Find a composite function, inverse, and range for a rational function

Expert intro: This question tests how well you can move between a function, its inverse, and its domain or range. Once you pin down the input restrictions, the rest follows in a fairly routine way.

Detailed walkthrough

Given

$g(x)=\frac{2x+5}{x-3},\quad x\ge 5$

(a) Find $g(g(5))$

First find $g(5)$:

$g(5)=\frac{2(5)+5}{5-3}=\frac{15}{2}$

Now substitute this into $g$ again:

$g\!\left(\frac{15}{2}\right)=\frac{2\left(\frac{15}{2}\right)+5}{\frac{15}{2}-3}$

$=\frac{15+5}{\frac{9}{2}}=\frac{20}{\frac{9}{2}}=\frac{40}{9}$

So

$g(g(5))=\frac{40}{9}$

(b) Find $g^{-1}(x)$, stating its domain

Let

$y=\frac{2x+5}{x-3}$

Swap $x$ and $y$:

$x=\frac{2y+5}{y-3}$

Solve for $y$:

$x(y-3)=2y+5$

$xy-3x=2y+5$

$xy-2y=3x+5$

$y(x-2)=3x+5$

$y=\frac{3x+5}{x-2}$

Hence

$g^{-1}(x)=\frac{3x+5}{x-2}$

Because the domain of $g$ is $x\ge 5$, the domain of $g^{-1}$ is the range of $g$. First find $g(5)=\frac{15}{2}$, and as $x\to\infty$, $g(x)\to 2$ from above. So the range of $g$ is

$2<y\le \frac{15}{2}$

Therefore the domain of $g^{-1}$ is

$2<x\le \frac{15}{2}$

(c) State the range of $g(x)$

The range is

$2<g(x)\le \frac{15}{2}$

If your course expects the endpoint behaviour rather than a fully justified interval, write the values as above and explain why $2$ is not reached.

💡 Pitfall guide

A common slip is to forget that the domain restriction $x\ge 5$ changes the range. Without that restriction, the function would behave differently.

Another trap is writing the inverse domain as all real numbers except $2$. That is the algebraic restriction of the inverse formula, but the actual domain must also reflect the original range.

🔄 Real-world variant

If the domain had been $x>5$ instead of $x\ge 5$, then the maximum value $g(5)=\frac{15}{2}$ would not be included, so the range would become

$2<y<\frac{15}{2}$

If you only needed the inverse formula and not the restricted domain, the algebraic result would still be

$g^{-1}(x)=\frac{3x+5}{x-2}$

but you would then need to attach the correct domain from the original function.

🔍 Related terms

inverse function, domain and range, composite function