Single Variable Calculus, Chapter 4, 4.5, Section 4.5, Problem 52

Use the guidelines of curve sketching to sketch the curve $\displaystyle y = \frac{(x + 1 )^3}{(x - 1)^2}$ then find the equation of slant asymptote.

A. Domain. We know that $f(x)$ is a rational function that is defined for all values of $x$ except for the value that will make the function undefined. In our case, it's $x = 1$. Therefore, the domain is $(- \infty, 1) \bigcup (1, \infty)$

B. Intercepts. Solving for $y$ intercept, when $x = 0$,

$\displaystyle y = \frac{(0 + 1)^3}{(0 - 1)^2} = 1$,

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


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


C. Symmetry. The function is not symmetric to either $y$ axis and origin by using symmetry test.

D. Asymptote. For vertical asymptote, we set $y$-axis on origin by using symmetry test.


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


For horizontal asymptote, since $lim_{x \to \pm \infty} f(x) = \pm \infty$, the function has no horizontal asymptote.

For slant asymptote, by using long division,







Thus, we can rewrite $f(x)$ as $\displaystyle y = x + 5 + \frac{12 x - 4}{x^2 - 2x + 1}$, so...

So, $\displaystyle \lim_{x \to \pm \infty} f(x) - (x + 5) = \frac{12x - 4}{x^2 - 2x + 1} = \frac{\displaystyle \frac{21}{x} - \frac{4}{x^2}}{\displaystyle 1 - \frac{2}{x} + \frac{1}{x^2}} = 0$

Therefore, the equation of slant asymptote is $y = x + 5$.

E. Intervals of increase or decrease,

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


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{(x - 1)^2 (3(x + 1)^2) - (x + 1)^3 (2 (x - 1)) }{[(x - 1)^2]^2}
\\
\\
f'(x) =& \frac{(x - 1)^2 (x + 1)^2 [3 (x - 1) - 2 (x + 1)]}{(x - 1)^4} = \frac{(x + 1)^2 [x - 5]}{(x - 1)^3}

\end{aligned}
\end{equation}
$


When $f'(x) = 0$, then


$
\begin{equation}
\begin{aligned}

0 =& \frac{(x + 1)^2 [x - 5 ]}{( x - 1)^3}

\end{aligned}
\end{equation}
$


We have,

$(x + 1)^2 = 0$ and $x - 5 = 0$

The critical numbers are

$x = 1$ and $x = 5$

Hence, the intervals of increase and decrease are..

$
\begin{array}{|c|c|c|}
\hline\\
\text{Interval} & f'(x) & f \\
x < -1 & + & \text{increasing on } (- \infty, -1) \\
-1 < x < 1 & + & \text{increasing on} (-1, 1) \\
1 < x < 5 & - & \text{decreasing on (1, 5)} \\
x > 5 & + & \text{increasing on } (5, \infty) \\
\hline

\end{array}
$


F. Local Maximum and Minimum Values

Since $f'(x)$ changes from negative to positive at $x = 5, f(5) = 13.5$ is a local minimum. The function has no local minimum since the function is not defined on $x = 1$.

G. Concavity and inflection point

If $f'(x) = \displaystyle \frac{(x + 1)^2 [x - 5]}{(x - 1)^3}$, then


$
\begin{equation}
\begin{aligned}

f''(x) =& \frac{(x - 1)^3 [2(x + 1) (x - 5) + (x + 1)^2] - [(x + 1)^2 (x - 5)] (3(x - 1)^2) }{[(x - 1)^3]^2}

\end{aligned}
\end{equation}
$


Which can be simplified as,

$\displaystyle f''(x) = \frac{24(x + 1)}{(x - 1)^4}$

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

$\displaystyle 0 = 24(x + 1)$

$x = -1$

So 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{Downward} \\
-1 < x < 1 & + & \text{Upward}\\
x > 1 & + & \text{Downward}\\
\hline
\end{array}
$

H. Sketch the graph.