Prove that (fog)oh=fo(goh) where f, g, h are three functions

If f,g,h are real valued function defined on R, then prove that (f+g)oh=foh+goh. what can you say about fo(g+h)? Justify your answer.

If f(x)=(g(x))+(g(-x))/2+2/([h(x)+h(-x)]^(-1)) where g and h are differentiable functions then f'(0)

Find the composition fog and gof and test whether fog = gof when f and g are functions or R given by f(x) = sin x, g(x), g(x) =x^5

Let h(x)=(fog)(x)+K where K is any constant . If d/dx(h(x))=-(sinx)/(cos^2(cosx)then compule the valu of j(0) where j(x) =int_g(x)^f(x) f(t)/g(t) dt, where f and g are trigonometric function.

Let f(x) = [x] , g(x)= |x| and f{g(x)} = h(x) ,where [.] is the greatest integer function . Then h(-1) is

Let f(x) = [x] , g(x)= |x| and f{g(x)} = h(x) ,where [.] is the greatest integer function . Then h(-1) is

Consider the function f(x), g(x), h(x) as given below. Show that (fog)oh=fo(goh) in each case. f(x)=x^(2), g(x)=2x and h(x)=x+4

Consider the function f(x), g(x), h(x) as given below. Show that (fog)oh=fo(goh) in each case. f(x)=x-4, g(x)=x^(2) and h(x)=3x-5

Consider the function f(x), g(x), h(x) as given below. Show that (fog)oh=fo(goh) in each case. f(x)=x-1, g(x)=3x+1 and h(x)=x^(2) .

Let f and g be two functions such that gof=L_(A) and fog=L_(B). Then; f and g are bijections and g=f^(-1)