Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 56

Show that $\displaystyle \lim_{x \to \pm \infty} [f(x) -x^2 ] = 0$ if $\displaystyle f(x) = \frac{x^3 + 1}{x}$. Use this fact sketch the graph of $f$.

By using long division,







We can rewrite $f(x)$ as $\displaystyle f(x) = x^2 + \frac{1}{x}$, so..

$\displaystyle \lim_{x \to \pm \infty} [f(x) - x^2] = \frac{1}{x} = \frac{1}{\infty} = 0$

It means that the function is asymptotic to $y = x^2$

Now, using the guidelines of curve sketching

A. Domain,

The domain of the function is $(- \infty, 0) \bigcup (0, \infty)$

B. Intercepts,

Solving for $y$-intercept, when $x = 0$

$\displaystyle y = \frac{0^3 + 1}{0} = \frac{1}{0}$

$y$ intercept does not exist

Solving for $x$-intercept when $y = 0$


$
\begin{equation}
\begin{aligned}

0 =& \frac{x^3 + 1}{x}
\\
\\
0 =& x^3 + 1
\\
\\
x =& -1

\end{aligned}
\end{equation}
$


C. Symmetry,

The function is not symmetric to either $y$ axis or origin by using symmetry test

D. Asymptotes,

For vertical asymptote, we set the denominator equal to 0, that is $x = 0$.

For horizontal asymptote, since $\lim_{x \to \pm \infty} = \pm \infty$, we can say that the function has no horizontal asymptote.

E. Intervals of Increase or Decrease,

If $\displaystyle f(x) = \frac{x^3 + 1}{x}$, then by using Quotient Rule..

$\displaystyle f'(x) = \frac{x (3x^2) - (x^3 + 1)(1) }{(x)^2}= \frac{3x^3 - x^3 - 1}{x^2} = \frac{2x^3 - 1}{x^2}$

when $f'(x) = 0$,


$
\begin{equation}
\begin{aligned}

0 =& 2x^3 - 1
\\
\\
x^3 =& \frac{1}{2}

\end{aligned}
\end{equation}
$


The critical number is,

$x = 0.7937$

Hence, the intervals of increase or decrease are..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
x < 0 & - & \text{decreasing on } (- \infty, 0) \\
0 < x < 0.7937 & - & \text{decreasing on } (0, 0.7937) \\
x > 0.7937 & + & \text{increasing on } (0.7937, \infty)\\
\hline
\end{array}
$

F. Local Maximum and Minimum Values,

Since $f'(x)$ changes from negative to positive at $x = 0.7937, f(0.7937) = 1.89$ is a local minimum.

G. Concavity and Inflection Points

If $\displaystyle f'(x) = \frac{2x^3 - 1}{x^2}$, then by using Quotient Rule..


$
\begin{equation}
\begin{aligned}

f''(x) =& \frac{x^2 (6x^2) - (2x^3 - 1)(2x) }{(x^2)^2}
\\
\\
f''(x) =& \frac{6x^4 - 4x^4 + 2x}{x^4} = \frac{2x^4 + 2x}{x^4} = \frac{2x (x^3 + 1)}{x^4} = \frac{2(x^3 + 1)}{x^3}

\end{aligned}
\end{equation}
$


when $f''(x) = 0$,


$
\begin{equation}
\begin{aligned}

0 =& 2(x^3 + 1)
\\
\\
x^3 + 1 =& 0
\\
\\
x =& -1

\end{aligned}
\end{equation}
$


The inflection point is at $f(-1) = 0$

Hence, the concavity is..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f''(x) & \text{Concavity} \\
x < -1 & + & \text{Upward} \\
-1 < x < 0 & - & \text{Downward} \\
x > 0 & + & \text{Upward}\\
\hline
\end{array}
$

H. Sketch the graph.