A Piecewise-defined Function If the brokerage firm in Example decides to keep the commissi...

A Piecewise-defined Function If the brokerage firm in Example decides to keep the commission charges unchanged for purchases up to and including $600, but to charge only 1.5% plus $15 for gold purchases exceeding $600, express the brokerage commission as a function of the amount x of gold purchased.

Example

A Piecewise-Defined Function A leading brokerage firm charges a 6% commission on gold purchases in amounts from $50 to $300. For purchases exceeding $300, the firm charges 2% of the amount purchased plus $12.00. Let x denote the amount of gold purchased (in dollars) and let f(x) be the commission charge as a function of x.

(a) Describe f(x).

(b) Find f(100) and f(500).

SOLUTION

(a) The formula for f(x) depends on whether 50 ≤ x ≤ 300 or 300 < x. When 50 ≤ x ≤ 300, the charge is .06x dollars. When 300 < x, the charge is .02x + 12. The domain consists of the values of x in one of the two intervals [50, 300] and (300,∞). In each of these intervals, the function is defined by a separate formula:

f(x) =

Note that an alternative description of the domain is the interval [50,∞). That is, the value of x may be any real number greater than or equal to 50.

(b) Since x = 100 satisfies 50 ≤ x ≤ 300, we use the first formula for f(x): f(100) = .06(100) = 6. Since x = 500 satisfies 300 < x, we use the second formula for f(x): f(500) = .02(500) + 12 = 22.

In calculus, it is often necessary to substitute an algebraic expression for x and simplify the result, as illustrated in the following example.