anonymous
  • anonymous
use f(x)= (3x+1)/ (Absolute value of x+2) to find lim x> infinity, lim x> - infinity, and all horizontal asymptotes??
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\lim_{x \rightarrow \infty}\frac{ 3x+1 }{ \left| x+2 \right|}=\lim_{x \rightarrow \infty}\frac{ 3x+1 }{ x+2 }=3\]\[\lim_{x \rightarrow -\infty}\frac{ 3x+1 }{ \left| x+2 \right|}=\lim_{x \rightarrow -\infty}\frac{ 3x+1 }{-(x+2) }=-3\]Those are precisely the horizontal asymptotes: y=3 and y=-3 There is a vertical asymptote also in x=-2, what makes denominator=0
Hero
  • Hero
\[f(x) = \frac{3x + 1}{|x+2|}\] Separate into two \[f(x)=\frac{3x + 1}{x + 2}\] \[f(x) = -\frac{3x + 1}{x + 2}\] Re-write both as: \[f(x) = 3 - \frac{5}{x + 2}\] \[f(x) = \frac{5}{x + 2} - 3\] To find the horizontal asymptotes, let y = f(x): \[y = 3 - \frac{5}{x + 2}\] \[y = \frac{5}{x + 2} - 3\] Swap x and y: \[x = 3 - \frac{5}{y + 2}\] \[x = \frac{5}{y + 2} - 3\] Then isolate y in each: \[y = \frac{5}{3-x} - 2\] \[y = \frac{5}{x + 3} -2\]
Hero
  • Hero
In the first case we notice that \[x \ne 3\] In the second case we notice that \[x \ne -3\] The domain of the inverse function is the range of the original function. So 3 and -3 are the horizontal asymptotes of the original function.

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anonymous
  • anonymous
so whatever f(x) = will be the horizontal retricemptote??
anonymous
  • anonymous
retricemptote*?
Hero
  • Hero
@CarlosGP showed you how to find the horizontal asymptote by using limits. I showed you how to find it by using the inverse. The horizontal asymptotes occur where \[f(x) \ne 3\]\[f(x) \ne -3\]
Hero
  • Hero
In the first case we notice that \[x \ne 3\] In the second case we notice that \[x \ne -3\] The domain of the inverse function is the range of the original function. So 3 and -3 are the horizontal asymptotes of the original function.

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