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Slope of tangent line

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    find the slope of the tangent line whose g(x)=x^2-4 at point (1,-3)

    2. Relevant equations

    lim f(x+Δχ) -F(c)/ (Δχ)

    3. The attempt at a solution
    g(x)= x^2-4
    G(1+Δχ)= (1+Δχ)^2-4 ==> Δχ^2+2Δχ-3

    lim (Δχ^2+2Δχ-3) - (-3)/(Δχ) = 0
    but in the book the answer is 2 so what could I've done wrong?
     
  2. jcsd
  3. Feb 27, 2009 #2

    Mark44

    Staff: Mentor

    Try to keep your letters straight. You have g and G and f and F and x and X and c. There will come a time when this will get you in trouble.

    Your last expression (which by the way isn't equal to 0), when simplified a bit, is
    [itex][\Delta x ^2 + 2 \Delta x - 3 + 3]/\Delta x[/itex]

    = [itex](\Delta x ^2 + 2 \Delta x)/\Delta x[/itex]

    Factor [itex]\Delta x [/itex] from both terms in the numerator, and cancel with the one in the denominator, then take the limit as [itex]\Delta x[/itex] goes to zero.
     
  4. Feb 27, 2009 #3

    lanedance

    User Avatar
    Homework Helper

    the limt looks ok until you jump to 0, you still have a deltaX on the denominator, which would tend towrds infinty while the top will tend towards zero. so at teh moment you limit is undetermined until you clean it up a bit more...

    so you need to cancel deltaX as much as possible before taking the limit
     
  5. Feb 28, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I assume you mean g(1+ Δx)

    That limit is NOT 0. If you simply set Δx= 0 you get 0/0 so you have to be more careful.(Δχ^2+2Δχ-3) - (-3)= (Δx)^2+ 2Δx so [(Δχ^2+2Δχ-3) - (-3)]/(Δχ) = (Δx^2+ 2Δx)/Δx= Δx + 2. Take the limit, as Δx goes to 0 of Δx+ 2.
     
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