Question-42

maxima and minima

Consider the function defined on \(\displaystyle \mathbb{R} \backslash \{0\}\):

\[ \begin{equation*} f( x) =x+\frac{1}{x} \end{equation*} \]

\(\displaystyle f\) has a local maximum at \(\displaystyle x=1\)
\(\displaystyle f\) has a global maximum at \(\displaystyle x=1\)
\(\displaystyle f\) has a local minimum at \(\displaystyle x=1\)
\(\displaystyle f\) has a global minimum at \(\displaystyle x=1\)
\(\displaystyle f\) has a local maximum at \(\displaystyle x=-1\)
\(\displaystyle f\) has a global maximum at \(\displaystyle x=-1\)
\(\displaystyle f\) has a local minimum at \(\displaystyle x=-1\)
\(\displaystyle f\) has a global minimum at \(\displaystyle x=-1\)

Hint

First and second derivatives

Answer

\(\displaystyle f\) has a local maximum at \(\displaystyle x=1\)
\(\displaystyle f\) has a global maximum at \(\displaystyle x=1\)
\(\displaystyle f\) has a local minimum at \(\displaystyle x=1\)
\(\displaystyle f\) has a global minimum at \(\displaystyle x=1\)
\(\displaystyle f\) has a local maximum at \(\displaystyle x=-1\)
\(\displaystyle f\) has a global maximum at \(\displaystyle x=-1\)
\(\displaystyle f\) has a local minimum at \(\displaystyle x=-1\)
\(\displaystyle f\) has a global minimum at \(\displaystyle x=-1\)

Solution

We have:

\[ \begin{equation*} f^{\prime }( x) =1-\frac{1}{x^{2}} \end{equation*} \]

Setting this to zero, we get \(\displaystyle x=\pm 1\) as the two critical points. Now for the second derivative:

\[ \begin{equation*} f^{\prime \prime }( x) =\frac{2}{x^{3}} \end{equation*} \]

We see that \(\displaystyle f^{\prime \prime }( 1) >0\) and \(\displaystyle f^{\prime \prime }( -1) < 0\). This implies that \(\displaystyle x=1\) corresponds to a local minimum and \(\displaystyle x=-1\) corresponds to a local maximum. These two points cannot corresponds to global extrema since:

\[ \begin{equation*} \lim\limits _{x\rightarrow \infty } \ f( x) =\infty ,\ \ \ \lim\limits _{x\rightarrow -\infty } \ f( x) =-\infty \end{equation*} \]