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Calculus help

  1. Oct 9, 2006 #1
    can you guys help me how to do this problem?

    f(x) = 3x^2 - 1 (3,8)
    Find slope and tangent line,

    I have answer 6, is that right? and can you guys show the step to get the answer, thanks =) (tangent line i already know how to do)

    oh and this one,

    y = x^2 - 4
    Find instantaneous rate of change [3,5]

    (if there was one point, i could do it, but i dont know how to do with 2 point,) can you show steps too,


    thanks alot
     
  2. jcsd
  3. Oct 9, 2006 #2
    For the first one, you find the derivative first.
    [tex]f(x) = 3x^2-1 \ \longrightarrow \ f'(x) = 6x [/tex] then you substitude for x. That doesn't give 6. Now that you have the slope you can find the tangent line.

    For the second line, I think that would be the gradient of the secant line through those points.
     
  4. Oct 9, 2006 #3
    for this one,
    y = x^2 - 4
    Find instantaneous rate of change [3,5]

    i dont really get what you are saying, sorry,

    so how do I solve it

    urg
     
  5. Oct 9, 2006 #4
    @3, y = 5
    @5, y = 11

    Now you have 2 points, P1(3,5) & P2(5,11).
    What is the gradient of the line that passes through these points ?
     
  6. Oct 9, 2006 #5
    @5 , do you mean 21 ?

    and i dont think i learn gradient yet,

    in class, when we finding instantaneous rate of change, we use limit x approaching a number, then solve it,

    is there another way to find the instantaneous rate of change in this problem?
     
  7. Oct 9, 2006 #6
    yeah i mean 21, lol sry. Gradient = slope
     
  8. Oct 9, 2006 #7
    so the instant.. rate of change is 8 ?

    aha
     
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