What is the value of k + c if f(x) is everywhere differentiable. ?

Equation:

f(x) = [ 2kx^2 - x > 3 ]

[ x^3 + cx x <or equal to, 3 ]

Thanks!

1 Answer

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  • Anonymous
    1 decade ago
    Best Answer

    Make the 2 parts of f(x) equal, and the 2 parts of the derivative equal.

    f'(x) = 4kx - 1 for x > 3

    f'(x) = 3x^2 + c for x < 3

    2k(3)^2 - 3 = (3)^3 + c(3)

    and

    4k(3) - 1 = 3(3)^2 + c

    18k - 3 = 27 + 3c

    and

    12k - 1 = 27 + c

    18k = 30 + 3c

    6k = 10 + c

    and

    12k = 28 + c

    c = 6k - 10

    12k = 28 + 6k - 10

    6k = 18

    k = 3

    c = 6(3) - 10

    c = 8

    k + c = 3 + 8 = 11

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