12
$\begingroup$

If I know the graph of $f(x)$, how do I draw the graph of $\frac{1}{f(x)}$?

$\endgroup$
20
$\begingroup$

If you wish to sketch $y=\frac{1}{f(x)}$ given the graph of $y=f(x)$, some things to note are:

  • $f(x)$ and $\frac{1}{f(x)}$ have the same sign
  • $y=\frac{1}{f(x)} \implies \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{f'(x)}{f(x)^2}$; so $\frac{1}{f(x)}$ increases as $f(x)$ decreases and vice versa (minimum turning points become maximum turning points, etc.)
  • $x$-intercepts of $y=f(x)$ correspond to vertical asymptotes of $y=\frac{1}{f(x)}$
  • As $f(x) \to \infty$, $\frac{1}{f(x)}\to0$
  • If $f(x) > 1$, $0 < \frac{1}{f(x)} < 1$; similarly if $0 < f(x) < 1$, $\frac{1}{f(x)}>1$

This should make some sense with an example:

graph

Note that the dotted horizontal lines represent $y=1$ and $y=-1$. The dashed vertical lines are asymptotes.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ben, thanks very much for such detailed answer and picture!! $\endgroup$ – AGamePlayer Dec 24 '12 at 2:26
9
$\begingroup$

Any points where it's at $1$ or $-1$ will remain the same, so figuring out those will help.

Maxima will become minima and vice-versa, with $y$ value as the inverse of whatever they were before. Roots will become asymptotes (thanks to $1/0$), and vice-versa. Positive/negative will stay the same. I would find all the major features of the original function (roots, maxima, etc.) and invert those, connecting the graph smoothly. For example:

$y=x$. Has a root at $0$, which will become an asymptote. Reaches $1$ at $x=1$ and $-1$ at $x=-1$, and these points will stay the same. Positive/negative will stay the same. $y=x$ blows up to $\pm\infty$ on both sides, so inverting it should have it decrease asymptotically to zero in each case. It has no relative maxima/minima, so $1/x$ shouldn't either. This all taken together roughly suggests the proper hyperbolic shape of $1/x$. Similar ideas should work for more general functions.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.