Section 3.4 The Domain and Range of Functions 195 Solutions 4a. It doesn’t make sense to sell a negative number of donuts or a partial donut. Only whole donuts can be sold. So, the reasonable domain for this situation is d = 0, 1, 2, 3, . . . , or the domain is the set of whole numbers, denoted by =50, 1, 2, 3, . . .6 . Note: It is not practical for a bakery to make infinitely many donuts in a month. There is an upper limit to the domain but we need more information to determine what that is. 4b. Since the expressions x and 10 - 2x represent the lengths of the sides of a rectangle, it is appropriate that these values be positive. So, to find the domain we have to solve the compound inequality, x > 0 and 10 - 2x > 0. x > 0 and 10 - 2x > 0 10 - 2x - 10 > 0 -10 -2x > -10 22x 22 , 210 22 x < 5 The intersection of the sets x . 0 and x , 5 is the interval (0, 5). So, the domain of A(x) is (0, 5). Student Check 4 Determine an appropriate domain for each situation. a. The monthly cost of a cell phone plan is C(x) = 69.99 + 0.45x, where x is the number of minutes over 4000. b. The area of the rectangle is A(x) = x(18 - 2x). Troubleshooting Common Errors A common error associated with the domain and range of functions is shown. Objective 5 Example A problem and an incorrect solution are given. Provide the correct solution and an explanation of the error. Find the domain and range of the given function. 2 2 4 –2 –6 –4 –2 –4 x y Incorrect Solution Correct Solution and Explanation The domain is -2, ∞) and the is 2, ∞). The graph of the function extends indefinitely to the left of x 5 22. So, the domain is (-∞, ∞). The graph also extends indefinitely below y = 2. So, the range is (-∞, ∞). Set each side greater than zero. Subtract 10 from each side. Simplify. Divide each side by -2 and reverse the inequality symbol. Simplify. x 18 – 2x Objective 5 ▶ Troubleshoot common errors. omain he range i ∞
hendricks_intermediate_algebra_1e_ch1_3
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