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AP Calculous

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let f be the function defined by
|x-1|+2 for X<1
ax^2-Bx, for X>or equal to 1. where a and b are constants

a)if a=2 and b=3 is f continious for all x? justify your answer

b)describe all the values of a and b for which f is a continious function

c) For what values of a and b is f both continious and differentiable?

  • AP Calculous - ,

    f(x) is both functions.. I don't know why the system didn't let me keep the spaces to show that

  • AP Calculous - ,

    First off, stop misspelling "calculus".

    Now, if a=2 and b=3, we have
    f(x) = 2x^2 - 3x
    f(1) = -1

    But lim x->1- = |1-1| + 2 = 2

    So, f is not continuous at x=1.

    If f is to be continuous, it needs to be continuous at x=1. It is continuous everywhere else already.

    So, ax^2 - bx must = 2 at x=1
    a - b = 2

    So, there are any number of parabolas which will make f continuous at x=1.

    6x^2 - 4x
    -3x^2 + 5x

    Now, for f to be also differentiable, the slopes must match at x=1

    The slope of |x-1| is -1 when approaching from the left.

    ax^2 - (a-2)x must also have slope -1 at x=1

    f'(x) = 2ax - (a-2)
    f'(1) = 2a - a + 2 = -1
    a = -3


    f(x) = |x-1| + 2 for x<1
    f(x) = -3x^2 + 5x for x>=1

    is both continuous and differentiable everywhere.

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