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