For what values of a and b that make the function f continuous everywhere.
f(x)={(x^2-4)/(x-2) if x<2
ax^2-bx-18 if 2<x<3
10x-a+b if x>3
There are three segments of functions, and two possibly discontinuous points, at x=2 and x=3.
Equate the adjacent segments to find the value of a and b that make the segments have the same values at x=2 and x=3.
Finally, check that the limits approaching from each side (2- and 2+, 3- and 3+) of the combined function are identical. They should be if the function is continuous.
To find the values of "a" and "b" that make the function f continuous everywhere, we need to make sure that the function is continuous at the points where the different pieces of the function meet.
The first piece of the function is defined for x < 2. To find the limit as x approaches 2 from the left, we can substitute the value into the function:
lim(x->2-) [(x^2 - 4)/(x - 2)] = lim(x->2-) [(x - 2)(x + 2)/(x - 2)] = lim(x->2-) (x + 2) = 2 + 2 = 4
The second piece of the function is defined for 2 < x < 3. To find the limit as x approaches 2 from the right, we can substitute the value into the function:
lim(x->2+) [ax^2 - bx - 18] = a(2)^2 - b(2) - 18 = 4a - 2b - 18
For the function to be continuous at x = 2, the limits from both sides must be equal. Therefore, we have the equation:
4 = 4a - 2b - 18
Now let's consider the second transition point, x = 3. To ensure continuity, we need to make sure that the limits from both sides are equal:
For x > 3, the limit is:
lim(x->3+) [10x - a + b] = 10(3) - a + b = 30 - a + b
For the function to be continuous at x = 3, the limits from both sides must be equal. Therefore, we have the equation:
30 - a + b = 4a - 2b - 18
Now we have a system of two equations with two unknowns:
1) 4 = 4a - 2b - 18
2) 30 - a + b = 4a - 2b - 18
We can solve this system of equations to find the values of "a" and "b" that make the function continuous everywhere.