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Posted by on Friday, November 9, 2012 at 10:51am.

It is given that a differentiable function f(x)=4x^3+kx^2-36x-15 (k is a constant) is decreasing on -3/2<x<2 and increasing on 2<x<5. Find the value of k and the turning point(s) of the curve y=f(x).

  • Calculus - , Friday, November 9, 2012 at 11:28am

    you know that
    f'(x) has a root at x=2.
    It is negative for -3/2 < x < 2
    It is positive for 2<x<5
    Nothing is said about any domain outside (-3/2,5)

    But, since you want a cubic f(x), f'(x) must be positive for x < -3/2, so it has another root at x = -3/2

    f'(x) = a(2x+3)(x-2) = a(2x^2 - x - 6)
    now we can get somewhere
    f(x) = a(2/3 x^3 - 1/2 x^2 - 6x) + c
    or, changing a and c (which are arbitrary) and clearing fractions,
    f(x) = a(4x^3 - 3x^2 - 36x) + c

    Now, we want f(x) as described above, so a = 1 and
    f(x) = 4x^3 - 3x^2 - 36x - 15

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