- #1
kald13
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Homework Statement
I am to explain all intercepts, critical numbers, extrema, inflection points, and asymptotes of the function f(x)=(x-4)/x^3.
2. The attempt at a solution
a) The y-intercept does not exist, as the domain of the function is all real numbers except x=0. Solving the numerator for 0 yields an x-intercept at x=4
b) f'(x) is easily calculated to be (12-2x)/x^4; solving the numerator for 0 yields a critical number of x=6, and x is undefined at 0, yielding our second critical number; Using values from the intervals shows f'(x)>0 on (-infinity, 0] and [0, 6); f'(x)<0 on (6, infinity), indicating what appears to be a local maximum at x=6. Plugging this into f(x):
f(6)=(6-4)/6^3
f(6)=2/216 or 1/108
I note that this is a positive number, where f(2) is negative, meaning the line has crossed the x-axis between 2 and 6 (specifically at x=4, as predicted)
f(2)=(2-4)/2^3
f(2)=-2/8 or -1/4
Inflection points and second derivative information aside, I now come to the point where I am trying to predict the behavior of the function as x approaches positive and negative infinity.
c) I know there is a vertical asymptote at x=0 due to the fact that the limit of the function does not exist as x approaches 0, but what about horizontal asymptotes?
limit as x approaches infinity of (x-4)/x^3
This is indeterminate, infinity divided by infinity, so using L'Hospital's rule I take the derivative of the numerator and denominator for
limit as x approaches infinity of 1/(3x^2)
I factor out constants:
(1/3) limit as x approaches infinity of 1/x^2
And I know that the limit of 1/x is 0. But here's my problem: I already know that my function has crossed the x-axis at 4, so how can there be a horizontal asymptote at y=0?