Solution to (b) Given f(x) = x 2 , g(x) = 2x and h(x) = x - … - Sikademy
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Archangel Macsika

(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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Here's the Solution to this Question

(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


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