Posted by **Anonymous** on Saturday, September 24, 2011 at 2:54pm.

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 -
**Reiny**, Saturday, September 24, 2011 at 3:16pm
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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