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

therefore ((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))

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

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

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

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