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Worked Examples · Example 3

Q.If f(x)=2x+1f(x)=2x+1 and g(x)=x2−3g(x)=x^2-3, find (f∘g)(x)(f\circ g)(x) and (g∘f)(x)(g\circ f)(x), and verify both at x=2x=2.

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Finding (f∘g)(x)(f\circ g)(x): by definition (f∘g)(x)=f(g(x))(f\circ g)(x)=f(g(x)), so gg is applied first.

g(x)=x2−3,g(x)=x^2-3,

f(g(x))=f(x2−3)=2(x2−3)+1=2x2−6+1=2x2−5.f(g(x))=f(x^2-3)=2(x^2-3)+1=2x^2-6+1=2x^2-5.

So (f∘g)(x)=2x2−5(f\circ g)(x)=2x^2-5.

Finding (g∘f)(x)(g\circ f)(x): here ff is applied first.

f(x)=2x+1,f(x)=2x+1,

g(f(x))=g(2x+1)=(2x+1)2−3=(4x2+4x+1)−3=4x2+4x−2.g(f(x))=g(2x+1)=(2x+1)^2-3=(4x^2+4x+1)-3=4x^2+4x-2.

So (g∘f)(x)=4x2+4x−2(g\circ f)(x)=4x^2+4x-2.

Verification at x=2x=2 (dual check):

  • g(2)=22−3=1g(2)=2^2-3=1, then f(1)=2(1)+1=3f(1)=2(1)+1=3. Formula: (f∘g)(2)=2(2)2−5=8−5=3(f\circ g)(2)=2(2)^2-5=8-5=3. ✓
  • f(2)=2(2)+1=5f(2)=2(2)+1=5, then g(5)=52−3=22g(5)=5^2-3=22. Formula: (g∘f)(2)=4(2)2+4(2)−2=16+8−2=22(g\circ f)(2)=4(2)^2+4(2)-2=16+8-2=22. ✓ …

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