anonymous
  • anonymous
17x/(x^4 + 1)^1/4...what are the horizontal asymptotes. only the denominator is raised to 1/4
Mathematics
schrodinger
  • schrodinger
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Mertsj
  • Mertsj
Since the lead exponents are equal, the horizontal asymptote is the ratio of the coefficients of the leading terms.
anonymous
  • anonymous
what mertsj said, with the sophistication that as x gets large \[\sqrt[4]{x^4+1}\] behaves like \[\sqrt[4]{x}=x\] so you might as well pretend you have \[\frac{17x}{x}=17\]
anonymous
  • anonymous
oh i see. how do i find y2

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Mertsj
  • Mertsj
What do you mean, y2?
anonymous
  • anonymous
y1= and y2= where y1>y2
Mertsj
  • Mertsj
And don't forget there are two fourth roots of one so there are two horizontal asymptotes, y = 17 and y=-17
anonymous
  • anonymous
thats what i meant, thanks
anonymous
  • anonymous
\[y=\frac{17x}{(x^4+1)^{\frac{1}{4}}}\] Rearrange to get x in term of y: \[y(x^4+1)^{\frac{1}{4}} =17x\] \[y^4(x^4+1) = 83521x^4\] \[\frac{(x^4+1)}{x^4} = \frac{83521}{y^4}\] \[\frac{1}{x^4} = \frac{83521}{y^4}-1=\frac{83521 - y^4}{y^4}\] So \[x =\pm \left(\frac{y^4}{83521 - y^4}\right)^{\frac{1}{4}}\] So the value of y where the denominator is zero is the value at which the horizontal asymtote occurs... \[y^4 = 83521\] so \[y = \pm 17\]
anonymous
  • anonymous
that's how you prove it the long way round I think...

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