Question

Let f, g and h be functions from R to R. Show that

`(f + g)oh = foh + goh`

`(f.g)oh = (foh).(goh)`

Answer 1

To prove:

(f + g)oh = foh + goh

Consider:

`((f+g)oh)(x)`

= (f +  g)(h(x))

`= f(h(x)) + g(h(x))`

= (foh)(x) + (goh) (x)

= {(foh) + (goh)} (x)

:. ((f+g)oh) (x) = {(foh) +(goh) } (x)           ∀x ∈ R

Hence (f + g)oh =  foh + goh 

To prove

`(f.g)oh = (foh).(goh)`

Consider

`((f.g)oh) (x)`

`= (f . g)(h(x))`

`= f(h(x)).g(h(x))`

`=(foh)(x).(goh)(x)`

`={(foh).(goh)}(x)`

`:. ((f.g)oh)(x)  = {(foh).(goh)}(x)`   ∀x ∈ R

Hence `(f.g) oh = (foh).(goh)`