Question

Find the local maximum and minimum values of the function \(f\) using First and Second Derivative Test and explain which method is preferable.

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

Short answer

The local maximum is \(f(2) = 4\) and local minimum is \(f(0) = 0\).

The Second Derivative Test is preferable as it requires less testing and it consumes less time.

Step-by-step solution

Given data

The given function is, \(f(x) = \frac{{{x^2}}}{{{x^2} - 1}}\).

Concept of first and second derivative test

First Derivative Test: "Suppose that\(c\)is a critical number of a continuous function\(f\).

(a) If\({f^\prime }\)changes from positive to negative at\(c\), then\(f\)have a local maximum at\(c\).

(b) If\({f^\prime }\)changes from negative to positive at\(c\), then\(f\)has a local minimum at\(c\).

(c) If\({f^\prime }\)is positive to the left and right of\(c\), or negative to the left and right of\(c\), then\(f\)has no local maximum or minimum at\(c\).

Second Derivative Test: "Suppose\({f^{\prime \prime }}\)is continuous near\(c\).

(a) If\({f^\prime }(c) = 0\)and\({f^{\prime \prime }}(c) > 0\), then\(f\)has a local minimum at\(c\).

(b) If\({f^\prime }(c) = 0\)and\({f^{\prime \prime }}(c) < 0\), then\(f\)has a local maximum at\(c\).

Obtain the critical numbers

Obtain the derivative of \(f(x)\) as follows:

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{{(x - 1){{\left( {{x^2}} \right)}^\prime } - {x^2}{{(x - 1)}^\prime }}}{{{{(x - 1)}^2}}}\\ &= \frac{{(x - 1)2x - {x^2}(1)}}{{{{(x - 1)}^2}}}\\ &= \frac{{2{x^2} - 2x - {x^2}}}{{{{(x - 1)}^2}}}\\ &= \frac{{x(x - 2)}}{{{{(x - 1)}^2}}}\end{aligned}\)

Set \({f^\prime }(x) = 0\) and find the critical numbers as follows:

\(\begin{aligned}{c}\frac{{x(x - 2)}}{{{{(x - 1)}^2}}} &= 0\\x(x - 2) = 0\\x &= 0,\;x - 2 = 0\\x &= 0,\;x = 2\end{aligned}\)

The critical numbers are \(x = 0,2\) and identify that \({f^\prime }(x)\) is undefined when \(x = 1\).

In interval notation, \(( - \infty ,0) \cup (0,1) \cup (1,2) \cup (2,\infty )\).

Use First Derivative Test

Use the first derivative test and find the nature of \(f(x)\) based on the above intervals in figure 1 as follows:

Interval	\(f'\left( x \right) = \frac{{x\left( {x - 2} \right)}}{{{{\left( {x - 1} \right)}^2}}}\)	\(f\left( x \right)\)
\(x < 0\)	\(\begin{aligned}{l}f'\left( { - 1} \right) &= \frac{{ - 1\left( { - 1 - 2} \right)}}{{{{\left( { - 1 - 1} \right)}^2}}}\\f'\left( x \right) > 0\end{aligned}\)	Increasing on \(\left( { - \infty ,0} \right)\)
\(0 < x < 1\)	\(\begin{aligned}{l}f'\left( {\frac{1}{2}} \right) &= \frac{{\frac{1}{2}\left( {\frac{1}{2} - 2} \right)}}{{{{\left( {\frac{1}{2} - 1} \right)}^2}}}\\ &= \frac{{\frac{1}{2}\left( { - \frac{3}{2}} \right)}}{{{{\left( { - \frac{1}{2}} \right)}^2}}}\\f'\left( x \right) < 0\end{aligned}\)	Decreasing on \(\left( {0,1} \right)\)
\(1 < x < 2\)	\(\begin{aligned}{l}f'\left( {\frac{3}{2}} \right) &= \frac{{\frac{3}{2}\left( {\frac{3}{2} - 2} \right)}}{{{{\left( {\frac{3}{2} - 1} \right)}^2}}}\\ &= \frac{{\frac{3}{2}\left( { - \frac{1}{2}} \right)}}{{{{\left( { - \frac{1}{2}} \right)}^2}}}\\f'\left( x \right) < 0\end{aligned}\)	Decreasing on \(\left( {1,2} \right)\)
\(x > 2\)	\(\begin{aligned}{l}f'\left( 3 \right) &= \frac{{3\left( {3 - 2} \right)}}{{{{\left( {3 - 1} \right)}^2}}}\\ &= \frac{{3\left( 1 \right)}}{{{{\left( 2 \right)}^2}}}\\f'\left( x \right) > 0\end{aligned}\)	Increasing on \(\left( {2,\infty } \right)\)

Table 1

From the above table, it is observed that \(f(x)\) is decreasing on \((0,\;1) \cup (1,\;2)\).

