Question

Discuss the asymptotic behavior of \(f\left( x \right) = \frac{{\left( {{x^4} + 1} \right)}}{x}\), in

the same manner as in Exercise 78. Then use your results to

help sketch the graph of \(f\) .

Short answer

To sketch a graph we need to follow some procedure. Those are as follows:

1. Finding the domainof the function

2. Intercepts of the function (both \(x\) and \(y\)intercepts)

3. Checking the symmetricity of the function.

4. Finding the asymptotes of the function.

Formula for slant asymptote:

if \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {f\left( x \right) - y\left( x \right)} \right) = 0\), then \(y\left( x \right) = mx + c\) is the slant asymptote.

5. Finding the interval of increasing or decreasing.

6. Finding local maximaand local minima.

7. Concavity and finding inflection points.

Step-by-step solution

Sketching a Graph of a function

To sketch a graph we need to follow some procedure. Those are as follows:

1. Finding the domainof the function

2. Intercepts of the function (both \(x\) and \(y\)intercepts)

3. Checking the symmetricity of the function.

4. Finding the asymptotes of the function.

Formula for slant asymptote:

if \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {f\left( x \right) - y\left( x \right)} \right) = 0\), then \(y\left( x \right) = mx + c\) is the slant asymptote.

5. Finding the interval of increasing or decreasing.

6. Finding local maximaand local minima.

7. Concavity and finding inflection points.

Sketch the Graph of given Function

The given function is \(f\left( x \right) = \frac{{{x^4} + 1}}{x}\).

Domain of the function is \(\mathbb{R} - \left\{ 0 \right\}\).

The function has no \(x\) and\(y\)- intercept.

The function odd function so the function is symmetric about origin.

Now we have

\(\mathop {\lim }\limits_{x \to {0^ - }} \frac{{{x^4} + 1}}{x} = - \infty \)and \(\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^4} + 1}}{x} = \infty \).

Hence \(x = 0\) is a vertical asymptote.

Also,

\(\)\(\begin{array}{c}\mathop {\lim }\limits_{x \to \pm \infty } \left( {f\left( x \right) - {x^3}} \right) = \mathop {\lim }\limits_{x \to \pm \infty } \left( {\frac{{{x^4} + 1}}{x} - {x^3}} \right)\\ = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{{x^4} + 1 - {x^4}}}{x}\\ = \mathop {\lim }\limits_{x \to \pm \infty } \frac{1}{x}\\ = 0\end{array}\)

Hence \(y = {x^3}\) is a slant asymptote.

Now \(f'\left( x \right) = \frac{{4{x^4} - {x^4} - 1}}{{{x^2}}} = \frac{{3{x^4} - 1}}{{{x^2}}}\).

\(f'\left( x \right) > 0\), when \(\left| x \right| > \sqrt(4){{\frac{1}{3}}}\). Hence the intervals of increasing are \(\left( { - \infty , - \sqrt(4){{3\frac{1}{{}}}}} \right)\) \(\left( {\sqrt(4){{\frac{1}{3}}},\infty } \right)\).

And \(f'\left( x \right) < 0\), when \(\left| x \right| < \sqrt(4){{\frac{1}{3}}}\). Hence the intervals of decreasing are \(\left( { - \sqrt(4){{\frac{1}{3}}},0} \right)\) and \(\left( {0,\sqrt(4){{\frac{1}{3}}}} \right)\).

We have

\(\begin{array}{c}f''\left( x \right) = \frac{{12{x^5} - 2\left( {3{x^4} - 1} \right)x}}{{{x^4}}}\\ = \frac{{6{x^4} + 2}}{{{x^3}}}\\ = 6x + \frac{2}{{{x^3}}}\end{array}\)

Now \(f''\left( x \right) > 0\) when \(x > 0\). Hence the function is concave up on \(\left( {0,\infty } \right)\).

And the function is concave down on \(\left( { - \infty ,0} \right)\)

There is no inflection point.

There local maxima for this function is \(f\left( { - \sqrt(3){{\frac{1}{3}}}} \right) = \frac{{ - 4}}{{{3^{\frac{5}{4}}}}}\).

The local minima is \(f\left( {\sqrt(4){{\frac{1}{3}}}} \right) = \frac{4}{{{3^{\frac{5}{4}}}}}\).