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replace f(x) with it "value" into the 2nd equation.... x^4 +2x +1 -4x // x^3 -1
the top and bottom differ by 1 degree so you asymptote with be a slant or oblique at y=x
i saw the words corectly in my head, but my fingers are forming a coop....
I still can't find solution prof. :/
I know, i should simplify equation and plug in 1.But can't simplify.
You can't place 1 for x into the equation as you might with other problems because in \[f(x)= (x^4-2x+1)/(x^3-1)\] placing 1 in for x would result in a zero in the denominator. You have to first get rid of the x^3-1 in the denominator
I know , but i can't get rid of the x^3-1 denominator, i have tried to simplify but couldn't.Can you show how ?
i thought it said as x _. infinty...im going blind i tellya ....BLIND!!
x->....and my hand eye brain coordination is getting shot....
x^3 - 1 is a difference of cubes right? x^3 - 1^3
(x-1)(x^2 +x +1) is its factor
Yeah ,it equals (x-1)(x^2+x+1)
facotr the top and see if x-1 cancles
You need to simplify using the differnce of cubes: \[f(x)= (x^4-2x+1)/((x-1)(x^2+x+1))\] You c\]an then use long division on the numerator (divide the whole top by x-1 to get [f(x)= (x^3-x^2+x-1)/(x^2+x+1)\
yeah that way....lol
The bottom part glitched up but you get the point.
Cheers! I got it ツ
I get: (x-1)(x^3 +x^2+x -1)
Yeah (x-1)(x^3 +x^2+x -1)
2/3 as the limit? as x->1 ?
X^2 would have to be negative because you're dividing by (x-1). Once you multiply (x-1) by x^3 to get rid of the x^4, you would get -x^3
it aint negative....i double checked that ;)
I made Polynomial long division and it's positive , but thanks Athenian
1 | 1 0 0 -2 1 0 1 1 1 -1 ------------- 1 1 1 -1 0 x^3 +x^2 +x -1
yeah, good job Athen....really, that was great work :)