(pen)o (g)Choosing a Branch(pen)
(pen)-> A (g)piecewise function(pen) uses different rules on input intervals.
(pen)-> Choose the branch by the (g)input(pen), not by the output.
(pen)-> First compare x with each branch's interval condition.
(pen)-> Substitute into only the rule whose condition holds.
(pen)-> An included input must have exactly one output.
(pen)-> Check boundary inputs carefully for < versus <=.
(pen)-> An input outside all listed intervals is outside the domain.
(pen)-> A (g)branch condition(pen) may test x, or it may name an interval.
(pen)-> With x >= 2 listed, every larger x is already (g)covered(pen).
(pen)o (pu)Worked Piecewise Example(pen)
(pen)-> Illustrative example using three (g)distinct input intervals(pen).
(pen)-> Define f(x) = 2x for x < 0.
(pen)-> Define f(x) = 3 for 0 <= x < 2.
(pen)-> Define f(x) = -x + 5 for x >= 2.
(pen)-> f(-2) = 2(-2) = -4 using the first branch.
(pen)-> f(1) = 3 using the middle branch.
(pen)-> f(4) = -4 + 5 = 1 using the last branch.
(pen)-> f(0) uses the (pu)middle branch(pen) because 0 satisfies 0 <= 0.
(pen)-> f(2) uses the (pu)last branch(pen) because 2 satisfies x >= 2.
(pen)o (br)Restricted Graphs(pen)
(pen)-> Graph each rule only on its (br)assigned input interval(pen).
(pen)-> For the example, draw y = 2x only to the left of x = 0.
(pen)-> Mark (0, 0) open; mark (0, 3) closed.
(pen)-> Draw y = 3 from x = 0 up to, but not including, x = 2.
(pen)-> The middle branch has an open endpoint at (2, 3).
(pen)-> The last branch includes (2, 3); the final point is (br)closed(pen).
(pen)-> Thus f(0) = 3 and f(2) = 3, each with one output.
(pen)-> f is (br)increasing(pen), flat, then decreasing across its domain.
(pen)o (y)Jersey Total Cost(pen)
(pen)-> Let n be the number of jerseys; C(n) is (y)total cost(pen).
(pen)-> The domain is (y)positive integers(pen), not all real numbers.
(pen)-> C(n) = 88n for 1 <= n <= 5.
(pen)-> C(n) = 75n for 6 <= n <= 11.
(pen)-> C(n) = 65n for 12 <= n <= 29.
(pen)-> C(n) = 56n for n >= 30.
(pen)-> These are all-units discounts: one rate for the whole order.
(pen)-> The (y)tiers(pen) leave no gap: 1-5, 6-11, 12-29, then 30 and up.
(pen)-> Reading the (y)domain(pen) first stops you from using 0 jerseys.
(pen)o (hp)Jersey Calculations(pen)
(pen)-> For 11 jerseys, choose the 6 through 11 tier.
(pen)-> C(11) = 75(11) = 825 dollars.
(pen)-> For 25 jerseys, choose the 12 through 29 tier.
(pen)-> C(25) = 65(25) = 1625 dollars, not 1885.
(pen)-> C(5) = 440, while C(6) = 450.
(pen)-> C(11) = 825, while C(12) = 780: total cost can (hp)fall(pen).
(pen)-> C(29) = 1885, while C(30) = 1680.
(pen)-> (hp)Rates(pen) step down: 88, 75, 65, then 56 per jersey.
(pen)-> At n = 30 the (hp)total drops(pen) the most: 205 dollars.
(pen)o (hp)Rates and Model Checks(pen)
(pen)-> An all-units price changes for every item at a tier boundary.
(pen)-> A (hp)progressive rate(pen) applies each rate only to its own band.
(pen)-> Do not add earlier tier costs in this all-units model.
(pen)-> Unit price is constant within tiers and jumps between them.
(pen)-> Total cost rises within tiers; it is (hp)not a step function(pen).
(pen)-> Plot separate integer points; do not fill gaps in the domain.
(pen)-> Tier prices jump; the integer-domain total has tier changes.
(pen)-> (hp)Compare(pen) the two models at n = 12 before choosing one.