find the constants a and b so that the function is continuous on the entire real line. F(x)= 3, x≤-1
ax+b, -1<x<2
-3,x≥2 Help???
To find the constants a and b so that the function is continuous on the entire real line, we need to make sure that the function values match at the points of transition.
First, let's consider the transition from x ≤ -1 to -1 < x < 2. At x = -1, the function should have the same value from both sides. So, we set the constant function to be equal to the linear function at x = -1:
3 = a(-1) + b
Next, let's consider the transition from -1 < x < 2 to x ≥ 2. At x = 2, the function should have the same value from both sides. So, we set the linear function to be equal to the constant function at x = 2:
a(2) + b = -3
Now, we have a system of two equations with two unknowns (a and b). We can solve this system of equations to find the values of a and b.
First equation: 3 = a(-1) + b
Rearranging, we get: -a + b = 3 ----(1)
Second equation: a(2) + b = -3 ----(2)
To solve the system of equations, we can either use substitution or elimination method. Let's use the elimination method.
Multiply equation (1) by 2 to eliminate 'a':
-2a + 2b = 6 ----(3)
Now, we can subtract equation (3) from equation (2) to eliminate 'b':
(a(2) + b) - (-2a + 2b) = -3 - 6
This simplifies to:
3a = -9
Divide both sides by 3:
a = -3
Substitute the value of a into equation (1):
-(-3) + b = 3
3 + b = 3
b = 0
So, the constants a and b that make the function continuous on the entire real line are a = -3 and b = 0.