find the value of the constants a and b to ensure the following functions is continuous for all real valuesof x
f(x) = (x^2 - 4) / (x-2) ....... x<2
f(x) = ax^2 - bx + 3 ........ 2
f(x) = 2x - a + b ..............x >orequal 3
Basically find A and B i think its telling us
f(x) = (x^2 - 4) / (x-2) ....... x<2
f(x) = ax^2 - bx + 3 ........ 2
f(x) = 2x - a + b ..............x >orequal 3
Basically find A and B i think its telling us
-
(x²-4)/(x-2) = (x+2)(x-2)/(x-2) = (x+2) when x<2.
So the limit of (x²-4)/(x-2) as x approaches 2 from below is 2+2=4.
So for continuity we require 2²a - 2b + 3 = 4.
So our first equation is 4a - 2b = 1 . . . equation 1
For continuity at x=3 we require 3²a - 3b + 3 = 2(3) - a + b
9a - 3b + 3 = 6 - a + b
10a - 4b = 3 . . . equation 2
If you multiply equation 1 by 2 you have
8a - 4b = 2
Subtract this from equation 2.
2a = 1
a = ½
Substitute this into equation 1.
2 - 2b = 1
2b = 1
b = ½
So the solution is a = b = ½
So the limit of (x²-4)/(x-2) as x approaches 2 from below is 2+2=4.
So for continuity we require 2²a - 2b + 3 = 4.
So our first equation is 4a - 2b = 1 . . . equation 1
For continuity at x=3 we require 3²a - 3b + 3 = 2(3) - a + b
9a - 3b + 3 = 6 - a + b
10a - 4b = 3 . . . equation 2
If you multiply equation 1 by 2 you have
8a - 4b = 2
Subtract this from equation 2.
2a = 1
a = ½
Substitute this into equation 1.
2 - 2b = 1
2b = 1
b = ½
So the solution is a = b = ½