 112
 6
 Homework Statement
 If ##f : R \Rightarrow R##and ##g:R \Rightarrow R## defined by ##f(x) = x## and ##g(x)=(x3)## then ##{g(f(x)) : 1.6 < x < 1.6}##
 Homework Equations

Given ##f(x)## and ##g(x)## for ## 1.6 < x < 1.6## we get ##0\leq f(x)<1.6##
Now for ##0\leq f(x)< 1## ##\implies## ##g(f(x))=3## since ## 3\leq f(x)3<2##
Again for, ##1\leq f(x)< 1.6## ##\implies## ##g(f(x))=2## since ## 2 \leq f(x)3 < 1.4##
Thus the required set is {3,2}
My attempt :
Given ##f(x)## and ##g(x)## for ## 1.6 < x < 1.6## we get ##0\leq f(x)<1.6##
Thus, for ##f(g(x))## we get ## 3 \leq g(f(x)) < 1.4##
Thus the required set should be the interval ##[3, 1.4)##?
My Questions :
1. What have I missed since my answer does not match the given solution.
2. In the given solution why is the ##f(x)## inequality broken into ##0\leq f(x)< 1## and ##1\leq f(x)< 1.6##
3. After splitting the inequality I don't understand how the result is not an interval but two unique integers.
Given ##f(x)## and ##g(x)## for ## 1.6 < x < 1.6## we get ##0\leq f(x)<1.6##
Thus, for ##f(g(x))## we get ## 3 \leq g(f(x)) < 1.4##
Thus the required set should be the interval ##[3, 1.4)##?
My Questions :
1. What have I missed since my answer does not match the given solution.
2. In the given solution why is the ##f(x)## inequality broken into ##0\leq f(x)< 1## and ##1\leq f(x)< 1.6##
3. After splitting the inequality I don't understand how the result is not an interval but two unique integers.
Last edited by a moderator: