Question

Use the guidelines of this section to sketch the curve.

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

Short answer

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

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

Here, on simplifying the functions, we get:

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

So, for theDomain, we have:

\(\begin{array}{l}D = \left\{ {x\left| {x \ne 0} \right.} \right\}\\ \Rightarrow x \in \left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)\end{array}\)

Hence, the domain obtained is \(\left\{ {\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)} \right\}\).

Solving for Intercepts

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

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

Also, putting\(y = 0\)in the given function, we have:

\({x^2} + x + 1 = 0 \to {\rm{no real solution}}\)

Hence, there is no intercept here.

Checking the Symmetry of the Curve

Since, the given function is related as:

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

Hence, there is no symmetry.

Solving for Asymptotes

Now, for asymptotes, we have:

\(y = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{{x^2} + x + 1}}{{{x^2}}} = 1\)

Here, this is a Horizontal Asymptote.

Also,

\(y = \mathop {\lim }\limits_{x \to {0^ \pm }} \frac{{{x^2} + x + 1}}{{{x^2}}} = \pm \infty \)

This is a Vertical Asymptote.

Hence, the asymptotes are: \(x = 0{\rm{ and }}y = 1\).

Solving for Increasing-Decreasing

Let us differentiate the given function as:

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

Here,

\(\begin{array}{l}f'\left( x \right) > 0 \Rightarrow {\rm{Increasing on }}\left( { - 2,0} \right)\\f'\left( x \right) < 0 \Rightarrow {\rm{Decreasing on }}\left\{ {\left( { - \infty , - 2} \right) \cup \left( {0,\infty } \right)} \right\}\end{array}\)

Hence, these are the required intervals.

Solving for Maxima and Minima

For interval\(\left\{ {\left( { - \infty , - 2} \right) \cup \left( {0,\infty } \right)} \right\}\), the local minimum value obtained is\(f\left( { - 2} \right) = \frac{3}{4}\), but there is no local maximum value of the function.

Hence, these are the required values of minima as well as maxima.

Examining Concavity and Inflection Point

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

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

Here, we have:

\(\begin{array}{l}{\rm{for }}x < - 3\,\,\, \to \,\,\,y'' < 0\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y{\rm{ is concave down}}\\{\rm{for }} - 3 < x < 0\,\,\, \to \,\,\,y'' > 0\,\,\, \to \,\,\,\,\,\,y{\rm{ is concave up}}\\\\{\rm{and at }}x = - 3\\y = \frac{{{{\left( { - 3} \right)}^2} - 3 + 1}}{{{{\left( { - 3} \right)}^2}}} = \frac{7}{9}\end{array}\)

Hence, the function is Concave upon\(\left\{ {\left( { - 3,0} \right) \cup \left( {0,\infty } \right)} \right\}\)and Concave Downon\(\left( { - \infty , - 3} \right)\).

And the Inflection Point is \(\left( { - 3,\frac{7}{9}} \right)\).

Sketching the Graph

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

Hence, this the required graph.