Graphs of Continuous Functions, calculus lesson

A graph is often the first thing you see of a function, and the eye spots a break in a curve instantly. Reading a graph well still takes care. A tiny circle can decide whether a function is continuous, and some breaks are invisible on a screen while some smooth functions look broken.

Key terms

Open and closed points

Definition The graph of a function $f$ with domain $D$ is the set of points $\{(x, f(x)) \mid x \in D\}$ . In a drawing, a closed (filled) dot at $(c, y_0)$ means that point is on the graph, so $f(c) = y_0$ . An open (hollow) dot at $(c, y_0)$ means that point is not on the graph, even though a piece of the curve runs right up to it. Above any one $x$ there is at most one closed dot, because a function has only one value at each input.

In plain words A filled dot shows where the function actually is. A hollow dot shows where the curve was heading but did not land.

Example Suppose a graph shows the line $y = x + 1$ with a hollow dot at $(1, 2)$ and a filled dot at $(1, 4)$ . Then $f(1) = 4$ , while $\lim_{x \to 1} f(x) = 2$ . The limit and the value disagree, so $f$ is not continuous at $1$ .

One-sided continuity on a graph

Definition Let $f$ be defined on $[c, c + r)$ for some $r > 0$ . Then $f$ is continuous from the right at $c$ when $\lim_{x \to c^+} f(x) = f(c)$ . On a graph this means the piece of curve just to the right of $x = c$ runs into the closed dot above $c$ . Continuity from the left is the same with the piece just to the left of $c$ and $\lim_{x \to c^-} f(x) = f(c)$ .

In plain words Follow the curve toward $x = c$ from one side. If you arrive exactly at the filled dot, the function is continuous from that side.

Example A step graph has the segment $y = 0$ for $0 \le x < 2$ ending in a hollow dot at $(2, 0)$ , and the segment $y = 3$ for $x \ge 2$ starting with a filled dot at $(2, 3)$ . Coming from the right you land on the filled dot, so $f$ is continuous from the right at $2$ . Coming from the left you reach height $0$ , not $3$ , so it is not continuous from the left.

Endpoints of a graph

Definition If the domain of $f$ is a closed interval $[a, b]$ , the graph begins above $a$ and ends above $b$ . The function is continuous at $a$ when $\lim_{x \to a^+} f(x) = f(a)$ and at $b$ when $\lim_{x \to b^-} f(x) = f(b)$ . Nothing outside $[a, b]$ is checked, because there is no graph there. If an endpoint is not in the domain, as in $(a, b]$ , the graph shows a hollow dot or simply stops, and continuity at that endpoint is not a question at all.

In plain words At the edge of the domain the curve only needs to reach its endpoint dot from the inside.

Example The graph of $f(x) = \sqrt{4 - x^2}$ is the upper half of the circle of radius $2$ , from $(-2, 0)$ to $(2, 0)$ . Both ends are filled dots, and the curve runs into each of them from inside, since

\lim_{x \to 2^-} \sqrt{4 - x^2} = 0 = f(2)

and similarly at $-2$ . So $f$ is continuous on $[-2, 2]$ , even though the curve stops at both ends.

Spotting discontinuities on a graph

Definition At a point $c$ where $f$ is defined on both sides, read three things from the graph. These are the height the curve approaches from the left, the height it approaches from the right, and the closed dot above $c$ , if any. Then

a hole, with or without a filled dot elsewhere above $c$ , and both sides approaching one height, is a removable discontinuity,
two sides approaching two different finite heights is a jump discontinuity,
a side that climbs or falls without bound along a vertical asymptote is an infinite discontinuity,
a side that keeps swinging up and down ever faster near $c$ is an oscillating discontinuity,
both sides approaching the height of the filled dot above $c$ is continuity at $c$ .

In plain words Ask where the left side goes, where the right side goes, and where the dot is. If all three agree, the function is continuous there.

Example A graph has these features.

Point	Left side heads to	Right side heads to	Filled dot	Verdict
$x = -1$	$2$	$2$	none	removable
$x = 1$	$1$	$3$	$(1, 3)$	jump, right continuous
$x = 3$	$\infty$	$-\infty$	none	infinite
$x = 5$	$0$	$0$	$(5, 0)$	continuous

What a graph cannot prove

Definition A plotted graph shows $f$ at finitely many sampled inputs, usually joined by line segments, drawn with lines of positive width. Continuity at $c$ is a statement about all inputs arbitrarily close to $c$ . So a plot is evidence, not proof. It can hide a discontinuity at a single point, it can draw a steep continuous function as if it jumped, and it can smear fast oscillations into a solid band. To decide continuity, use the formula and the definition.

In plain words A picture can suggest the answer. Only the algebra and limits settle it.

Example A plot of the function

f(x) = \frac{x^2 - 1}{x - 1}

looks exactly like the line $y = x + 1$ , with no visible gap, yet $f$ is undefined at $1$ . In the other direction, the function

g(x) = \frac{x}{\sqrt{x^2 + 10^{-6}}}

is continuous everywhere, since the denominator is never zero. On the window $-1 \le x \le 1$ its plot looks like a jump from $-1$ to $1$ at $0$ .

$x$	$-0.01$	$-0.001$	$0$	$0.001$	$0.01$
$g(x)$	$-0.995$	$-0.707$	$0$	$0.707$	$0.995$

Almost the whole climb from $-1$ to $1$ happens within $0.01$ of the origin. On a plot of width $2$ that strip is only a few pixels wide, so the steep part looks like a vertical jump.

Check yourself

A graph shows a curve approaching height $4$ from both sides of $x = 2$ , a hollow dot at $(2, 4)$ , and a filled dot at $(2, 1)$ . What is $f(2)$ , what is $\lim_{x \to 2} f(x)$ , and is $f$ continuous at $2$ ?

Show the answer

The value is $f(2) = 1$ and the limit is $\lim_{x \to 2} f(x) = 4$ . They differ, so $f$ is not continuous at $2$ . The discontinuity is removable.

A function on $[0, 3]$ has a graph that starts with a hollow dot at $(0, 1)$ and a filled dot at $(0, 2)$ , and the curve to the right of $0$ approaches height $1$ . Is $f$ continuous on $[0, 3]$ ?

Show the answer

No. At the left endpoint only right continuity matters, and $\lim_{x \to 0^+} f(x) = 1 \neq 2 = f(0)$ .

Why can no screen plot show the discontinuity of $\frac{x^2 - 1}{x - 1}$ at $x = 1$ unless the software marks it on purpose?

Show the answer

Only one point, $(1, 2)$ , is missing, and the plotted line has positive width, so the hole is covered by the neighboring samples.

Graphs of Continuous Functions

Theory, part 31

Practice, part 32

Key terms

Open and closed points

One-sided continuity on a graph

Endpoints of a graph

Spotting discontinuities on a graph

What a graph cannot prove

Check yourself

How this lesson connects

Builds on