Composing f and g: f(x)=2x-5, g(x)=(x+5)/2 — (f∘g)(x) = (g∘f)(x) = x

Overview

  • This is an example of showing that two composite functions are equal.
  • The goal is to show that f composed with g and g composed with f both yield the identity function on the domain: (f ∘ g)(x) = (g ∘ f)(x) = x for every x in the appropriate domain.

Given functions

  • ( f(x) = 2x - 5 )
  • ( g(x) = \dfrac{x + 5}{2} )

Objective

  • Show that ( (f \circ g)(x) = (g \circ f)(x) = x ) for all x in the domain of ( f \circ g ) and ( g \circ f ).
  • Recall: ( (f \circ g)(x) = f(g(x)) ).

Computation: (f ∘ g)(x)

  • Start with ( (f \circ g)(x) = f(g(x)) ).
  • Since ( g(x) = \dfrac{x + 5}{2} ), substitute:
    [ f\left( \dfrac{x + 5}{2} \right) ]
  • Compute: [ f\left( \dfrac{x + 5}{2} \right) = 2 \cdot \dfrac{x + 5}{2} - 5 ]
  • Simplify step by step:
    • ( 2 \cdot \dfrac{x + 5}{2} = x + 5 )
    • Therefore, ( f\left( \dfrac{x + 5}{2} \right) = (x + 5) - 5 )
    • ( (x + 5) - 5 = x )
  • Final result: [ (f \circ g)(x) = x ]
  • Expressed in LaTeX: (f∘g)(x)=f(g(x))=f(x+52)=2⋅x+52−5=(x+5)−5=x.(f \circ g)(x) = f(g(x)) = f\left( \dfrac{x + 5}{2} \right) = 2 \cdot \dfrac{x + 5}{2} - 5 = (x + 5) - 5 = x.

Computation: (g ∘ f)(x)

  • Start with ( (g \circ f)(x) = g(f(x)) ).
  • Since ( f(x) = 2x - 5 ), substitute: ( g(2x - 5) ).
  • Compute: [ g(2x - 5) = \dfrac{(2x - 5) + 5}{2} ]
  • Simplify:
    • ( (2x - 5) + 5 = 2x )
    • So, [ \dfrac{2x}{2} = x ]
  • Final result: [ (g \circ f)(x) = x ]
  • Expressed in LaTeX: (g∘f)(x)=g(f(x))=g(2x−5)=(2x−5)+52=2x2=x.(g \circ f)(x) = g(f(x)) = g(2x - 5) = \dfrac{(2x - 5) + 5}{2} = \dfrac{2x}{2} = x.

Domain considerations

  • Both f and g are linear functions defined for all real numbers, so:
    • ( \text{Dom}(f) = \textbf{R} )
    • ( \text{Dom}(g) = \textbf{R} )
  • Therefore, the domains of the compositions are:
    • ( \text{Dom}(f \circ g) = \textbf{R} )
    • ( \text{Dom}(g \circ f) = \textbf{R} )
  • Consequently, the equality ((f \circ g))(x) = ((g \circ f))(x) = x holds for all real numbers: ( x \in \mathbb{R} ).

Conclusion

  • We conclude that:
    • ( (f \circ g)(x) = x ) and ( (g \circ f)(x) = x ) for all real x.
    • Thus, ( (f \circ g) = (g \circ f) = \text{Identity on } \mathbb{R} ).

Connections to inverse functions

  • This result shows that f and g are inverse functions of each other:
    • Since ( f(g(x)) = x ) and ( g(f(x)) = x ), it follows that ( g = f^{-1} ) and ( f = g^{-1} ).
  • For this pair: ( f(x) = 2x - 5 ) and ( g(x) = \dfrac{x + 5}{2} ) satisfy the inverse relationship.

Quick checks and common points

  • Double-check with a sample x, e.g., ( x = 0 ):
    • ( g(0) = \dfrac{0 + 5}{2} = \dfrac{5}{2} )
    • ( f(g(0)) = f(5/2) = 2 \cdot \dfrac{5}{2} - 5 = 5 - 5 = 0 )
    • ( f(0) = -5 ) and ( g(-5) = \dfrac{-5 + 5}{2} = 0 )
    • Both compositions yield 0 when x = 0, consistent with the general result.

Summary of key formulas

  • Given: ( f(x) = 2x - 5 ), ( g(x) = \dfrac{x + 5}{2} )
  • Composition results:
    • ( (f \circ g)(x) = f(g(x)) = \boxed{x} )
    • ( (g \circ f)(x) = g(f(x)) = \boxed{x} )
  • Domain: ( x \in \mathbb{R} ) for both compositions.