The function is continuous everywhere if it is continuous at every point of its domain. In other words, it is continuous at every point where it is defined.
The function is continuous at a point `x = x_0` if the limits of this function as `x` approaches `x_0` from both left and right are equal and equal to the value of the function at this point, `f(x_0)` .
Since the given function is piece-wise and defined on three separate intervals, the possible points of discontinuity are the ends of these intervals. In other words, the points x = 2 and x = 3.
Consider the function at the interval `(-oo, 2)` . The "piece" of the function on this interval is a rational function
`f(x) = (x^2 - 4)/(x -2)` . This function is not defined at x = 2, because this value of x makes the denominator 0. The equivalent expression for this function is
`f(x) = (x^2 - 4)(x - 2) = ((x-2)(x+2))/(x - 2) = x+2` , `x ne 2` . The graph of this looks like a straight line with a hole at the point (2, 4).
Since x = 2 is where the interval ends, in order to make the entire function continuous the second "piece" of the function has to be such that its graph will "fill in" the hole. This means
`f(x) = ax^2 - bx + 3` has to equal 4 at x = 2.
`f(2) = a*2^2 - b*2 + 3 = 4`
4a - 2b = 1.
The "pieces" of the function on two other intervals are a quadratic function `ax^2 - bx + 3` and a linear function 2x - a + b, and they are continuous wherever they are defined. For the entire function to be continuous at x = 3, the values of these two "pieces" have to be equal:
`a*3^2 - b*3 + 3 = 2*3 - a + b`
`9a - 3b + 3 = 6 - a + b`
`10a - 4b-3 = 0`
Together with the equation 4a - 2b = 1 obtained above, this can be solved for a and b as a system:
4a - 2b = 1
10a - 4b = 3
Multiply the first equation by -2:
-8a + 4b = -2
Add it to the second equation: 10a - 4b = 3
This results in 2a = 1, or a = 1/2. Then, from the first equation,
`4*1/2 - 2b = 1`
2 - 2b = 1
2b = 1 and b = 1/2.
Therefore, for the given function to be continuous everywhere, a has to be 1/2 and b has to be 1/2.