Continuity and piecewise-defined functions

Continuity at a point

Informally, a function is continuous at $x = a$ if you can draw its graph through $x = a$ without lifting your pen.

Graphs can fail at a point in three ways: a hole (as in $\dfrac{x^2 - 9}{x - 3}$ at $x = 3$), a jump (where two pieces don't meet), or an asymptote (as in $\dfrac{1}{x}$ at $x = 0$).

Piecewise-defined functions

A piecewise-defined function uses different rules on different parts of its domain:

$$f(x) = \begin{cases} x^2, & x < 1 \\ 3 - 2x, & x \ge 1 \end{cases}$$

To evaluate, first decide which piece the input belongs to: $f(-2) = 4$ (top piece) and $f(3) = -3$ (bottom piece).

Left of x = 1: y = x². From x = 1 on: y = mx + c. Adjust m and c until there is no jump.

Try this

• Keep $m = -2$. Which value of $c$ joins the pieces?
• Find another line that joins up and leaves no corner. (Hint: try $m = 2$.)

Example 1●●

Is $f(x)$ above continuous at $x = 1$?

Left piece: $1^2 = 1$.

Approaching from the left, use x².

Right piece: $3 - 2(1) = 1$, and $f(1) = 1$.

At and to the right of 1, use 3 − 2x.

Both pieces give $1$, so $f$ is continuous at $x = 1$.

Example 2●●●

Find $k$ so that $g(x) = \begin{cases} kx + 1, & x \le 2 \\ x^2 - 3, & x > 2 \end{cases}$ is continuous.

$2k + 1 = 2^2 - 3 = 1$

The two pieces must give the same value at x = 2.

$k = 0$

Flashcards

Continuous at x = a

No break: $f(a)$ exists and both sides approach $f(a)$.

Three ways to be discontinuous

A hole, a jump, or an asymptote.

Pieces join up

Substitute the boundary value into both rules and set them equal.

Practice

1. $f(x) = \begin{cases} 2x + 1, & x \ge 0 \\ x^2, & x < 0 \end{cases}$. Find $f(4)$.

2. With the same $f$, find $f(-3)$.

3. Is that $f$ continuous at $x = 0$?

   YesNo, it jumps from $0$ to $1$No, it has an asymptoteIt is not defined at $0$

4. $y = \dfrac{1}{x - 2}$ is discontinuous at:

   $x = 0$$x = -2$$x = 2$nowhere

5. Find $a$ so that $\begin{cases} x + a, & x < 1 \\ 5x, & x \ge 1 \end{cases}$ is continuous.

6. Which function is continuous for all real $x$?

   $y = |x|$$y = \dfrac{1}{x}$$y = \dfrac{x^2 - 1}{x - 1}$$y = \sqrt{x}$

7. $\dfrac{x^2 - 4}{x - 2}$ has what kind of discontinuity at $x = 2$?

   a jumpan asymptotenonea hole

8. $|x|$ written piecewise is $x$ for $x \ge 0$ and what for $x < 0$? Enter the value of that piece at $x = -2$.

Try again