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I get two different values when calculating the tangent of a curve.

  1. Mar 10, 2014 #1
    1. (a) Find the slope of the tangent line to the curve y=x-x^3 at point (1,0)
    (i) using the 1st definition of a limit: lim(x->a)- (f(x)-f(a))/(x-a)
    (ii) using the 2nd equation of a limit: lim(h->a)- (f(a+h)-f(a))/h​


    3. The attempt at a solution
    In my attempt I got two different values (the same integer value, except one's a negative).

    i) (0-(x-x^3))/(x-1)= (-x+x^3)/(x-1)

    Then I factored an x out of the numerator which gives me a difference of squares.

    (x(x^2-1))/(x-1) = (x(x-1)(x+1))/(x-1)

    Then (x-1) cancels out

    =x(x+1)

    Plug in lim(x->1)

    =1(1+1)=2

    Then I tried the other equatin:

    lim(h->0)- (f(a+h)-f(a))/h

    since 1 is the limit, we put 1 in as the a and it equals:

    ((1+h-(h+3)^3)-(1-1^3))/h

    and 1 minus 1 cubed=0 so we drop the minus sign at the end along with the zero and get:

    =(1+h-(h^3+3h^2+3h+1))/h

    =(1+h-h^3-3h^2-3h-1)/h

    =(-h^3-3h^2-2h)/h

    =(-h^2-3h-2)

    Then we plug in the lim(h->0)

    =(-0^2-3(0)-2

    lim(h->0)=-2

    So as you can see, by doing both equations I get 2 and -2, but only one is tangent to the curve and the other one crosses the graph but never touches it again. I re-did these problems three times to check my calculations and nothing seems off.
     
    Last edited: Mar 10, 2014
  2. jcsd
  3. Mar 10, 2014 #2

    SteamKing

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    You mixed up what should be f(x) and what should be f(1) in the numerator above.

    Again, you have an incorrect substitution in the second form of the derivative. Be more careful in inserting the correct values in the formulas.
     
  4. Mar 10, 2014 #3
    Oh, thanks! That should straighten out the problem.

    The "h+3" was a typo. I ended up with the same result. Appreciate it!
     
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