If f- X- Y- g- Y- Z and h-Z- S are functions- then hogof-hogof-

Proof: Here, ho(gof)(x)=h{gof (x)}=h[g{f(x)}] , ∀ x∈ X . and (hog)of(x)=(hog){f(x)}=h[g{f(x)}] , ∀ x ∈ X . Hence, ho(gof)=(hog)of . ...

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Standard XII

Mathematics

Definition of Functions

If f: X→ Y,...

Question

If $f : X \to Y$ , $g : Y \to Z$ and $h : Z \to S$ are functions, then $h o (g o f) = (h o g) o f$ .

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Solution

Proof: Here, $h o (g o f) (x) = h {g o f (x)} = h [g {f (x)}]$ , $\forall x \in X$ . and $(h o g) o f (x) = (h o g) {f (x)} = h [g {f (x)}]$ , $\forall x \in X$ . Hence, $h o (g o f) = (h o g) o f$ .

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If f:X→Y, g:Y→Z and h:Z→S are functions, then ho(gof)=(hog)of.

Proof: Here, ho(gof)(x)=h{gof(x)}=h[g{f(x)}], ∀x∈X. and (hog)of(x)=(hog){f(x)}=h[g{f(x)}], ∀x∈X. Hence, ho(gof)=(hog)of.

If $f : X \to Y$ , $g : Y \to Z$ and $h : Z \to S$ are functions, then $h o (g o f) = (h o g) o f$ .

Proof: Here, $h o (g o f) (x) = h {g o f (x)} = h [g {f (x)}]$ , $\forall x \in X$ . and $(h o g) o f (x) = (h o g) {f (x)} = h [g {f (x)}]$ , $\forall x \in X$ . Hence, $h o (g o f) = (h o g) o f$ .