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Question 2.8.8: Showing That (g ° f )(x) Is Not Equivalent to (f ° g) (x) Le...

Showing That (g ° f )(x) Is Not Equivalent to (f ° g) (x)

Let ƒ(x) = 4x + 1 and g(x) = 2x² + 5x. Show that (g ∘ ƒ)(x) ≠ (ƒ ∘ g)(x).

(This is sufficient to prove that this inequality is true in general.)

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First, find (g ∘ ƒ)(x). Then find (ƒ ∘ g)(x).

\begin{array}{ll}(g ∘ ƒ)(x)\\= g(ƒ(x))&\text{ By definition}\\[0.5 cm]= g(4x +1 )& ƒ(x) = 4x + 1\\[0.5 cm]= 2(4x + 1)² + 5(4x + 1)&g(x) = 2x² + 5x\\[0.5 cm]= 2(16x² + 8x + 1) + 20x + 5     &\text{ Square 4x + 1 and apply}\\& \text{the distributive property.}\\[0.5 cm]= 32x² + 16x + 2 + 20x + 5&\text{Distributive property}\\[0.5 cm]= 32x² + 36x + 7&\text{Combine like terms.}\end{array}

(ƒ ∘ g)(x)

= ƒ (g(x))                              By definition

= ƒ (2x² + 5x)                      g(x) = 2x² + 5x

= 4 (2x² + 5x) + 1               ƒ(x) = 4x + 1

= 8x² + 20x + 1                   Distributive property

Thus, (g ∘ ƒ)(x) ≠ (ƒ ∘ g)(x).

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