Let f, g and h be functions from R to R. Show that
`(f +g ) oh =foh+goh`
`(f. g) oh = ( fog`). (goh) `

If f, g, h are functions from R to R such that f(x) = x^(2)+1, g(x) =sqrt(x^(2)-1) AA x in R and h(x) = {{:(0, x le 0),(x, x ge 0):} , then AA x in R, ho(fog)(x)=

If f, g, h are functions defined from R to R such that f(x) =x^(2)-1, g(x)=sqrt(x^(2)+1), h(x)={(0" if "x lt 0),(x" if "x ge 0):} then [ho (fog)](x) is defined by

Let f(x) and g(x) be differentiable functions for 0 le x le 1 such that f(0) = 2, g(0) = 0, f(1) = 6 . Let there exist a real number c in (0,1) such that f ’(c) = 2 g ’(c), then g(1) =

Let f and g be two diffrentiable functions on R such that f'(x) gt 0 and g'(x) lt 0 for all x in R . Then for all x

If f and g are two decreasing functions such that fog exists then fog is

Let f(x) and g(x) be two differentiable functions in R such that f(2) = 8, g(2)=0 f(4) = 10, g(4) = 8 and for atleast one x in ( 2,4), g' (x) = kf'(x) , then k is

Let f(x) and g(x) be two continous functions defined from Rrarr R , such that f(x_(1)) gt f(x_(2)) and g(x_(1)) lt g(x_(2)) AA x_(1) gt x_(2) , then find the solution set of f(g(alpha^(2) - 2 alpha) gt f(g(3 alpha - 4) .