Expert: Paul Klarreich - 10/13/2007

Question
for f(x) =2x^(5/3) - 5x^(4/3)how can I prove whether or not there is a vertical asymptote?  Thank you so much for your help!
Andrew

Answer
Questioner:   Andrew
Category:  Calculus
Private:  No
 
Subject:  Calculus 1
Question:  for f(x) = 2x^(5/3) - 5x^(4/3)how can I prove whether or not there is a vertical asymptote?  Thank you so much for your help!
Andrew
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Hi, Andrew,

I think you want to determine what values of x, if any, make f(x) undefined.  Normally this would mean either:
(1) a fraction with a denominator of zero.
(2) an imaginary number -- a square root of a negative number.  Cube roots don't count -- a cube root of a negative number is negative.
(3) other special things, like  trig or logarithm functions that we don't have here.

Here's a little algebra on this function

f(x) = 2x^(5/3) - 5x^(4/3)

f(x) = 2x^(4/3)x^(1/3) - 5x^(4/3)  <<  5/3 = 4/3 + 1/3

f(x) = x^(4/3)(2x^(1/3) - 5)    << factor x^(4/3)

Now the first factor is no problem.  It is defined and continuous everywhere.  So is the second factor.  So I don't think there are any vertical asymptotes.

Note: Sometimes functions like this have vertical TANGENTS, where f'(x) = 0.  (This one does not.)  This is not the same thing.