Thread: Describing the function
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Old January 1st, 2009, 08:39 PM
chabmgph chabmgph is offline
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Quote:
Originally Posted by OnMyWayToBeAMathProffesor View Post
I see my mistake, so let me just show all my answer to makes sure the others are correct.

a) as \lim_{x \to 0}f(x) f(x)=\frac{1}{4}. I got this answer after i followed what Jhevon said and factored out certain terms
Looks good.

Quote:
b)\frac{x^3-x}{x^3-4x}=\frac{x(x^2-1)}{x(x^2-4)}=\frac{x^2-1}{x^2-4} so x^2-1=0 thus x=1, x=-1 correct?
Yes. Since x = 0 is not in the domain, that zero is thrown out.

Quote:
c) x^3-4x=0 equals x(x^2-4)=0 thus zeros at x=0 x=-2 x=2 correct?
Like Jhevon had mentioned, a "hole" is not an asymptote. So only x = 2, x = -2 are the vertical asymptotes. And note that the question also ask for horizontal asymptote.

Quote:
d)even because f(x)=f(-x)
Correct.
Last edited by chabmgph; January 1st, 2009 at 08:44 PM. Reason: typo
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