Describing the function

calc_help123 · #1 $Old$ December 24th, 2008, 01:57 PM

F is the function defined by f(x) = (x^3-x)/(x^3-4x)

A) find the limit of f(x), as x approaches 0
B) FInd the zeroes of f(x)
C)Write an equation for the vertical and horizontal asymptotes of the graph of f(x)
D)Describe the symmetry of f(x)

Krizalid · #2 $Old$ December 24th, 2008, 02:18 PM

Any idea how to tackle every point you want to do?

Jhevon · #3 $Old$ December 24th, 2008, 02:36 PM

As Krizalid said, what have you tried?

to get you started

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Originally Posted by calc_help123 $View Post$

F is the function defined by f(x) = (x^3-x)/(x^3-4x)

A) find the limit of f(x), as x approaches 0

try factorizing the numerator and denominator before taking the limit. see anything?

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B) FInd the zeroes of f(x)

for a rational function, the zeroes occur where the numerator is zero, provided the denominator is not zero at the same time.

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C)Write an equation for the vertical and horizontal asymptotes of the graph of f(x)

for rational functions, vertical asymptotes occur where the function is undefined, this does not include "holes" however. do you know what "holes" are?

as for horizontal asymptotes, find $\lim_{x \to \infty}f(x)$ and $\lim_{x \to - \infty}f(x)$ . if either of those are finite, those are your horizontal asymptotes

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D)Describe the symmetry of f(x)

what types of symmetry are you familiar with? do you know what "odd" and "even" mean when describing functions. is this function either of those? something else?

calc_help123 · #4 $Old$ December 24th, 2008, 04:22 PM

For A I know how to find the limit using (f(x+h)-f(x))/h but I know there was an easier way to find it using the derivative and the derivative i got was
{[-x^3+x]/[(x^3-4x)^2(3x^2-4)]} + (3x-1)/(x^3-4x)
but I dont know where to go.

Also for D I'm not certain but i think if f(x)=f(-x) its even
and if -f(x)=f(-x) its odd?

calc_help123 · #5 $Old$ December 24th, 2008, 04:24 PM

Also for asymptotes I know the vertical ones are when the denominator is 0 but what are the ways to find the horizontal ones analytically. I recall that it has something to do with the powers of the numerator and denominator but I'm not quite sure what it is exactly

Jhevon · #6 $Old$ December 24th, 2008, 04:25 PM

Quote:

Originally Posted by calc_help123 $View Post$

For A I know how to find the limit using (f(x+h)-f(x))/h but I know there was an easier way to find it using the derivative and the derivative i got was
{[-x^3+x]/[(x^3-4x)^2(3x^2-4)]} + (3x-1)/(x^3-4x)
but I dont know where to go.

that is the limit for finding the derivative. the first question has NOTHING to do with that. take my advice for what to do. you simply want $\lim_{x \to 0}f(x)$

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Also for D I'm not certain but i think if f(x)=f(-x) its even
and if -f(x)=f(-x) its odd?

ok, so, do either of those fit this function?

do you know what kind of symmetry an odd function possesses? an even function?

Jhevon · #7 $Old$ December 24th, 2008, 04:33 PM

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Originally Posted by calc_help123 $View Post$

Also for asymptotes I know the vertical ones are when the denominator is 0

correct...provided the numerator is not zero at the same time

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but what are the ways to find the horizontal ones analytically. I recall that it has something to do with the powers of the numerator and denominator but I'm not quite sure what it is exactly

i told you exactly how to find the horizontal asymptotes. the rules you seem to be referring to can be found here (post #2).

OnMyWayToBeAMathProffesor · #8 $Old$ December 31st, 2008, 11:04 PM

so for a) $\lim_{x \to 0}f(x)$ $f(x)=-\infty$ ?

a)the zeros are x=1 and x=-1 this is because $\frac{x^3-x}{x^3-4x}=\frac{x(x^2-1)}{x(x^2-4)}=\frac{x^2-1}{x^2-4}$ so $x^2-1=0$ , $x=+/- 1$ , correct?

b and c were easy, but d is still a little confusing, so far from what i have, it is an even function, correct?

Happy New Year!

mr fantastic · #9 $Old$ January 1st, 2009, 12:06 AM

Quote:

Originally Posted by OnMyWayToBeAMathProffesor $View Post$

so for a) $\lim_{x \to 0}f(x)$ $f(x)=-\infty$ ? Mr F says: This is not correct. Go back to what Jhevon said in post #3 about the common factor.

a)the zeros are x=1 and x=-1 this is because $\frac{x^3-x}{x^3-4x}=\frac{x(x^2-1)}{x(x^2-4)}=\frac{x^2-1}{x^2-4}$ so $x^2-1=0$ , $x=+/- 1$ , correct?

b and c were easy, but d is still a little confusing, so far from what i have, it is an even function, correct?

Happy New Year!

Post #4 contains the ideas needed to answer d).

OnMyWayToBeAMathProffesor · #10 $Old$ January 1st, 2009, 12:21 AM

so it is even because $f(5)=f(-5)$ . thanks a lot!

mr fantastic · #11 $Old$ January 1st, 2009, 12:24 AM

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Originally Posted by OnMyWayToBeAMathProffesor $View Post$

so it is even because $f(5)=f(-5)$ . thanks a lot!

It's even because f(-x) = f(x).

OnMyWayToBeAMathProffesor · #12 $Old$ January 1st, 2009, 04:01 PM

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.

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?

c) $x^3-4x=0$ equals $x(x^2-4)=0$ thus zeros at $x=0$ $x=-2$ $x=2$ correct?

d)even because $f(x)=f(-x)$

thanks again for all your wonderful help.

chabmgph · #13 $Old$ January 1st, 2009, 08:39 PM

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.

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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.

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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.

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d)even because $f(x)=f(-x)$

Correct.

OnMyWayToBeAMathProffesor · #14 $Old$ January 1st, 2009, 08:57 PM

I'm sorry, the horizontal asymptote is y=1 due to the ratio of the numerator and denominator. thanks for catching that.