Math (Calculus) (mean value theorem emergency)

Consider the graph of the function f(x)=x^2-x-12

a) Find the equation of the secant line joining the points (-2,-6) and (4,0).
I got the equation of the secant line to be y=x-4

b) Use the Mean Value Theorem to determine a point c in the interval (-2,4) such that the tangent line at c is parallel to the secant line.
I took the derivative of the original question to be f(x)=2x-1 but don't know what to do from there

c) Find the equation of the tangent line through c.

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asked by Ray

  1. b) Using the mean value theorem gave me f'(c)=1 and solving for c gave me 1 as well.

    c) Tangent line y=x-13

    I checked with a graphing calculator and my answers look right

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    posted by Ray

  2. your secant equation is correct
    checking the other parts:
    f '(x) = 2x - 1 ---> you had that

    setting that equal to 1, since the slope of our secant is 1
    2x-1 = 1
    2x = 2
    x = 1
    so all you have to do is sub that back into the original
    f(1) = 1 - 1 - 12 = -12

    So the point where the slope of the secant is equal to the slope of the tangent is (1, -12)
    the equation of the tangent at that point is
    y+12 = 1(x-1)
    y = x - 13 ----> you had that

    good job.

    You seem to be able to take care of the mechanics of the problem, but perhaps understanding what all that means seems a bit fuzzy.

    I suggest you take a look at KhanAcademy video of this, especially the first two parts.
    (These videos done by Sal Khan are extremely well done)

    https://www.khanacademy.org/math/ap-calculus-ab/derivative-applications-ab/mean-value-theorem-ab/v/mean-value-theorem-1

    The second part fits your problem extremely close

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    posted by Reiny

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