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anonymous

  • 5 years ago

f(x)=x^4+2x+1 lim x goes to 1 (f(x)-4x)/(x^3-1)=?

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  1. amistre64
    • 5 years ago
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    replace f(x) with it "value" into the 2nd equation.... x^4 +2x +1 -4x // x^3 -1

  2. amistre64
    • 5 years ago
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    the top and bottom differ by 1 degree so you asymptote with be a slant or oblique at y=x

  3. amistre64
    • 5 years ago
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    i saw the words corectly in my head, but my fingers are forming a coop....

  4. anonymous
    • 5 years ago
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    I still can't find solution prof. :/

  5. anonymous
    • 5 years ago
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    I know, i should simplify equation and plug in 1.But can't simplify.

  6. anonymous
    • 5 years ago
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    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

  7. anonymous
    • 5 years ago
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    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 ?

  8. amistre64
    • 5 years ago
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    i thought it said as x _. infinty...im going blind i tellya ....BLIND!!

  9. amistre64
    • 5 years ago
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    x->....and my hand eye brain coordination is getting shot....

  10. amistre64
    • 5 years ago
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    x^3 - 1 is a difference of cubes right? x^3 - 1^3

  11. amistre64
    • 5 years ago
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    (x-1)(x^2 +x +1) is its factor

  12. anonymous
    • 5 years ago
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    Yeah ,it equals (x-1)(x^2+x+1)

  13. amistre64
    • 5 years ago
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    facotr the top and see if x-1 cancles

  14. anonymous
    • 5 years ago
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    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)\

  15. amistre64
    • 5 years ago
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    yeah that way....lol

  16. anonymous
    • 5 years ago
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    The bottom part glitched up but you get the point.

  17. anonymous
    • 5 years ago
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    Cheers! I got it ツ

  18. amistre64
    • 5 years ago
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    I get: (x-1)(x^3 +x^2+x -1)

  19. anonymous
    • 5 years ago
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    Yeah (x-1)(x^3 +x^2+x -1)

  20. amistre64
    • 5 years ago
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    2/3 as the limit? as x->1 ?

  21. anonymous
    • 5 years ago
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    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

  22. anonymous
    • 5 years ago
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    Yep 2/3,

  23. amistre64
    • 5 years ago
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    it aint negative....i double checked that ;)

  24. anonymous
    • 5 years ago
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    I made Polynomial long division and it's positive , but thanks Athenian

  25. amistre64
    • 5 years ago
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    1 | 1 0 0 -2 1 0 1 1 1 -1 ------------- 1 1 1 -1 0 x^3 +x^2 +x -1

  26. amistre64
    • 5 years ago
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    yeah, good job Athen....really, that was great work :)

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