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Finding the slope of the tangent line using the f(a+h)-f(a)/h form?

  1. Jan 26, 2014 #1
    Find the slope of the tangent line to the curve f(x)=x-x^3 at the point (1,0)

    So, I went through and plugged (a+h) in everywhere I saw "x"

    (a+h)-(a+h)^3

    Factoring:

    (a+h-a^3-3a^2h-3ah^2-h^3)

    I'm really stuck on what to do next...I don't see anything you can cancel/pull out? I know you need to cancel the h in the denominator, but I can't pull an h out because of the "a"s...

    Thank you!
     
  2. jcsd
  3. Jan 26, 2014 #2

    Dick

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    That's just the f(a+h) part. Subtract the f(a) part before you start trying to factor.
     
  4. Jan 26, 2014 #3
    But isn't the f(a) part 0?
     
  5. Jan 26, 2014 #4

    Ray Vickson

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    Yes, in this case. But you have not used that information yet in your calculation of f(a+h).
     
    Last edited: Jan 26, 2014
  6. Jan 26, 2014 #5
    So can I plug in 0 for a? and then factor out h, and then plug in 0 for h?
     
  7. Jan 26, 2014 #6

    Ray Vickson

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    Don't ask; just decide for yourself.
     
  8. Jan 26, 2014 #7
    But you can't because then you're going to get 0/0, right?
     
  9. Jan 26, 2014 #8

    CAF123

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    What do a and h represent in your equation?
     
  10. Jan 26, 2014 #9
    a would represent the "lim as x->a" so it would be the "lim as x->1"...h is what you are trying to factor out.
     
  11. Jan 26, 2014 #10

    Dick

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    If you are saying a=1 in this problem, that would be right.
     
  12. Jan 26, 2014 #11
    But I calculated the slope another way, and the answer is -2. When I do it this way, you'll get 0/0 though if you plug in 1 for a because you would then plug in 0 for the h's and have no number left?
     
  13. Jan 26, 2014 #12

    Dick

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    Show your work. You should be able to cancel an h and then let h->0 and get -2 this way too.
     
  14. Jan 26, 2014 #13
    I put the first steps on how I factored it out above


    Then,

    a+h-a^3-3a^2h-3ah^2-h^3

    Plug in 1:
    h-3h-3h^2-h^3

    pull out an h, I get h(-3-3h-h^2)

    Then you'll get 0/0?
     
  15. Jan 26, 2014 #14

    Dick

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    i) your expression will have an h in the denominator as well. Cancel it with the one in the numerator before taking h->0. ii) Your algebra isn't correct h-3h-3h^2-h^3 isn't equal to h(-3-3h-h^2). Fix it.
     
  16. Jan 26, 2014 #15

    Ray Vickson

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    There is something fundamental that you are not "getting". Here is how it should go.

    (1) First you calculate [f(a+h)-f(a)]/h, where h is NOT zero. Simplify it as much as you can.
    (2) Then (and only then), let h → 0 to see what is the limit.

    You never actually set h = 0. Taking h → 0 is not the same as setting h = 0.
     
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