Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)
x^2-4/x-2 x<2
f(x)= ax^2-bx+5 2 is less than or equalx<3
2x-a+b x is greater than or equal to 3
a=
b=
a=1/2
b=1/2
To find the values of a and b that make f continuous everywhere, we need to ensure that the function is continuous at the points where the different cases meet.
At x = 2, we need to make sure that the two cases have the same value for f(x) in order for the function to be continuous. So we set the two expressions equal to each other and solve for a and b:
x^2 - 4 / x - 2 = ax^2 - bx + 5
To simplify this equation, we start by multiplying both sides by (x - 2):
(x - 2)(x^2 - 4) = (ax^2 - bx + 5)(x - 2)
Expanding both sides of the equation gives us:
x^3 - 2x^2 - 4x + 8 = ax^3 - 2ax^2 - bx^2 + 2bx + 5x - 10
Combining like terms gives us:
0 = (a - 1)x^3 + (2a - 2 - b)x^2 + (5 + 2b - 4)x + (-10 - 8)
To ensure continuity at x = 2, the coefficient of x^2, x, and the constant term on the right-hand side of the equation must all be zero.
First, let's set the coefficient of x^2 to zero:
2a - 2 - b = 0
Next, let's set the coefficient of x to zero:
5 + 2b - 4 = 0
Lastly, let's set the constant term to zero:
-10 - 8 = 0
Solving these three equations will give us the values of a and b that make f continuous everywhere.
Let's solve them one by one:
1. 2a - 2 - b = 0
Rearranging this equation, we have:
2a - b = 2
2. 5 + 2b - 4 = 0
Rearranging this equation, we have:
2b = -1
b = -1/2
3. -10 - 8 = 0
Rearranging this equation, we have:
-18 = 0
The third equation, -18 = 0, is not possible, meaning this system of equations is inconsistent. Therefore, there are no values of a and b that make f continuous everywhere.
Hence, there are no valid solutions for a and b.