# How do you find the values of 'a' and 'b' that make f(x) continuous everywhere given #(x^2 - 4)/(x - 2)# if x < 2, #ax^2 - bx + 3# if 2 < x < 3 and #2x - a + b# if #x>=3#?

##### 1 Answer

Jul 25, 2016

I'm guessing you have a piece-wise function:

#f(x) = {((x^2 - 4)/(x-2)",", x < 2),(ax^2 - bx + 3",", 2 < x < 3),(2x - a + b",", x >= 3):}#

The way the function is defined, we already know that *everywhere*.

**discontinuity** (a hole) at

**real** and **finite**, but only if