$$ f(x) = 1 + \\frac{7}{x} - \\frac{9}{x^{2}} $$\n(a) find the vertical asymptote(s). (enter your answers as…

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

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

Answer

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\rightarrow\pm\infty), (\lim_{x\rightarrow\pm\infty}(1+\frac{7}{x}-\frac{9}{x^{2}})=1) (since (\lim_{x\rightarrow\pm\infty}\frac{a}{x^{n}} = 0) for (n>0,a\neq0)).

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<x<\frac{18}{7}), let (x = 1), (y'=\frac{-7(1)+18}{1^{3}}=11>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}))