Exercise · ZW-COMPOSITION-AND-INVERSE-007 · ★★☆ Intermediate

Is \(g\circ f=f\circ g\) always true?

Choose one option.

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  • D. No, in general it does not commute

Explanation

With \(f(x)=x+1\) and \(g(x)=2x\), we get \(g\circ f=2x+2\) and \(f\circ g=2x+1\), which differ.

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Review the concept: Composition and inverse

The composition \(g\circ f\) applies \(f\) first and then \(g\). The inverse \(f^{-1}\) exists when \(f\) is bijective and undoes its action.

\[(g\circ f)(x)=g(f(x)), \quad (g\circ f)^{-1}=f^{-1}\circ g^{-1}\]

More practice problems: composition and inverse

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