Composition of Functions

1. Composition of Functions Chapter 1, Section 2 -Christina McInnis

2. Real World Application • On February 8, a store will sell any item for $50 less than the listed price. On any day in February, the store will give a discount of 15% to any customer who can prove that he/she contributed to a local charity. • OK, let x be the listed price of an item in the store. • If P(x) is the price you will pay for an item on Feb. 8, then • P(x) = x - 50. • If D(x) is a price discounted at 15% , then, in February, the amount you will pay is • D(x) = 0.85x • You have two functions, P and D, which depend on the value of x, the listed price. If you don't know x, you can't calculate P(x) or D(x). And, note that if you do know x and "follow the rules," you get one value for P(x), and one value for D(x). P is a rule that says "take the listed price, than subtract $50." D is a rule that says "take the listed price, then take 85% of it." • P(x) = x - 50 and D(x) = 0.85x.

3. If f(x)= x/2 -7 and g(x)=x+6 find [f o g] (x) note: [f o g] (x) = f(g(x)) Step 1 :substitute x+6 for g(x) f(g(x)) f(x+6) Step 2 :substitute g(x) for ever x value in f(x) f(x)= x/2 -7 x+6/2 -7 x/2 + 6/2 -7 x/2 + 3 – 7 x/2 -4 Answer: x/2 -4

4. If f(x)= x/2 -7 and g(x)=x+6 find [g o f] (x) note: [g o f] (x) = g(f(x)) Step 1 :substitute x/2 -7 for f(x) g(f(x)) g(x/2 -7) Step 2 :substitute f(x) for ever x value in g(x) g(x)= x+6 x/2 -7 +6 x/2 -1 Answer: x/2 -1