Question

$$ f(x) = 1 + \frac{7}{x} - \frac{9}{x^{2}} $$
(a) find the vertical asymptote(s). (enter your answers as a comma - separated list.)
$$ x = $$
find the horizontal asymptote(s). (enter your answers as a comma - separated list.)
$$ y = $$
(b) find the interval(s) of increase. (enter your answer using interval notation.)
$$ $$
find the interval(s) of decrease. (enter your answer using interval notation.)
$$ $$
(c) find the local maximum and minimum values.
local maximum value $$ $$
local minimum value $$ $$
(d) find the interval(s) on which ( f ) is concave up. (enter your answer using interval notation.)
$$ $$
find the interval(s) on which ( f ) is concave down. (enter your answer using interval notation.)
$$ $$
find the inflection point.
$$ (x,y) = ( $$ )

Explanation:

Step1: Find vertical asymptote

Vertical asymptote occurs where denominator is \(0\) (after simplifying if needed). For \(y = 1+\frac{7}{x}-\frac{9}{x^{2}}=\frac{x^{2}+7x - 9}{x^{2}}\), denominator \(x^{2}=0\) when \(x = 0\).

Step2: Find horizontal asymptote

For \(y=1+\frac{7}{x}-\frac{9}{x^{2}}\), as \(x
ightarrow\pm\infty\), \(\lim_{x
ightarrow\pm\infty}(1+\frac{7}{x}-\frac{9}{x^{2}})=1\) (since \(\lim_{x
ightarrow\pm\infty}\frac{a}{x^{n}} = 0\) for \(n>0,a
eq0\)).

Step3: Find derivative for increasing/decreasing

First, \(y'=-\frac{7}{x^{2}}+\frac{18}{x^{3}}=\frac{-7x + 18}{x^{3}}\). Set \(y'=0\), then \(-7x+18 = 0\Rightarrow x=\frac{18}{7}\).

• Test intervals:
• For \(x<0\), let \(x=-1\), \(y'=\frac{-7(-1)+18}{(-1)^{3}}=\frac{25}{-1}=-25<0\).
• For \(0
• For \(x>\frac{18}{7}\), let \(x = 3\), \(y'=\frac{-7(3)+18}{3^{3}}=\frac{-3}{27}<0\).

So function is increasing on \((0,\frac{18}{7})\) and decreasing on \((-\infty,0)\cup(\frac{18}{7},\infty)\).

Step4: Find local maxima/minima

Since function changes from increasing to decreasing at \(x = \frac{18}{7}\), \(y(\frac{18}{7})=1+\frac{7}{\frac{18}{7}}-\frac{9}{(\frac{18}{7})^{2}}=1+\frac{49}{18}-\frac{9\times49}{324}=\frac{324 + 882-441}{324}=\frac{765}{324}=\frac{85}{36}\).

Step5: Find second - derivative for concavity

\(y''=\frac{14}{x^{3}}-\frac{54}{x^{4}}=\frac{14x-54}{x^{4}}\). Set \(y'' = 0\), then \(14x-54=0\Rightarrow x=\frac{27}{7}\).

• Test intervals:
• For \(x<\frac{27}{7}\), let \(x = 1\), \(y''=\frac{14(1)-54}{1^{4}}=-40<0\).
• For \(x>\frac{27}{7}\), let \(x = 4\), \(y''=\frac{14(4)-54}{4^{4}}=\frac{56 - 54}{256}=\frac{2}{256}>0\).

Function is concave down on \((-\infty,\frac{27}{7})\) and concave up on \((\frac{27}{7},\infty)\). At \(x=\frac{27}{7}\), \(y(\frac{27}{7})=1+\frac{7}{\frac{27}{7}}-\frac{9}{(\frac{27}{7})^{2}}=1+\frac{49}{27}-\frac{9\times49}{729}=\frac{729+1323 - 441}{729}=\frac{1611}{729}=\frac{179}{81}\).

Answer:

(a) \(x = 0\), \(y = 1\)
(b) Increasing: \((0,\frac{18}{7})\), Decreasing: \((-\infty,0)\cup(\frac{18}{7},\infty)\)
(c) Local maximum value: \(\frac{85}{36}\), Local minimum value: None
(d) Concave up: \((\frac{27}{7},\infty)\), Concave down: \((-\infty,\frac{27}{7})\), Inflection point: \((\frac{27}{7},\frac{179}{81})\)