If `f:R rarr R` and `g:R rarr R` are defined by `f(x)=x-[x]` and `g(x)=[x]` is `x in R` then `f{g(x)}`

If f:R rarr R and g:R rarr R is given by f(x)=|x| and g(x)=[x] for each x in R then {x in R:g(f(x))<=f(g(x))}

If f:R rarr R and g:R rarr R given by f(x)=x-5 and g(x)=x^(2)-1, Find fog

If f:R rarr R and g:R rarr R are define by f(x)=3x-4 and g(x)=2+3x , then (g^(-1)of^(-1))(5)=

f:R rarr Rf:R rarr R is defined by f(x)=x^(2)-3x+2 find

Let f:R rarr R and g:R rarr R be defined by f(x)=x^(2) and g(x)=x+1. Show that fog != gof

If f:R rarr R and g:R rarr R defined f(x)=|x| and g(x)=[x-3] for x in R[.] denotes the greatest integer function

If f:R rarr R,g:R rarr R are defined by f(x)=3x-4 and g(x)=2+3x ,then (g^(-1) of^(-1))(5)=

Let f:R rarr R and g:R rarr R be defined by f(x)=x+1 and g(x)=x-1. Show that fog =gof=I_(R)

If f:R rarr R and g:R rarr R defined by f(x)=2x+3 and g(x)=x^(2)+7 then the values of x for which f(g(x))=25 are