Q.If $f:R\to R$ is defined by $f(x)=3x-5$, prove that $f$ is a bijection and find its inverse.

Composition. Given $f:X\to Y$ and $g:Y\to Z$, the composition $g\circ f:X\to Z$ is $(g\circ f)(x)=g(f(x))$ -- apply $f$ first, then $g$. More generally $g\circ f$ is defined whenever the range of $f$ is contained in the domain of $g$, even if their stated co-domain/domain don't literally match. Composition is, in general, not commutative: $f(x)=3x-4,\ g(x)=x^2+3$ give $(g\circ f)(x)=9x^2-24x+19$ but $(f\circ g)(x)=3x^2+5$ -- different. If $f,g$ are both one-to-one, then $g\circ f$ is one-to-one (chain the injectivity of each). The converse fails: $f$ and $g\circ f$ both one-to-one does not force $g$ to be one-to-one (a counterexample needs $g$ to collapse two points outside the range of $f$, which $g\circ f$ never "sees"). …