How do you find domain and range of inverse functions f(x) = 1/(x-2)?

Jul 27, 2015

Most important limitation is that the denominator may not be $0$

Explanation:

So the domain is limited by $x \ne 2$

As $x \to 2$ the fraction gets larger and larger, whether you go to $2$ from above or below. In "the language":

${\lim}_{x \to {2}^{-}} f \left(x\right) = - \infty$ and ${\lim}_{x \to {2}^{+}} f \left(x\right) = + \infty$

If $x$ gets larger the fraction will become smaller, but never quite reaching $0$, so the range is $f \left(x\right) \ne 0$ or:

${\lim}_{x \to - \infty} f \left(x\right) = 0$ and ${\lim}_{x \to + \infty} f \left(x\right) = 0$
graph{1/(x-2 [-10, 10, -5, 5]}

$x = 2 \mathmr{and} f \left(x\right) = 0$ are called asymptotes .

Jul 27, 2015

The domain of an inverse function is the range of the function and vice versa.

Explanation:

For $f \left(x\right) = \frac{1}{x - 2}$, the domain of $f$ is all reals except $2$, so the range of ${f}^{-} 1$ is all reals except $2$.

The range of $f$ is all reals except $0$, so the domain of ${f}^{-} 1$ is all reals except $0$.

Notice that is we solve $y = \frac{1}{x - 2}$ for $x$, we get:

$y \left(x - 2\right) = 1$

$x y - 2 y = 1$

$x y = 2 y + 1$

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

We can see from this that for the original function, $f$, we can get every number for $y$ except $0$.
That is the range of $f$ and the domain of ${f}^{-} 1$