Snhu
MODULE FIVE PROBLEM SET
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Jaime Rowland
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 :ZZ,where f(x)=x2 g:ZZ,where g(x)=|x|2
= x is great than or equal to 0 or -x f x is less than 0 Then -x
x2xisgreaterthanorequalto0or(-x)2xislessthanzero|x|2x2xisgreaterthanorequalto0orx2xislessthanzeroHer x2for911xZThusF :Z Zandg:Z ZF(x)=x2andg(x)=|x|2Since|x|2= x2Vxz=> f(x)=g(x)VxzSofandgaretwoequalequations
f :ZZ>Z,where f(x,y)=|x+y|. |f|+|x|=(fx)6aI0qMZZXZ:6
=(T-)6zzxz:b0=(T-fos0=|0|=1(I-)+1|=(T-)fzzxz 1+-1=2sog1,-1)=2Henceg1,-1)g1,-1Fandgarenotequal functions
snhu
PROBLEM 2
The domain and target set of functions f and g is R. The functions are defined as:
(b)fx)=2x+3
g(x)=5x+7
(a) fog?
fog = f[g(x)] f[g(x)]=f[5x+7] To get f(5x+7), we will replace the variable x in f(x) with 5x+7 as shown; f(x)=2x+3 f(5x+7)=2(5x+7)+3 f(5x+7) = 10x+14+3 f(5x+7)= 10x+17 Hence fog = 10x+17 (b) go f?
gof = g[f(x)] g[f(x)]= g[2x+3] To get g(2x+3), we will replace the variable x in g(x) with 2x+3 as shown; g(x)=5x+7 g(2x+3) =5(2x+3)+7 g(2x+3) = 10x+15+7 g(2x+3)=10x+22 Hence gof = 10x+22 (c) (f og)-1?
For (fog)1 (inverse of (fog)) Given (fog) = 10x+17 To find the inverse, first we will replace (fog) with variable y to have; y =10x+17 Then we will interchange variable y for x: x= 10y+17 We will then make y the subject of the formula; 10y = x-17 y = x-17/10 Hence the inverse of the function (fog)1 = (x-17)/10 (d) f-iog-1?
For the function flogl We need to get the inverse of function f(x) and g(x) first. For f-1(x): Given f(x)= 2x+3