Warm-up

Arithmetic Combinations (f+g)(x) = f(x) + g(x) (f-g)(x) = f(x) – g(x) (fg)(x) = f(x) ∙ g(x) (f/g)(x) = f(x) ; g(x) ≠0 g(x) The domain for these combinations is all values of x common to both domains, with the additional restriction that g(x)≠0 for f/g.

Evaluate each of these combinations for f(x) = 4x – 12 and g(x) = x 2 – 9. (f+g)(x) = x 2 + 4x – 21, domain is all reals compare f(2) + g(2) to (f+g)(2) (f-g)(x) = -x 2 + 4x – 3, domain is all reals (f·g)(x) = 4x 3 -12x x + 108, domain is all reals (f/g)(x) = 4/(x + 3); domain is all reals, x ≠ ±3

Look out for the domain… If f(x) = 1/(x – 5) and g(x) = √x, what is the domain of (f+ g)(x)? The domain of f(x) is all reals but 5. The domain of g(x) is all non-negative numbers. So the domain of the sum is [0,5) U (5, ∞).

Let f(x) = √x and g(x) = √(x + 3) Find the domain of g/f and f/g. The domain of f(x) is x ≥ 0. The domain of g(x) is x ≥ -3 The intersection of these is x ≥ 0. However…

We have (f/g)(x) = √x √(x+ 3) And in this case the domain is [0,∞). But (g/f)(x) = √(x+ 3) √x In this case the domain is (0, ∞). Can you see why there is a difference?

Composition of Functions The composition of the function f with g is (f ° g)(x) = f(g(x)). The domain of the function is all x that are in the domain of g such that g(x) is in the domain of f.

Let f(x) = x 2 + 2x and g(x) = x + 5. Find the f(g(1)) and g(f(1)). f(g(1)) = 48 g(f(1)) = 8 Find f(g(x)) and g(f(x)). f(g(x)) = x x + 35 g(f(x)) = x 2 + 2x + 5

Let f(x) = x 2 – 25 and g(x) = √(25 – x 2 ). Simplify f(g(x)). Find the domain of the f(g(x)). You might think that the domain is all real numbers, but it is [-5, 5] because that is the domain of g(x). Try graphing y = (√(25 – x 2 )) 2 – 25 and see what happens.

Let f(x) = √x and g(x) = x + 5. What is the domain of f(g(x))? The domain of g(x) is all real numbers. But the domain of f(x) is x ≥0. So only values of x such that g(x), which is x + 5, are greater than or equal to 0 work. So x must be greater than or equal to -5. Or just look at the composition function (before simplifying) to find its domain.

Decomposing Functions Write each function as the composition of two or more functions. a)f(x) = (2x+ 1) 2 – 7 let g(x) = 2x + 1, let h(x) = x 2 – 7 then h(g(x)) = (2x + 1) 2 – 7 (This answer is not unique.) b)f(x) = (x – 3) 2 + 2(x- 3) + 1 let g(x) = x – 3, let h(x) = x 2 + 2x + 1 then h(g(x)) = (x – 3) 2 + 2(x- 3) + 1