snhu MODULE FIVE PROBLEM SET This document is proprietary to Southern New Hampshire University. It and the problems within may not be posted on any non-SNHU website. James Lightner 1
snhu Directions: Type your solutions into this document and be sure to show all steps for arriving at your solution. Just giving a final number may not receive full credit. PROBLEM 1 Indicate whether the two functions are equal. If the two functions are not equal, then give an element of the domain on which the two functions have different values. (a) f: Z-> Z, where f(x)=x2. g : Z -> Z, where g(x) = |x|2. . The 2 functions are equivalents because they have the same domain and target. · x2 = | x |2 · Squaring a number makes it a positive and the absolute value that is squared is also a positive. · (1)2 = |(1)|2 = 1 (b) · (-1)2 = |(-1)|2 = 1 f : ZxZ -> Z, where f (x, y) = |x + y|. g : Z x Z > Z, where g (x, y) = |x | + ly|. . These to outcomes are not equivalent. When x or y is ¡ 0, the result is not the same. · |(1) + (-2)| =|- 1| = 1 · |(1)|+|(-2)|=1+2=3
snhu PROBLEM 2 The domain and target set of functions f and g is R. The functions are defined as: . f(x) = 2x + 3 · g(x) = 5x + 7 (a) fog? . f(g(x))=2(5x+7)+3 · 10x + 14+3 = 10x + 17 · fog= 10x + 17 (b) gof? . g(f(x))=5(2x+3)+7 · 10x + 15 +7 = 10x + 22 · gof= 10x + 22 (c) (f o g)-1? · (f o g) = y . y= 10x + 17 . Solve for x. · y- 17 = 10x . y-17 = x 10 . (fog) -1 = 2-17 (d) f-1 og-1? 10 · f(x) = y . y = 2x + 3 · y - 3 = 2x . y-3 = x · f-1(x) = 2-3 · g(x) = y . y = 5x + 7 · y - 7 = 5x . y-7 5 = x · g-1(x) = " · f-1(g-1(x)) = (e) g-1 o f-1? · 9-1(f-1(x)) = 2,-7 = 1-17