x^{ - 1 - 1}}\] (Here we applied the power rule of differentiation)
\[ = 1.{x^0} - {x^{ - 2}}\]
\[ = 1 - \dfrac{1}{{{x^2}}}\]
Hence, we get$f'\left( x \right) = 1 - \dfrac{1}{{{x^2}}}$
Now, we need to equate the resultant derivative to zero to obtain the value of$x$ .
That is$1 - \dfrac{1}{{{x^2}}} = 0$
\[ \Rightarrow \dfrac{1}{{{x^2}}} = 1\]
\[ \Rightarrow {x^2} = 1\]
\[ \Rightarrow x = \pm 1\]
$ \Rightarrow x = 1andx = - 1$
The obtained values of $x$are usually known as the stationary points. We need to find whether the resultant stationary points are extremum points.
Now, we need to find the second derivative of the given function.
That is$\dfrac{d}{{dx}}f'\left( x \right) = \dfrac{d}{{dx}}\left( {1 - \dfrac{1}{{{x^2}}}} \right)$
$f''\left( x \right) = \dfrac{d}{{dx}}\left( 1 \right) - \dfrac{d}{{dx}}\left( {\dfrac{1}{{{x^2}}}} \right)$
                $ = 0 - \dfrac{d}{{dx}}\left( {{x^{ - 2}}} \right)$
     $ = - \left( { - 2} \right){x^{ - 2 - 1}}$ (Here we applied the power rule of differentiation)
$ = 2{x^{ - 3}}$
$ = \dfrac{2}{{{x^3}}}$
Hence, we get$f''\left( x \right) = \dfrac{2}{{{x^3}}}$
Now, we need to apply the stationary points in the above equation.
At the stationary point$x = 1$, $f''\left( x \right) = \dfrac{2}{{{1^3}}}$
$ \Rightarrow f''\left( x \right) = 2$ and it is greater than zero
At the stationary point$x = - 1$, $f''\left( x \right) = \dfrac{2}{{ - {1^3}}}$
$ \Rightarrow f''\left( x \right) = - 2$ and it is less than zero.
If we get$f''\left( x \right) < 0$, then we can say that$x = - 1$is a point of local maxima. And, if$f''\left( x \right) > 0$, then $x = 1$is a point of local minima
Hence, we get a local minimum at$x = 1$ and a local maximum at $x = - 1$ .

So, the correct answer is “Option B”.

Note: First, we need to find the first derivative of the given function and then we need to equate it with zero to obtain the stationary points$x = {x_0}$. Then, we need to find the second derivative of the given function. Now, we shall apply the obtained stationary points in the previous equation.
       If we get$f''\left( x \right) < 0$, then we can say that$x = {x_0}$is a point of local maxima. And, if$f''\left( x \right) > 0$, then $x = {x_0}$is a point of local minima.