Question

1 –54 Use the guidelines of this section to sketch the curve.

28. \(y = \frac{x}{{\sqrt {{x^2} - 1} }}\).

Short answer

Graph of the given curve is:

Step-by-step solution

Guidelines for sketching a curve Step 1: Steps for sketching a curve

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.

(A) Determine Domain

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

The domain of the function will be defined when \(\left\{ {x|{x^2} - 1 \ne 0} \right\}\).

So, the domain of the given function is \(\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)\).

(B) Determine Intercepts

Find \(y\)-intercepts by substituting 0 for \(x\) into \(y = \frac{x}{{\sqrt {{x^2} - 1} }}\).

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

So, there is no \(y\)-intercept as per the domain.

Find \(x\)-intercepts by substituting 0 for \(y\) into \(y = \frac{x}{{\sqrt {{x^2} - 1} }}\).

\(\begin{array}{l}0 = \frac{x}{{\sqrt {{x^2} - 1} }}\\0 = x\end{array}\)

So, there is no \(x\)-intercepts as per the domain.

(C) Determine symmetry

For the given function \(y = \frac{x}{{\sqrt {{x^2} - 1} }}\), \(f\left( { - x} \right) = - x\), so the function is symmetric about the origin.

(D) Find Asymptotes 

Determine \(\mathop {\lim }\limits_{x \to \pm \infty } y\) for Horizontal asymptotes.

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

So, \(x = \pm 1\).

Determine \(\mathop {\lim }\limits_{x \to {1^ + }} y\) and \(\mathop {\lim }\limits_{x \to {1^ - }} y\) for Vertical asymptotes.

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

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

So, the Vertical asymptote is \(x = \pm 1\).

(E) Find Intervals of Increase or Decrease 

Find the first derivative of the given function with respect to \(x\).

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

For \(y' = 0\), there is no \(x\), as \(y' = 0\) does not exist.

Draw a table for the interval of increasing and decreasing.

Interval	\(y'\)	Behaviour of \(y\)
\(\left( { - \infty - 1} \right)\)	-	Decreasing
\(\left( {1,\infty } \right)\)	-	Decreasing

(F) Find Local Minimum and Maximum values 

From the obtained table and the condition of local maxima and minima, it can be said that there is no maximum or minimum values as there is no sign of change.

(G) Determine Concavity and Points of Inflection 

Find \(y''\).

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

For \(y'' = 0\), \(x = 0\), but it does not lie in the domain of the function, so neglect it.

Draw a table for Concavity for different intervals.

Interval	Sign of \(y''\)	Behavior of \(y\)
\(\left( { - \infty , - 1} \right)\)	-	Concave downward
\(\left( {1,\infty } \right)\)	-	Concave upward

There is no inflection point.

(H) Draw Graph

Using all the obtained information from steps 2 to 8, draw the graph of the given function.