Sketching reciprocal functions

Sketch y = 1/f(x) from the graph of y = f(x): sign, zeroes and vertical asymptotes, horizontal asymptotes, turning points, domain and range.

From f(x) to 1/f(x)

Every feature of $y = \dfrac{1}{f(x)}$ can be read from $y = f(x)$, point by point, because each $y$-value is replaced by its reciprocal.

On $y = f(x)$	On $y = \dfrac{1}{f(x)}$
positive / negative	same sign
a zero at $x = a$	a vertical asymptote $x = a$
$f(x) \to \pm\infty$ (at a vertical asymptote or as $x \to \pm\infty$)	$\dfrac{1}{f(x)} \to 0$
a horizontal asymptote $y = c$, with $c \ne 0$	a horizontal asymptote $y = \dfrac{1}{c}$
$f(x) = 1$ or $f(x) = -1$	the same point (it is fixed)
increasing	decreasing (and vice versa)
a local minimum at $(a, b)$, with $b \ne 0$	a local maximum at $\left(a, \dfrac{1}{b}\right)$ (and vice versa)

y = 1/(x² − k) from y = x² − k

Blue dashed: y = f(x) = x² − k. Orange: its reciprocal. The two graphs cross wherever f(x) = ±1.

Try this

Slide $k$ from $4$ down to $-3$. When do the vertical asymptotes disappear, and what is the range of the reciprocal then?

Example 1●●

Sketch $y = \dfrac{1}{x^2 - 4}$ using $f(x) = x^2 - 4$.

Vertical asymptotes $x = 2$ and $x = -2$.

The zeroes of f are ±2.

The reciprocal is positive for $|x| > 2$ and negative for $|x| < 2$.

f is positive outside ±2 and negative between them.

The reciprocal has a local maximum at $\left(0, -\tfrac14\right)$.

f has a minimum (0, −4).

Horizontal asymptote $y = 0$, approached from above.

f → ∞ as x → ±∞.

Example 2: domain and range●●

Find the domain and range of $y = \dfrac{1}{x^2 + 1}$.

$x^2 + 1$ is never zero, so the domain is all real $x$.

Range: $0 < y \le 1$, with maximum $1$ at $x = 0$.

The range of x² + 1 is y ≥ 1. Reciprocals of numbers from 1 to ∞ run from 1 down towards 0.

Flashcards

Zero of $f$ at $x = a$

Vertical asymptote of $\dfrac{1}{f}$ at $x = a$.

$f(x) o infty$

$\dfrac{1}{f(x)} \to 0$

Local minimum $(a, b)$ of $f$

Local maximum $\left(a, \dfrac{1}{b}\right)$ of $\dfrac{1}{f}$

Fixed points

Where $f(x) = 1$ or $f(x) = -1$.

Practice

The zeroes of $y = f(x)$ become, on $y = \dfrac{1}{f(x)}$:
Find the $y$-coordinate of the turning point of $y = \dfrac{1}{x^2 - 4}$.
The range of $y = \dfrac{1}{x^2 + 1}$ is:
Where $f$ is increasing (and not zero), $\dfrac{1}{f}$ is:
$y = f(x)$ has horizontal asymptote $y = 4$. What horizontal asymptote does $y = \dfrac{1}{f(x)}$ have? Enter the $y$-value.
$f(3) = -5$. Find the value of $\dfrac{1}{f(x)}$ at $x = 3$.
$y = f(x)$ has a local minimum at $(2, 3)$. Then $y = \dfrac{1}{f(x)}$ has:
Which points lie on both $y = f(x)$ and $y = \dfrac{1}{f(x)}$?
How many vertical asymptotes does $y = \dfrac{1}{x^2 - 5x + 6}$ have?
Where $y = f(x)$ is negative, $y = \dfrac{1}{f(x)}$ is: