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Last question i swear

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Use the limit definition to find the derivative at the indicated point.
    f(x)= -x^3+4x^2, at (-1,5)


    2. Relevant equations
    f(a)=lim x->1 (f(x)-f(a))/(x-(a))


    3. The attempt at a solution

    ok so i got:
    f'(-1)=(lim x->1) ((-x^3)+(4x^2)- f(-1))/(x-(-1))

    f(-1)= (lim x->1) ((-x^3)+(4x^2)- f(-1))/(x+1)

    my only question is .. what would f(-1) be..?

    is this right?:

    (1^3)-(4^2)
     
    Last edited: Mar 30, 2008
  2. jcsd
  3. Mar 30, 2008 #2
    No, incorrect. f(-1) is your function evaluated at -1. What is your function?
     
    Last edited: Mar 30, 2008
  4. Mar 30, 2008 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You mean f '(-1) or df(-1)/dx, not f(-1), on the left.

    Since f(x)= -x3+ 4x2, f(-1)= -(-1)3+ 4(-1)2. What is that? (the "4" is NOT squared, only x= -1.)
     
  5. Mar 30, 2008 #4
    ok so ultimately it would be..
    -1+ 4
    3..?
     
  6. Mar 30, 2008 #5
    Yes, so now put it all together and what is your limit?
     
  7. Mar 30, 2008 #6
    ok

    lim x->1 ((-x^3)+(4x^2)-3)/(x+1)

    but idk what to do next cuz if i input 1 for all the x's then id get
    0/2
     
  8. Mar 30, 2008 #7
    i was thinking i could just factor out the
    -x^3+4x^2-3 but when i do i dont get a (x+1) for one of my factors...
     
  9. Mar 30, 2008 #8
    There is no such :''Last question"
     
  10. Mar 30, 2008 #9
  11. Mar 30, 2008 #10
    so does neone know how to help me from here?

    (-x^3 + 4x^2 -3)/ (x+1)

    becuz i dont think thats the final answer.

    also i tried to make -1 one of the zeros..by using synthetic division
    and that doesnt make a zero so i dont know how to cancel out the denominator...-.-

    eeek plz help someone :)
     
  12. Mar 30, 2008 #11
    If you're trying to find the derivative, your equation is wrong.
    [itex]\frac{df}{dx}={\lim_{h\rightarrow{0}}}\frac{f(x+h)-f(x)}{h}[/itex]
     
  13. Mar 30, 2008 #12
    but when im givin a point then theres another formula you can use right?
     
  14. Mar 30, 2008 #13
    Find the derivative using that formula(there is an incredibly easy shortcut, but you were told to use the formula) and then plug in your point.
     
  15. Mar 30, 2008 #14
    ok ill try
    but for the f(x+h)

    would it be
    -x^3-h^3+4x^2+4h^2?
    and the for the -f(x)
    of course it would be
    -(-x^3 + 4x^2)

    then you would get -h^3+4h^2/(h)
    you could bring out an h
    youd have

    h(-h^2+4h)/h
    which would cancel out the 2 h's
    and youd be left with
    -h^2+4h

    and then plug them in?
     
  16. Mar 30, 2008 #15
    No. You plug (x+h) in for x when you are finding f(x+h).
     
  17. Mar 30, 2008 #16
    oh so it would be -(x+h)^3 + 4(x+h)^2 ?
     
  18. Mar 30, 2008 #17
    yes. That would be f(x+h)
     
  19. Mar 30, 2008 #18

    HallsofIvy

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    Science Advisor

    No, that is NOT correct. Your function is f(x)= -x3+ 4x2.
    f(-1)= -(-1)3+ 4(-1)2= 1+ 4= 5.
     
  20. Mar 30, 2008 #19
    so what do i do then???
    im so confused:?:
    like what equation do i need to use..and everything
     
  21. Mar 30, 2008 #20
    ok so its not 3 its 5...
    that changes everything...

    -x^3 +4x^2-5/(x+1)

    then i can factor out an x+1
    which the top and bottom x+1's cancel
    leaving you with
    -x^2+5x-5...

    is this right so far? so i can go ahead and plug in for the x's?
     
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