If `f:R rarr R, g:R rarr R`, given by `f(x)=[x], g(x)=[x]`, then find fog `(- 2/3)` and gof `(-2/3)`.

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 the function f:R rarr R , g:R rarr R defined by f(x)=3x-1 , g(x)=x^(2)+1 then find i) (fof)(x^(2)+1) ii) (fog )(2)

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

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,g:R rarr R defined as f(x)=sin x and g(x)=x^(2), then find the value of (gof)(x) and (fog)(x) and also prove that gof f= fog.

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 are two mappings such that f(x)=2x and g(x)=x^(2)+2 then find fog and gog.

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^(2) and g(x)=x+1. Show that fog != gof

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)=