Find the value, range, and inverse of a linear fractional function

Key takeaway: This question mixes direct substitution, domain restrictions, and inverse-function work. The key is to respect the given condition $x\ge 5$ while solving each part.

Given

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

(a) Find $g(5)$

Substitute $x=5$:

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

So

$\boxed{g(5)=\frac{15}{2}}.$

(b) State the range of $g$

Rewrite the function:

$g(x)=\frac{2x+5}{x-3}=2+\frac{11}{x-3}.$

Since $x\ge 5$, we have $x-3\ge 2$, so

$\frac{11}{x-3}>0.$

As $x$ increases, $\frac{11}{x-3}$ decreases towards $0$, so $g(x)$ decreases towards $2$ from above.

At $x=5$,

$g(5)=\frac{15}{2}.$

Therefore the range is

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

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

Start with

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

Swap $x$ and $y$:

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

Solve for $y$:

$x(y-3)=2y+5$ $xy-3x=2y+5$ $xy-2y=3x+5$ $y(x-2)=3x+5$ $y=\frac{3x+5}{x-2}.$

So

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

The domain of the inverse is the range of $g$, so

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

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Pitfalls the pros know 👇 A frequent mistake is to give the inverse domain as $x\ne 2$ just because the formula has a denominator. That is not enough here: the inverse must only accept values from the original range, so the correct domain is $2<x\le \frac{15}{2}$. Another easy slip is missing the restriction $x\ge 5$ when finding the range.

What if the problem changes? If the original domain were all real $x\ne 3$, then the range would exclude only the horizontal asymptote value $2$, so it would be $y\ne 2$. With the restriction $x\ge 5$, the range becomes a bounded interval instead. The inverse formula stays the same, but its domain changes with the original range.

Tags: inverse function, range, rational function