**(b) Given f(x) = x 2 , g(x) = 2x and h(x) = x - 2 be functions from R to R. Find (i) f ◦ g (ii) f ◦ h (iii) h ◦ f (iv) (f ◦ g) ◦ h (v) f ◦ (g ◦ h)**

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(i)

$f\circ g=f(g(x))=f(2x)=(2x)^2=4x^2$

(ii)

$f\circ h=f(h(x))=f(x-2)=(x-2)^2=x^2-4x+4$

(iii)

$h\circ f=h(f(x))=h(x^2)=x^2-2$

(iv)

$(f\circ g)\circ h=(f\circ g)(h(x))=(f\circ g)(x-2)=$

$=f(g(x-2))=f(2(x-2))=(2(x-2))^2=$

$=4x^2-16x+16$

(v)

$f\circ (g\circ h)=f((g\circ h)(x))=f(g(h(x)))=$

$=f(g(x-2))=f(2(x-2))=(2(x-2))^2=$

$=4x^2-16x+16$

(i)

$f\circ g=f(g(x))=f(2x)=(2x)^2=4x^2$

(ii)

$f\circ h=f(h(x))=f(x-2)=(x-2)^2=x^2-4x+4$

(iii)

$h\circ f=h(f(x))=h(x^2)=x^2-2$

(iv)

$(f\circ g)\circ h=(f\circ g)(h(x))=(f\circ g)(x-2)=$

$=f(g(x-2))=f(2(x-2))=(2(x-2))^2=$

$=4x^2-16x+16$

(v)

$f\circ (g\circ h)=f((g\circ h)(x))=f(g(h(x)))=$

$=f(g(x-2))=f(2(x-2))=(2(x-2))^2=$

$=4x^2-16x+16$

(i)

$f\circ g=f(g(x))=f(2x)=(2x)^2=4x^2$

(ii)

$f\circ h=f(h(x))=f(x-2)=(x-2)^2=x^2-4x+4$

(iii)

$h\circ f=h(f(x))=h(x^2)=x^2-2$

(iv)

$(f\circ g)\circ h=(f\circ g)(h(x))=(f\circ g)(x-2)=$

$=f(g(x-2))=f(2(x-2))=(2(x-2))^2=$

$=4x^2-16x+16$

(v)

$f\circ (g\circ h)=f((g\circ h)(x))=f(g(h(x)))=$

$=f(g(x-2))=f(2(x-2))=(2(x-2))^2=$

$=4x^2-16x+16$