Thus, the local maximum occurs at \(x = 0\) and the local minimum occurs at \(x = 2\).

Substitute \(x = 0\) in \(f(x)\) as follows:

\(\begin{aligned}{c}f(0) &= \frac{{{{(0)}^2}}}{{(0) - 1}}\\ &= 0\end{aligned}\)

Substitute \(x = 2\) in \(f(x)\) as follows:

\(\begin{aligned}{c}f(2) &= \frac{{{{(2)}^2}}}{{(2) - 1}}\\ &= \frac{4}{1}\\ &= 4\end{aligned}\)

Thus, the local maximum is \(f(0) = 0\) and local minimum is \(f(2) = 4\).

Use Second Derivative Test

Obtain the derivative of \({f^\prime }(x)\) as follows:

\(\begin{aligned}{c}{f^{\prime \prime }}(x) &= \frac{d}{{dx}}\left( {{f^\prime }(x)} \right)\\ &= \frac{d}{{dx}}\left( {\frac{{{x^2} - 2x}}{{{{(x - 1)}^2}}}} \right)\end{aligned}\)

Apply the Quotient Rule as follows:

\(\begin{aligned}{c}{f^{\prime \prime }}(x) &= \frac{{{{(x - 1)}^2}{{\left( {{x^2} - 2x} \right)}^\prime } - \left( {{x^2} - 2x} \right){{\left( {{{(x - 1)}^2}} \right)}^\prime }}}{{{{\left( {{{(x - 1)}^2}} \right)}^2}}}\\ &= \frac{{{{(x - 1)}^2}(2x - 2) - \left( {{x^2} - 2x} \right)2(x - 1)(1)}}{{{{(x - 1)}^4}}}\\ &= \frac{{{{(x - 1)}^2}(2x - 2) - \left( {{x^2} - 2x} \right)2(x - 1)(1)}}{{{{(x - 1)}^4}}}\\ &= \frac{{\left( {{x^2} + 1 - 2x} \right)2(x - 1) - \left( {{x^2} - 2x} \right)2(x - 1)}}{{{{(x - 1)}^4}}}\end{aligned}\)

Simplify further and obtain \({f^{\prime \prime }}(x)\) as follows:

\(\begin{aligned}{c}{f^{\prime \prime }}(x) &= \frac{{2(x - 1)\left( {{x^2} + 1 - 2x - {x^2} + 2x} \right)}}{{{{(x - 1)}^4}}}\\ &= \frac{{2(x - 1)}}{{{{(x - 1)}^4}}}\\ &= \frac{2}{{{{(x - 1)}^3}}}\end{aligned}\)

Find the critical numbers

Find the values of \({f^{\prime \prime }}(x)\) at the critical numbers as follows:

Substitute \(x = 0\) in \({f^{\prime \prime }}(x)\)as follows:

\(\begin{aligned}{c}{f^{\prime \prime }}(0) &= \frac{2}{{{{(0 - 1)}^3}}}\\ &= \frac{2}{{ - 1}}\\ &= - 2\end{aligned}\)

Substitute \(x = 2\) in \({f^{\prime \prime }}(x)\) as follows:

\(\begin{aligned}{c}{f^{\prime \prime }}(2) &= \frac{2}{{{{(2 - 1)}^3}}}\\ &= \frac{2}{1}\\ &= 2\end{aligned}\)

Here, \({f^{\prime \prime }}(2) = 2 > 0\), so \(f\) has local maximum at \(f(2) = 4\) and \({f^{\prime \prime }}(0) = - 2 < 0\), so \(f\) has local minimum at \(f(0) = 0\).

Therefore, the Second Derivative Test is preferable as it requires less testing and it consumes less time.