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Worked Examples · Example 3

Q.If f(x)=x+2f(x) = x+2 and g(x)=3x−1g(x) = 3x-1, find (f∘g)(x)(f \circ g)(x) and (g∘f)(x)(g \circ f)(x), and verify your results 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)=3x−1g(x) = 3x-1

f(g(x))=f(3x−1)=(3x−1)+2=3x+1f(g(x)) = f(3x-1) = (3x-1)+2 = 3x+1

So (f∘g)(x)=3x+1(f\circ g)(x) = 3x+1.

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

f(x)=x+2f(x) = x+2

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

So (g∘f)(x)=3x+5(g\circ f)(x) = 3x+5.

Verification at x=2x=2:

  • g(2)=3(2)−1=5g(2) = 3(2)-1 = 5, then f(5)=5+2=7f(5) = 5+2 = 7. Using the formula, (f∘g)(2)=3(2)+1=7(f\circ g)(2) = 3(2)+1 = 7 — agrees.
  • f(2)=2+2=4f(2) = 2+2 = 4, then g(4)=3(4)−1=11g(4) = 3(4)-1 = 11. Using the formula, (g∘f)(2)=3(2)+5=11(g\circ f)(2) = 3(2)+5 = 11 — agrees. …

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