Question

Define h(x)={x^2 if x>=-1
{ax+b if x<-1
If f(-2)=-1, how do i determine the values of a and b for which h(x) is continuous in the set of real numbers

Answer 1

lim(x -> -1+) x^2 = 1
ax+b = -aa+b when x=-1, so
lim(x -> -1-) = -a+b
So, we need -a+b = 1

f(-2) = -2a+b = -1

So, now we have two equations for a and b, giving us

a=2 and b=3

So,

f(x) = 2x+3 for x > -1

Now the limit from both sides at x = -1 is 1 and f(x) is continuous.

You can see on the graph below that the two curves intersect at x = -1, so you can slip from one to the other there without a break.

http://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2,+y%3D2x%2B3