Question

Use the guidelines of this section to sketch the curve.

19.\(y = \frac{{{x^3}}}{{{x^3} + 1}}\)

Short answer

The obtained graph for the function \(y = \frac{{{x^3}}}{{{x^3} + 1}}\) is:

Step-by-step solution

Steps for Sketching a Graph for any Function

There are following terms needed to examine for Sketching a Graph for any given Function:

1. Find the Domain of the Function.
2. Calculate the Intercepts.
3. Check for Symmetricity.
4. Find the Asymptotes.
5. Intervals of Increase and Decrease.
6. Evaluate the Maxima and Minima of the function.
7. Examine Concavity and the Point of Inflection.
8. Sketch the Graph.

Solving for Domain

The given function is\(y = \frac{{{x^3}}}{{{x^3} + 1}}\).

Here, for theDomain, we have:

\(\begin{array}{l}D = \left\{ {x\left| {x \ne - 1} \right.} \right\}\\ \Rightarrow x \in \left\{ {\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)} \right\}\end{array}\)

Hence, the function has a domain within \(\left\{ {\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)} \right\}\).

Solving for Intercepts

Now, for intercepts, putting\(f\left( x \right) = 0\)in the given function, we have:

\(\begin{array}{c}y = \frac{{{x^3}}}{{{x^3} + 1}} = 0\\x = 0\end{array}\)

Also at\(x = 0\),

\(\begin{array}{c}y\left| {_{x = 0}} \right. = \frac{{{x^3}}}{{{x^3} + 1}}\\y = 0\end{array}\)

Hence, these are the required intercepts.

Checking the Symmetry of the Curve

The given function is related as:

\(f\left( { - x} \right) \ne f\left( x \right)\)

Hence, the function has no symmetry.

Solving for Asymptotes

Now, for asymptotes, we have:

\(y = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{{x^3}}}{{{x^3} + 1}} = 1\)

Here, this is a Horizontal Asymptote.

And,

\(y = \mathop {\lim }\limits_{x \to - {1^ \pm }} \frac{{{x^3}}}{{{x^3} + 1}} = \mp \infty \)

This is a Vertical Asymptote.

Hence, the asymptotes are: \(y = 1{\rm{ and }}x = - 1\).

Solving for Increasing-Decreasing

Let us differentiate the given function as:

\(\begin{array}{c}y' = \frac{d}{{dx}}\left( {\frac{{{x^3}}}{{{x^3} + 1}}} \right)\\ = \frac{{\left( {{x^3} + 1} \right)\left( {3{x^2}} \right) - \left( {{x^3}} \right)3{x^2}}}{{{{\left( {{x^3} + 1} \right)}^2}}}\\ = \frac{{\left( {3{x^2}} \right)}}{{{{\left( {{x^3} + 1} \right)}^2}}}\end{array}\)

Here,

\(f'\left( x \right) > 0 \Rightarrow {\rm{Increasing on }}\left\{ {\left( { - \infty , - 1} \right) \cup \left( { - 1,\infty } \right)} \right\}\)

There exists no interval for decreasing function.

Hence, these are the required intervals.

Solving for Maxima and Minima

For interval obtained, there is no local minimum as well as the maximum value can be obtained.

Hence, there are no minimaas well as maxima.

Examining Concavity and Inflection Point

Now, again differentiating the derivative of the given function as:

\(\begin{array}{c}y'' = \frac{d}{{dx}}\left( {\frac{{\left( {3{x^2}} \right)}}{{{{\left( {{x^3} + 1} \right)}^2}}}} \right)\\ = \frac{{{{\left( {{x^3} + 1} \right)}^2}\left( {6x} \right) - \left( {3{x^2}} \right)2\left( {{x^3} + 1} \right)\left( {3{x^2}} \right)}}{{{{\left( {{x^3} + 1} \right)}^4}}}\\ = \frac{{6{x^2}\left( {1 - 2{x^3}} \right)}}{{{{\left( {{x^3} + 1} \right)}^3}}}\end{array}\)

Here, we have:

\(\begin{array}{l}{\rm{for }}0 < x < {\left( {\frac{1}{2}} \right)^{\frac{1}{3}}}\,\,{\rm{or}}\,\,x < - 1\,\,\, \to \,\,\,y'' > 0\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,y{\rm{ is concave up}}\\{\rm{for }} - 1 < x < 0\,\,{\rm{or}}\,\,x > {\left( {\frac{1}{2}} \right)^{\frac{1}{3}}}\,\, \to \,\,\,y'' < 0\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,y{\rm{ is concave down}}\end{array}\)

\(\begin{array}{l}{\rm{At }}x = {\left( {\frac{1}{2}} \right)^{\frac{1}{3}}} \Rightarrow y = \frac{1}{3}\\{\rm{at}}\,\,\,\,\,\,\,\,\,\,{\rm{ }}x = 0 \Rightarrow y = 0\end{array}\)

Hence, the function is Concave downon\(\left\{ {\left( { - 1,0} \right) \cup \left( {{{\left( {\frac{1}{2}} \right)}^{\frac{1}{3}}},\infty } \right)} \right\}\)and Concave-upon\(\left\{ {\left( { - \infty , - 1} \right) \cup \left( {0,{{\left( {\frac{1}{2}} \right)}^{\frac{1}{3}}}} \right)} \right\}\).

And, the Inflection Points are \(\left( {0,0} \right){\rm{ and }}\left( {{{\left( {\frac{1}{2}} \right)}^{\frac{1}{3}}},\frac{1}{3}} \right)\).

Sketching the Graph

Using all the above data for the given function, the graph sketched is:

Hence, this is the required graph.