Question

Use the guidelines of this section to sketch the curve.

\(y = \frac{x}{{\sqrt {{x^2} + 1} }}\)

Short answer

The obtained graph for the function \(y = \frac{x}{{\sqrt {{x^2} + 1} }}\) is:

Step-by-step solution

Steps for Sketching a Graph for any Function

There are following terms needed to examine forSketching a Graphfor any given Function:

1. Find the Domainof the Function.
2. Calculate the Intercepts.
3. Check forSymmetricity.
4. Find theAsymptotes.
5. Intervals ofIncrease and Decrease.
6. Evaluate theMaxima and Minimaof the function.
7. ExamineConcavityand the Point of Inflection.
8. Sketch the Graph.

Solving for Domain

The given function is:

\(y = \frac{x}{{\sqrt {{x^2} + 1} }}\)

Here, for theDomain, we have:

\(D = \left\{ {x\left| {x \in \mathbb{R}} \right.} \right\}\)

Hence, the function exists for all real numbers.

Solving for Intercepts

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

\(\begin{array}{c}y = \frac{x}{{\sqrt {{x^2} + 1} }} = 0\\x = 0\end{array}\)

Also at \(x = 0\),

\(\begin{array}{c}y\left| {_{x = 0}} \right. = \frac{x}{{\sqrt {{x^2} + 1} }}\\y = 0\end{array}\)

Hence, these are the required intercepts.

Checking the Symmetry of the Curve

The given function is an odd function as:

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

Hence, the function has symmetry about the origin.

Solving for Asymptotes

Now, for asymptotes, we have:

\(\begin{array}{c}y = \mathop {\lim }\limits_{x \to \pm \infty } \frac{x}{{\sqrt {{x^2} + 1} }}\\ = \mathop {\lim }\limits_{x \to \pm \infty } \frac{1}{{\sqrt {1 + \frac{1}{{{x^2}}}} }}\\ = \pm 1\end{array}\)

Here, this is a Horizontal Asymptote.

And, there exists no vertical asymptote.

Hence, the asymptotes are: \(y = \pm 1\).

Solving for Increasing-Decreasing

Let us differentiate the given function as:

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

Here,

\(f'\left( x \right) > 0 \Rightarrow {\rm{Increasing on }}\forall {\rm{x}} \in \mathbb{R}\).

There exists no interval for decreasing function.

Hence, these are the required intervals.

Solving for Maxima and Minima

For interval obtained, there exist no extreme values.

Hence, there are no minima as well as maxima.

Examining Concavity and Inflection Point

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

\(\begin{array}{c}y'' = \frac{d}{{dx}}\left( {\frac{1}{{{{\left( {{x^2} + 1} \right)}^{\frac{3}{2}}}}}} \right)\\ = - \frac{3}{2}\frac{{2x}}{{{{\left( {{x^2} + 1} \right)}^{\frac{5}{2}}}}}\\ = \frac{{ - 3x}}{{{{\left( {{x^2} + 1} \right)}^{\frac{5}{2}}}}}\end{array}\)

Here, we have:

\(\begin{array}{l}{\rm{for }}x < 0\,\,\, \to \,\,\,y'' > 0\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,y{\rm{ is concave up}}\\{\rm{for }}x > 0\,\, \to \,\,\,y'' < 0\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y{\rm{ is concave down}}\end{array}\)

And,

\({\rm{At }}x = 0 \Rightarrow y = 0\)

Hence, the function is Concave downon \(\left( {0,\infty } \right)\) and Concave-upon \(\left( { - \infty ,0} \right)\).

And, the Inflection Point is: \(\left( {0,0} \right)\).

Sketching the Graph

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

Hence, this is the required graph.