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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)
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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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