Question

(a) Determine the intervals on which the function \(f(x) = \frac{x}{{{x^2} + 1}}\) is increasing or decreasing.

(b) Determine the local maximum and minimum values of \(f(x) = \frac{x}{{{x^2} + 1}}\).

(c) Determine the intervals of concavity and the inflection points of\(f(x) = \frac{x}{{{x^2} + 1}}\).

Short answer

(a) The function \(f(x)\) is increasing on \(( - 1,\;1)\) and decreasing on \(( - \infty ,\;1) \cup (1,\; + \infty )\).

(b) The local maximum is \(f(1) = \frac{1}{2}\) and the local minimum is \(f( - 1) = \frac{{ - 1}}{2}\).

(c) The function \(f(x)\) is concave upward on \(( - \sqrt 3 ,\;0) \cup (\sqrt 3 ,\;\infty ),\;f(x)\) is concave downward on \(( - \infty , - \sqrt 3 ) \cup (0,\sqrt 3 )\) and the inflection points are\((0,\;0),\;\left( {\sqrt 3 ,\,\frac{{\sqrt 3 }}{4}} \right),\;\left( { - \sqrt 3 ,\;\frac{{ - \sqrt 3 }}{4}} \right)\).

Step-by-step solution

Given data

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

Concept of Increasing/decreasing test and concavity

Increasing/decreasing test:

(a) If\({f^\prime }(x) > 0\)on an interval, then\(f\)is increasing on that interval.

(b) If\({f^\prime }(x) < 0\)on an interval, then\(f\)is decreasing on that interval.

Definition of concavity:

"(a) If\({f^\prime }(x) > 0\)for all\(x\)in\(I\), then the graph of\(f\)is concave upward on\(I\).

(b) If \({f^\prime }(x) < 0\)for all\(x\)in\(I\), then the graph of\(f\)is concave downward on\(I\).”

Obtain the critical points 

(a)

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

\({f^\prime }(x) = \frac{d}{{dx}}\left( {\frac{x}{{{x^2} + 1}}} \right)\)

Apply the Quotient Rule as follows:

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

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

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

Hence, the critical point occurs at \(x = - 1\) and \(x = 1\) in the interval notation \(( - \infty ,\; - 1) \cup ( - 1,\;1) \cup (1,\;\infty )\).

Determine the nature of \({f^\prime }(x)\)

Determine the nature of \(f(x)\) in table 1 as follows:

Table 1

From the above table observe that, function \(f(x)\) is increasing on \(( - 1,\;1)\) and decreasing on \(( - \infty ,\;1) \cup (1,\; + \infty )\).

Determine the local maximum and local minimum

(b)

From part (a), the function \(f(x)\) is decreasing on \(( - 1,3)\).

So, the curve changes from increasing to decreasing at \(x = 1\) and the curve changes from decreasing to increasing at \(x = - 1\).

Thus, the local maximum occurs at \(x = 1\) and the local minimum occurs at \(x = - 1\).

Substitute \(x = 1\) in \(f(x)\) and obtain the local maximum as follows:

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

Substitute \(x = - 1\) in \(f(x)\) and obtain the local minimum as follows:

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

Thus, the local maximum is \(f(1) = \frac{1}{2}\) and local minimum is \(f( - 1) = \frac{{ - 1}}{2}\).

Determine the interval notation

(c)

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

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

Apply the Quotient Rule as follows:

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

The derivative \({f^{\prime \prime }}(x) > 0\) as follows:

Thus, the values of \(x\) are \(x < - \sqrt 3 ,0 < x < \sqrt 3 \). In interval notation\(( - \infty ,\; - \sqrt 3 ) \cup ( - \sqrt 3 ,\;0) \cup (0,\;\sqrt 3 ) \cup (\sqrt 3 ,\;\infty )\).

Determine the nature of \({f^{\prime \prime }}(x)\)

Determine the nature of in table 2 as follows:

Table 2

Hence, from the above table it can be concluded that the function \(f(x)\) is concave upward on \(( - \sqrt 3 ,0) \cup (\sqrt 3 ,\infty )\) and concave downward on \(( - \infty , - \sqrt 3 ) \cup (0,\sqrt 3 )\).

Identify that the curve changes from concave downward to concave upward at\(x = 0,x = \sqrt 3 ,x = - \sqrt 3 \).

Substitute \(x = 0,x = \sqrt 3 \) and \(x = - \sqrt 3 \) in \(f(x)\) and find the inflection points as follows:

Find the inflection points

When \(x = 0\) in \(f(x)\) and find the inflection points as follows:

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

When \(x = \sqrt 3 \)in \(f(x)\) and find the inflection points as follows:

\(\begin{aligned}{c}f(\sqrt 3 ) &= \frac{{\sqrt 3 }}{{{{(\sqrt 3 )}^2}}}\\ &= \frac{{\sqrt 3 }}{{3 + 1}}\\ &= \frac{{\sqrt 3 }}{4}\end{aligned}\)

When \(x = - \sqrt 3 \)in \(f(x)\) and find the inflection points as follows:

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

Thus, the inflection points are \((0,\,0),\;\left( {\sqrt 3 ,\;\frac{{\sqrt 3 }}{4}} \right),\;\left( { - \sqrt 3 ,\;\frac{{ - \sqrt 3 }}{4}} \right)\).