If `f: R to R and g : R to R` are given by by f(x)=cos x and `g(x)=3x^(2)`, then shown that `gof ne fog`.

If f: R to R and g: R to R are given by f(x)= cosx and g(x) =3x^(2) .Show that gof ne fog

If f:RtoRandg:RtoR are given by f(x)=cosxandg(x)=3x^(2) . Find gof and fog.

Find fog if f(x)= x-2 and g(x) = 3x

If functions f : R to R and g: R to R are given by f(x) = |x| and g(x) = [x] , ( where [x] is greatest function) find fog (-(1)/(2) ) and "gof" (- (1)/(2) )

Find gof and fog if f : R rarr R and g : R rarr R are given by f(x) = cosx and g(x) = 3x^(2)

If f:R rarr R and g : R rarr R are defined by f(x) = 3x + 2 and g(x) = x^2 - 3 , then the value of x such that g(f(x))=4 are

If f : R rarr R and g : R rarr R be two mapping such that f(x) = sin x and g(x) = x^(2) , then find the values of (fog) (sqrt(pi))/(2) "and (gof)"((pi)/(3)) .

If functions f: R -gt R and g : R -gt R arc given by f(x) = abs(x) and g(x) =abs(x) , (where abs(A) is indicatescates: integer function) find f o g(abs(-1/2)) and go f(-1/2) .

If f:R -> R, g:R -> 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 != fog .

If f : R to R be given by f (x) = (3 -x ^(3)) ^(1/3), then fof (x) is