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
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
To determine the values of a and b for which h(x) is continuous, we need to make sure that the two pieces of the function h(x) are equal when x is equal to -1.
First, let's check if h(x) is continuous when x >= -1.
For x >= -1, h(x) is defined as x^2. This is a continuous function for all x.
Next, let's check if h(x) is continuous when x < -1.
When x < -1, h(x) is defined as ax + b.
Given that f(-2) = -1, we can substitute x = -2 into h(x) to get:
-1 = a(-2) + b
Now, to ensure that h(x) is continuous at x = -1, the values of h(x) for when x >= -1 and x < -1 must be equal.
So, we equate the two expressions for h(x):
x^2 = ax + b
Substituting x = -1 into this equation:
(-1)^2 = a(-1) + b
1 = -a + b
Now, we have two equations:
-1 = a(-2) + b
1 = -a + b
From here, we can solve these two equations simultaneously to find the values of a and b.
Multiply the first equation by -1:
1 = 2a - b
Adding this equation to the second equation, we can eliminate b:
(1 = 2a - b) + (1 = -a + b)
2 = a
Substituting this value into either of the equations, let's use the second equation:
1 = -a + b
1 = -2 + b
b = 3
Therefore, the values of a and b that make h(x) continuous are a = 2 and b = 3.