- #1
susan__t
- 20
- 0
The question asks to find a value for a and b that makes f continuous everywhere.
f(x)=
[tex]\frac{x - 4}{x-2}[/tex] , where x<2
ax2 - bx + 3 , where 2<x<3
2x - a +b , where x > or = 3
I know that in order for a function to be continuous the limit as x approaches 2 must be equal from the left and right and f(2) must be defined. I don't know how I can possibly make it continuous if x never is equal to 2, then f(x) would not be possible and it would be a removable discontinuity.
There's no answer available at the back of my text or in the solutions manual so it seems a bit fishy to me
Any help solving this would be great (if there even is a solution)
f(x)=
[tex]\frac{x - 4}{x-2}[/tex] , where x<2
ax2 - bx + 3 , where 2<x<3
2x - a +b , where x > or = 3
I know that in order for a function to be continuous the limit as x approaches 2 must be equal from the left and right and f(2) must be defined. I don't know how I can possibly make it continuous if x never is equal to 2, then f(x) would not be possible and it would be a removable discontinuity.
There's no answer available at the back of my text or in the solutions manual so it seems a bit fishy to me
Any help solving this would be great (if there even is a solution)