" If "f:R rarr R," Where "f(x)=x^(2)" And "g:R rarr R,g(x)=2x+5," So prove that gof "!=" fog."

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

If f : R rarr R, f(x) = x^(2) + 2x - 3 and g : R rarr R, g(x) = 3x - 4 then the value of fog (x) is

If f : R rarr R, f(x) = x^(2) + 2x - 3 and g : R rarr R, g(x) = 3x - 4 then the value of fog (x) is

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

Let f : R rarr R, f(x)=x^2 and g : R rarr R,g(x)=sinx be two given functions. Is gof=fog ?

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 .

Let R be the set of real numbes.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.

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

If f : R rarr R, f(x) = x^(2) - 5x + 4 and g : R^(+) rarr R, g(x) = log x , then the value of (gof) (2) is

If f : R rarr R, f(x) = x^(2) - 5x + 4 and g : R^(+) rarr R, g(x) = log x , then the value of (gof) (2) is