Before the midterm

What does it take for a graph to have no break?

Make two pieces meet, see why a matching limit is not enough without the value, and watch the Intermediate Value Theorem promise a crossing only where there is no break.

Builds on limits.

From the left, lim x→2⁻ f(x): 0.50
From the right, lim x→2⁺ f(x): 3.00
The value, f(2): 3.00
Continuous at 2?: no

Step 1 of 5: Make the pieces meet

This $f$ is $kx - 1$ for $x < 2$ and $4 - (x - 3)^2$ for $x \ge 2$ , so $f(2) = 3$ . For most slopes $k$ the left piece arrives somewhere else, and the graph breaks at $2$ . Drag the hollow end of the left piece, or use the slider.

Find the slope $k$ that makes the two pieces meet.

Slope of the left piece, k0.75

1. Make the pieces meet
This $f$ is $kx - 1$ for $x < 2$ and $4 - (x - 3)^2$ for $x \ge 2$ , so $f(2) = 3$ . For most slopes $k$ the left piece arrives somewhere else, and the graph breaks at $2$ . Drag the hollow end of the left piece, or use the slider.
Find the slope $k$ that makes the two pieces meet.
$k = 2$ : the left piece arrives at $2 \cdot 2 - 1 = 3$ , which is $f(2)$ . The limit from each side equals the value there, and that is what continuity at a point asks.
$\lim_{x \to 2^-} (kx - 1) = 2k - 1 = 3 = f(2) \quad\Longrightarrow\quad k = 2$
The figure: The graph of f(x) = 2.00x − 1 for x < 2 and 4 − (x − 3)² for x ≥ 2. From the left it arrives at 3.00, from the right at 3, and f(2) = 3.00: the pieces meet at the value, with no break.
2. The value counts too
Keep $k = 2$ , but give $f$ a value of its own at the join: $f(2) = 1$ . The pieces still head for $3$ from both sides. A common slip checks only that, the amber ring, and calls $f$ continuous.
Question: Is $f$ continuous at $2$ now?
No. The limit is $3$ and the value is $1$ . Continuity asks for three things: $f(2)$ defined, the limit existing, and the two equal. This break is removable: change the one value $f(2)$ to $3$ and it is gone.
$f \text{ is continuous at } a \iff \lim_{x \to a} f(x) \text{ exists and equals } f(a)$
The figure: The graph of f(x) = 2.00x − 1 for x < 2 and 4 − (x − 3)² for x ≥ 2, except that f(2) = 1. From the left it arrives at 3.00, from the right at 3, and f(2) = 1.00: there is a break at x = 2.
3. A break no value mends
Back to one formula on each side, now with $k = 1$ : the left piece arrives at $1$ and the right one starts at $3$ .
Question: Could some choice of the one value $f(2)$ make $f$ continuous at $2$ ?
No. The left side heads for $1$ and the right for $3$ , so $\lim_{x \to 2} f(x)$ does not exist, and no value can equal a limit that is not there. A jump needs the pieces themselves changed; only a removable break, like the last one, is mended by one value.
The figure: The graph of f(x) = 1.00x − 1 for x < 2 and 4 − (x − 3)² for x ≥ 2. From the left it arrives at 1.00, from the right at 3, and f(2) = 3.00: there is a break at x = 2.
4. What no break buys
With $k = 2$ again, $f$ is continuous on all of $[0, 4]$ , and it runs from $f(0) = -1$ , below the line $y = 0$ , to $f(4) = 3$ , above it.
Question: Without solving anything: must $f$ be $0$ somewhere between $0$ and $4$ ?
Yes. A graph with no break cannot get from below the line to above it without crossing it: here at $x = 0.5$ . Drag the line to any height between $-1$ and $3$ and the graph crosses it every time. That is the Intermediate Value Theorem.
$f \text{ continuous on } [a, b],\ f(a) < N < f(b) \implies f(c) = N \text{ for some } c \in (a, b)$
The figure: The graph of f(x) = 2.00x − 1 for x < 2 and 4 − (x − 3)² for x ≥ 2. From the left it arrives at 3.00, from the right at 3, and f(2) = 3.00: the pieces meet at the value, with no break. The line y = 0.00 runs across, from f(0) = −1 to f(4) = 3; the graph meets it on [0, 4] at x = 0.50.
5. No promise across a jump
Now $k = 0.25$ . Still $f(0) = -1$ and $f(4) = 3$ , one below $0$ and one above. A common slip applies the theorem anyway and promises a zero, bridging the jump at $x = 2$ in amber.
Question: Does $f$ have a zero between $0$ and $4$ ?
No. The left piece only climbs to $-0.5$ at $x = 2$ , and the graph jumps from there to $3$ , over the line. The theorem needs $f$ continuous at every point of $[0, 4]$ , and at $2$ it is not; without that, it promises nothing.
The figure: The graph of f(x) = 0.25x − 1 for x < 2 and 4 − (x − 3)² for x ≥ 2. From the left it arrives at −0.50, from the right at 3, and f(2) = 3.00: there is a break at x = 2. The line y = 0.00 runs across, from f(0) = −1 to f(4) = 3; the graph meets it on [0, 4] nowhere. A dashed bridge marks the crossing the slip promises across the jump.

Step 1 of 5: Make the pieces meet

The lesson as text

1. Make the pieces meet

2. The value counts too

3. A break no value mends

4. What no break buys

5. No promise across a jump