Quick questions on Composite and inverse functions - TCE Mathematics Specialised (Tasmania)

The composite function $f \circ g$ is defined by $(f \circ g)(x) = f(g(x))$: apply $g$ first, then $f$. Order matters, because in general $f(g(x)) \ne g(f(x))$.

The inverse $f^{-1}$ undoes $f$, so $f^{-1}(f(x)) = x$. An inverse function exists only when $f$ is one-to-one, meaning no horizontal line crosses the graph more than once. If $f$ is many-to-one, you must restrict its domain to a piece on which it is one-to-one before an inverse exists.

To find an inverse: write $y = f(x)$, swap $x$ and $y$, then solve for $y$. State the domain of the inverse, which equals the range of the original.

A reliable check is to confirm $f(f^{-1}(x)) = x$. Here $f(f^{-1}(x)) = (\sqrt{x + 4})^2 - 4 = x + 4 - 4 = x$ on the stated domain. If the composition does not return $x$, an algebra error or a domain mistake has crept in.

Many TASC questions ask for the inverse of a rational function of the form $f(x) = \dfrac{ax + b}{cx + d}$. The method is always the same: write $y = f(x)$, swap $x$ and $y$, then solve the resulting equation for $y$ by clearing the denominator and collecting the $y$ terms on one side. The domain of the inverse is found from the range of the original, and for these functions the excluded value of the inverse is the horizontal asymptote of $f$ (namely $y = \tfrac{a}{c}$).