Let f(x) = x2 + 3x -7, and g(x) = 4x +5. Then:

(f + g)(3) = f(3) + g(3) = 11 + 17 = 28,
(f / g)(2) = f(2) / g(2) = 3 / 13,

We can evaluate the function f + g at any number by evaluating f and g separately and adding the results, as we did above for 3. However, we generally simplify the formula for f + g by combining similar terms, then use this new formula to evaluate the sum function.
(f + g)(x) = f(x) + g(x) = ( x2 + 3x -7 ) + ( 4x +5) = x2 + 7x - 2.
So, (f + g)(3) = 32 +7*3 -2 = 28, which agrees with our previous answer.

The domain of an arithmetic combination of two functions f and g comprises of any x that is in the domain of both f and g, with one exception. The quotient f/g is not defined at values of x where g is equal to 0.

For example, if f(x) = 2x + 1 and g(x) = x - 3, then the domains of f+g, f-g, and f x g are all real numbers. The domain of f/g is the set of all real numbers except 3, since g(3) = 0. Example: Let f(x) = x2 - 9, and 
g(x) = x - 3. Simplify the formula for f / g.

(f / g)(x) = f(x) / g(x) = (x2 - 9)/( x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3.

There is a technical point to be made about this example that is often ignored in precalculus classes:
(f / g)(x) is not equal to the function h(x) = x + 3, because f / g is not defined at x = 3, while h is. There is really no harm in thinking of f / g as being the same as h, but they are different functions. It is better to think about them as equivalent functions for all x's except x=3.
For a visual demonstration of  combinations of functions click the DEMO button