Finding the inverse of a rational function with x over x plus 2

Key concept: This problem tests inverse-function algebra for a rational expression, where the key step is clearing the denominator after swapping variables.

Step by step

Start with the original function

We are given

$f(x)=\frac{x}{x+2}$

To find the inverse, replace $f(x)$ with $y$ and then swap $x$ and $y$:

$y=\frac{x}{x+2}$

Swap the variables:

$x=\frac{y}{y+2}$

Solve for y carefully

Multiply both sides by $y+2$:

$x(y+2)=y$

Distribute $x$:

$xy+2x=y$

Move the $y$ terms to one side:

$2x=y-xy$

Factor out $y$ on the right side:

$2x=y(1-x)$

Now divide by $1-x$:

$y=\frac{2x}{1-x}$

So the inverse is

$f^{-1}(x)=\frac{2x}{1-x}$

Why this works

Inverse functions undo the original function. The original rule takes an input, divides it by the input plus 2, and the inverse must reverse that process.

A good algebra check is to substitute the inverse back into the original function and simplify. If the result returns the original variable, the inverse is correct.

Important domain idea

The original function cannot use $x=-2$, because that would make the denominator zero. The inverse also has a restriction: $1-x\neq 0$, so $x\neq 1$. Domain and range switch roles when you find an inverse.

Common algebra pattern

For rational inverses, the main challenge is clearing the denominator without losing terms. Multiply both sides by the full denominator before expanding. That keeps the equation accurate and avoids mistakes with fractions.

Pitfall alert

A common mistake is multiplying only one term by the denominator instead of the entire left side. In $x=\frac{y}{y+2}$, the factor $y+2$ must multiply the whole left side, giving $x(y+2)=y$. Another frequent error is stopping after obtaining an equation like $2x=y(1-x)$ and forgetting to isolate $y$. Also watch the domain: the inverse cannot include values that make its denominator zero, so you must state $x\neq 1$ for the inverse. Those restrictions matter just as much as the algebraic formula.

Try different conditions

If the function were changed to $f(x)=\frac{x}{x-2}$, the same swap-and-solve method would work, but the algebra would produce a different inverse. Starting from $x=\frac{y}{y-2}$, you would get $x(y-2)=y$, then $xy-2x=y$, and finally $y=\frac{2x}{x-1}$. If the numerator were $2x$ instead of $x$, such as $f(x)=\frac{2x}{x+2}$, the inverse would also change because the scaling affects every step of the isolation process.

Further reading

rational function, inverse function, domain and range