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Homework Help: Use calculus to find the slope of the tangent line

  1. Sep 24, 2006 #1
    I'm given f(x) = x^2 and g(x) = -x^2 + 2x - 5

    And it says

    Let A (a, f(a)) be the point of tangency on f(x) and B (b, g(b)) be the point of tangency on g(x). Use calculus to find the slope of the tangent line at each point. What must be true about these two slope expressions?

    I know we need to use the derivatives, but do I use the original form of the derivative f(x + deltax) - f(x)/delta x or the alternate form f(x) - f(c)/x-c?

    I would think to use the alternate form, because it gives us points of tangency... but when I try to use the alternate form on g(x), I can't seem to get an answer.

    For f(x): x^2 - a^2 / (x-a)
    (x+a)(x-a) / (x-a)
    = x+a

    I think I did that right. Now for g(x):
    -x^2+2x-5+b^2-2b+5/ (x-b)
    -x^2 + 2x + b^2 - 2b/ (x-b)

    And I'm basically stuck on what to do from here. Any ideas? Or am I supposed to be using the other derivative equation?
     
  2. jcsd
  3. Sep 24, 2006 #2
    f'(c) = lim{x->c} [f(x) - f(c)]/(x - c)

    for f(x) = x^2 ...

    f'(a) = lim{x->a} [x^2 - a^2]/(x - a)
    f'(a) = lim{x->a} [(x + a)(x - a)]/(x - a)
    f'(a) = lim{x->a} (x + a) = 2a


    for g(x) = -x^2 + 2x - 5 ...

    g'(b) = lim{x->b} [(-x^2 + 2x - 5) - (-b^2 + 2b - 5)]/(x - b)
    g'(b) = lim{x->b} [b^2 - 2b + 2x - x^2]/(x - b)
    g'(b) = lim{x->b} [b^2 - x^2 - 2(b - x)]/(x - b)
    g'(b) = lim{x->b} [(b - x)(b + x) - 2(b - x)]/(x - b)
    g'(b) = lim{x->b} [(b - x)[(b + x) - 2]]/(x - b)
    g'(b) = lim{x->b} [2 - (b + x)] = 2 - 2b
     
  4. Sep 24, 2006 #3
    Ah.. thank you, I get it now. It doesn't matter if you use the alternate form or not, heh.

    The next part says use the coordinates of A and B to write an expression for the slope AB in terms of a and b.

    So...

    f(a) - f(b)/ (a - b)

    I think that's what I'm supposed to do. The next part says to substitute things in so there's only one term to solve for. But I don't see what I can substitute.
     
  5. Sep 24, 2006 #4
    something is missing from your original post ... are the two tangent lines to f and g supposed to have some relationship (like being parallel, maybe)?
     
  6. Sep 24, 2006 #5
    Oh, sorry. The graph is supposed to have one tangent line that hits both f(x) and g(x).
     
  7. Sep 24, 2006 #6
    well ... the slopes have to be the same then, don't they?

    2a = 2 - 2b
    2a + 2b = 2
    a + b = 1

    now you have the required relationship between a and b.
     
  8. Sep 24, 2006 #7
    Oh! Wow, that answer now, seems so blantant ~_~

    So, a = 1-b

    Substitute that into the slope equation:

    (1-b) ^2 - (-b^2 + 2b -5)
    1 - 2b + b^2 + b^2 - 2b + 5
    = 2b^2 - 4b + 6

    Oookay. Then I need to solve for the equation to get b. The thing is I get two answers so how do I know which one is correct?

    To find f(b), I plug it back into 2-2b, right?
     
    Last edited: Sep 24, 2006
  9. Sep 24, 2006 #8
    who knows? ... maybe they're both good, and there are two different tangent lines that work.

    check out each solution, one at a time.
     
  10. Sep 24, 2006 #9
    Aw man, I just realized I did my math wrong.

    For the slope equation, it's supposed to be

    2b^2 - 4b + 6 / (1-2b)

    How would I solve for b if I'm not given that it equals anything?
     
  11. Sep 24, 2006 #10
    [f(1-b) - f(b)]/[(1-b) - b] = 2 - 2b

    [(1-b)^2 - b^2]/(1 - 2b) = 2 - 2b

    (1 - 2b)/(1 - 2b) = 2 - 2b

    1 = 2 - 2b

    2b = 1

    b = 1/2

    since a + b = 1, a = 1/2
     
  12. Sep 24, 2006 #11
    Why would it go to b^2? o_o;

    It's supposed to be g(b), and g(x) is -x^2 + 2x -5.

    lol, probably from my careless typing <.<

    *EDIT*

    So, I have 2b^2 - 4b + 6 / (1-2b) = 2-2b
    2b^2 - 4b + 6 = (2-2b)(1-2b)
    2b^2 - 4b + 6 = 2 - 6b + 4b^2
    2b^2 - 2b -4 = 0

    Factor that out to get

    (b - 2) (2b + 2)

    b = 2, -1

    And then plug those two points in the equation a+b = 1

    Solve for a, and then plug those values of a inside f(a) = a^2?
     
    Last edited: Sep 24, 2006
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