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

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 f : R to R :f (x) = x ^(2) g : R to R : g (x) = x + 5, then gof is:

If f: R to R and g: R to R be two functions defined as f(x)=x^(2) and g(x)=5x where x in R , then prove that (fog)(2) ne (gof) (2).

If f: R to R and g: R to R be two functions defined as f(x)=x^(2) and g(x)=5x where x in R , then prove that (fog)(2) ne (gof) (2).

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

If f : R to R , f (x) =x ^(3) and g: R to R , g (x) = 2x ^(2) +1, and R is the set of real numbers then find fog(x) and gof (x).

Let : R->R ; f(x)=sinx and g: R->R ; g(x)=x^2 find fog and gof .

Let f:R→R:f(x)=x^2 and g: R→R: g(x)=(x+1) . Show that (gof)≠(fog) .

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