If `f:R to R` be defined by `f(x)=3x^(2)-5` and `g: R to R ` by `g(x)= (x)/(x^(2)+1).` Then, gof is

Let f:R to be defined by f(x)=3x^(2)-5 and g : R to R by g(x)=(x)/(x^(2))+1 then gof is

Let f : R rarr R be defined by f(x) = 3x^2-5 and g : R rarr R by g(x) = (x)/(x^2+1) xx Then gof is

Let f: R rarr R be defined by f(x) = 3x^(2) - 5 and g : R rarr R, g(x) = x/(x^(2)+1) Then gof is ...........

Let f:R to R be defined by f(x)=x^(2) and g:R to R be defined by g(x)=x+5 . Then , (gof)(x) is equal to -

If f: R rarr R be defined by f(x) =e^(x) and g:R rarr R be defined by g(x) = x^(2) the mapping gof : R rarr R be defined by (gof)(x) =g[f(x)] forall x in R then

Let f : R rarr R be defined by f(x) = 3x - 2 ad g : R rarr R be defined by g(x) = (x+2)/(3) . Shwo that fog = I_R = gof

If f : R rarr R be defined by f(x)=2x-3 and g : R rarr R be defined by g(x)= (x+3)/2 , show that fog=gof=I_R

If f : R rarr R is defined by f(x)=x^2+2 and g : R-{1} rarr R is defined by g(x)=x/(x-1) , find (fog)(x) and (gof)(x) .

Let f : R rarr R be defined by f(x) = x^(2) + 3x + 1 and g : R rarr R is defined by g(x) = 2x - 3, Find, (i) fog (ii) gof