anonymous
  • anonymous
f(x)=x^4+2x+1 lim x goes to 1 (f(x)-4x)/(x^3-1)=?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
replace f(x) with it "value" into the 2nd equation.... x^4 +2x +1 -4x // x^3 -1
amistre64
  • amistre64
the top and bottom differ by 1 degree so you asymptote with be a slant or oblique at y=x
amistre64
  • amistre64
i saw the words corectly in my head, but my fingers are forming a coop....

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I still can't find solution prof. :/
anonymous
  • anonymous
I know, i should simplify equation and plug in 1.But can't simplify.
anonymous
  • anonymous
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
anonymous
  • anonymous
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 ?
amistre64
  • amistre64
i thought it said as x _. infinty...im going blind i tellya ....BLIND!!
amistre64
  • amistre64
x->....and my hand eye brain coordination is getting shot....
amistre64
  • amistre64
x^3 - 1 is a difference of cubes right? x^3 - 1^3
amistre64
  • amistre64
(x-1)(x^2 +x +1) is its factor
anonymous
  • anonymous
Yeah ,it equals (x-1)(x^2+x+1)
amistre64
  • amistre64
facotr the top and see if x-1 cancles
anonymous
  • anonymous
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)\
amistre64
  • amistre64
yeah that way....lol
anonymous
  • anonymous
The bottom part glitched up but you get the point.
anonymous
  • anonymous
Cheers! I got it ツ
amistre64
  • amistre64
I get: (x-1)(x^3 +x^2+x -1)
anonymous
  • anonymous
Yeah (x-1)(x^3 +x^2+x -1)
amistre64
  • amistre64
2/3 as the limit? as x->1 ?
anonymous
  • anonymous
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
anonymous
  • anonymous
Yep 2/3,
amistre64
  • amistre64
it aint negative....i double checked that ;)
anonymous
  • anonymous
I made Polynomial long division and it's positive , but thanks Athenian
amistre64
  • amistre64
1 | 1 0 0 -2 1 0 1 1 1 -1 ------------- 1 1 1 -1 0 x^3 +x^2 +x -1
amistre64
  • amistre64
yeah, good job Athen....really, that was great work :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.