If `f:R rarr R, f(x)=2x-1` and `g:R rarr R, g(x) = x^2` then `(gof)(x)` equals

If f: R rarr R, f(x)= x-2, g: R rarr R, g(x)= x + 2 then (f+ g) (x) = ……..

f : R rarr R , f(x) = cos x and g : R rarr R , g(x) = 3x^(2) then find the composite functions gof and fog.

If quad f:R rarr R,f(x)=2x;g:R rarr R,g(x)=x+1 then (fg)(2) equals

If f:R rarr R,f(x)=2x;g:R rarr Rg(x)=x+1, then (f*g)(2) equals

If f:R rarr R,f(x)=x^(3)+3, and g:R rarr Rg(x)=2x+1, then f^(-1) og ^(-1)(23) equals

f:R rarrR , f(x) =x^(2) , g : R rarr R , g(x) = 2^(x) , then {x|(fog)(x) = (gof)(x)} = ............

f: R rarr R , f(x) = x^(2)+2 and g :R rarr R , g (x) = x/(x-1) then find fog and gof.

Let :R rarr R;f(x)=sin x and g:R rarr Rg(x)=x^(2) find fog and gof.

If f:R rarr R where f(x)=x^(2) and g:R rarr R where g(x)=2x+5, then prove that gof=2x^(2)+5

Let R be the set of real numbers.If f:R rarr R:f(x)=x^(2) and g:R rarr R;g(x)=2x+1. Then,find fog and gof . Also,show that fog!=gof .