Average Cost, Cancellation, and a Removable Discontinuity
A small workshop models average cost per unit with C(x) = (2x² + 7x − 15)/(x + 5), where x is the number of units produced in a batch and C is in dollars. A new spreadsheet uses the simplified expression 2x − 3 for all x, including x = −5, because a formula-audit tool canceled the factor x + 5. At x = 10 the two forms agree, but at x = −5 the original expression gives 0/0 while the spreadsheet returns −13. The manager wants to know whether the simplified rule can be used without restriction and how to describe the graph near x = −5. The batch size is normally a nonnegative integer, but sensitivity testing includes negative x values as a mathematical check. The manager also asks for the average cost at x = 0 and x = 4 using the correct domain.
1. Factor the numerator and simplify C(x), stating the domain restrictions.
Read the worked answer
Factor 2x² + 7x − 15. The product of the leading coefficient and constant is 2·(−15) = −30; two numbers with product −30 and sum 7 are 10 and −3. Rewrite and factor: 2x² + 10x − 3x − 15 = 2x(x + 5) − 3(x + 5) = (2x − 3)(x + 5). Thus C(x) = [(2x − 3)(x + 5)]/(x + 5) = 2x − 3, provided x + 5 ≠ 0, so x ≠ −5. The domain is all real numbers except −5; for ordinary workshop use, x ≥ 0.
2. Explain why the spreadsheet returns −13 at x = −5 and how the graph should be described there.
Read the worked answer
At x = −5 the original denominator is zero, so C(−5) is undefined. The simplified rule gives 2(−5) − 3 = −13, but that value belongs to the hole in the graph, not to the function. The graph is the line y = 2x − 3 with a removable discontinuity, or hole, at (−5, −13). The simplified rule cannot be used to assign a value at x = −5 unless the function is explicitly extended to fill the hole. For the original average-cost model, x = −5 remains outside the domain.
3. Evaluate C(0) and C(4) using the correct expression and confirm that the simplified form agrees.
Read the worked answer
At x = 0: C(0) = (0 + 0 − 15)/(0 + 5) = −15/5 = −3, which agrees with 2(0) − 3 = −3. At x = 4: C(4) = (2·16 + 7·4 − 15)/(4 + 5) = (32 + 28 − 15)/9 = 45/9 = 5, which agrees with 2(4) − 3 = 5. Both values are valid because 0 and 4 are in the domain, unlike x = −5.