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Find k so that the following function is continuous on any interval:

f(x)= kx
0 (less than or equal to) x (less than) 2

2 (less than or equal to x)

i know the answer is 12 but i don't know how to arrive at that. could you please walk me through the steps? thanks.

  • Math/Calculus - ,

    kx and 3x^2 are continuous where they apply. What you have to to is make sure they both give the same value at x=2, where one functional form changes to the other.

    Thus require that 3x^2 = kx at x = 2.

    3*4 = 2k

    k = 6. The value of the function at x=2 is 12.

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