See below This is an example of a similar problem (See Week 4 Lecture page \"Solving Rational Inequalities\" for a starting point.) Suppose that a factory's cost (in dollars) for making x cars is C(x) = 3280x + 9600 Then the average cost per car is C'(x) = 3280 x +9600 x How many cars must the factory make each day to ensure the average cost per car is no more than $3600? Solution:since the average cost per car must be less than or equal to 3600 we may write 3280 x +9600 x 3600 now subtract 3600 from both sides this yields 3280 x +9600 x - 3600 0 now get a LCD for the expression on the left this yields 3280 x +96003600 x x 0 now add/subtract on left this yields - 320 x+9600 x 0 now calculate the critical values by setting the numerator and denominator to 0 this yields x = 0 and - 320x + 9600 = 0 to solve the second one first subtract 9600 from both sides this yields -320x = -9600 now divide both sides by - 320 yields final answer x = 30 so we have two critical values 0 and 30. We will use the "test point" method to solve the inequality. The first step is to use the two solutions we found to partition the x-axis into three open (test) intervals: Interval 1 (- , 0) Interval 2 (0, 30) Interval 3 (30, ) 1. Testing first interval using the test point - 1 and substituting - 1 for x into 320 x+ 9600 x yields 320(1)+ 9600 1 which becomes 9920 1 = - 9920 which is < 0 so the test point passes and the first interval does \"work\". 2. Testing 2nd interval using the test point 2 and substituting 2 for x into 320 x+ 9600 x yields 320(2)+9600 2 which becomes 8960 = 4480 which is not < 0 so the test point fails and the second interval does not \"work\". 2 3. Testing third interval using the test point 40 and substituting 40 for x into 320 x+ 9600 x yields 320(40)+ 9600 40 which becomes 3200 = - 80 which is < 0 so the test point passes and the third interval does \"work\". 40 final solution is then (-, 0) U (30, ) and since the number of cars cannot be negative the answer must be (30, ) or x 30