Question

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`?

Answer 1

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 `x in (-oo,2) uu (2,3) uu [3,oo)`. That is, `x ne 2` (but `x` can be `3`). However, because of that, there is no way for `f(x)` to be continuous everywhere.

`f(x)` would have a discontinuity (a hole) at `x = 2`, since even though you can cancel out `x - 2`, you still have `2 - 2 = 0`.

`:.` One might say that `f(x)` is continuous when `a,b` are real and finite, but only if `x ne 2`. So strictly speaking, `f(x)` cannot be continuous everywhere.