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

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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)

Theorem
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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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