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calculus

posted by on .

Let f be defined as follows:
h(x)=

-x-3 for x greater than or equal to 1

ax+b for -2<x<1

x(squared) for x less than or equal to -2

Find the values of "a" and "b" that make h(x) continous.

I'm a bit stumped on how to go about solving this one. Help would be greatly appreciated. :)

  • calculus - ,

    look at the transition points, they should be the same.

    when x = 1
    h(x) = -x - 3 = -1-3 = -4
    h(1) = a + b
    so a+b = -4

    h(-2) for ax+b = -2a+b
    h(-2) for x^2 = 4
    so -2a + b = 4

    a+b = -4
    -2a +b = 4
    subtract:
    3a = -8
    a = -8/3,
    back in a+b= -4
    -8/3+b = 4
    b = 20/3

    a = -8/3 , b = 20/3

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