AP Calculus AB · Lesson 2

Continuity and the Intermediate Value Theorem

Know Know it

A function $f$ is continuous at $x = a$ when three things hold: $f(a)$ is defined; $\lim_{x \to a} f(x)$ exists; and $\lim_{x \to a} f(x) = f(a)$ . Check all three. If any one fails, $f$ is discontinuous at $a$ .

Polynomials, roots, exponentials, logarithms, sine and cosine, and rational functions are continuous wherever they are defined. For a piecewise function the places to check are the breakpoints: compare the limit from the left piece, the limit from the right piece, and the value assigned at the breakpoint.

The three kinds of discontinuity
Kind	What the limits do	What the graph shows
Removable	The limit exists, but $f(a)$ is missing or different	A hole, perhaps with a dot elsewhere
Jump	Both one-sided limits exist but differ	A break from one height to another
Infinite	The outputs grow without bound on at least one side	A vertical asymptote

Worked example Easy

Let $f(x) = x^2 + 1$ for $x < 2$ and $f(x) = 3x - 1$ for $x \ge 2$ . Which statement describes $f$ at $x = 2$ ?

$f$ has a jump discontinuity at $x = 2$
$f$ has a removable discontinuity at $x = 2$
$f$ has a vertical asymptote at $x = 2$
$f$ is continuous at $x = 2$ Answer

How to solve it

Left-hand limit: $\lim_{x \to 2^-}(x^2 + 1) = 5$ .
Right-hand limit and value: $\lim_{x \to 2^+}(3x - 1) = 5$ and $f(2) = 3(2) - 1 = 5$ .
All three agree, so $f$ is continuous at 2.

Why each choice is right or wrong

A. Incorrect. A jump needs the one-sided limits to differ, but both pieces give 5 at $x = 2$ .
B. Incorrect. A removable discontinuity needs $f(2)$ to be missing or different from the limit, but $f(2) = 5$ equals the limit.
C. Incorrect. Both pieces are polynomials, which stay bounded near 2, so there is no vertical asymptote at $x = 2$ .
D. Correct. The left piece gives $2^2 + 1 = 5$ , the right piece gives $3(2) - 1 = 5$ , and $f(2) = 5$ from the right piece, so all three conditions hold.

Worked example Medium

Let $g(x) = \frac{x^2 - 36}{x - 6}$ for $x \ne 6$ , and $g(6) = k$ . For what value of $k$ is $g$ continuous at $x = 6$ ?

$0$
$6$
$12$ Answer
$36$

How to solve it

Factor: $x^2 - 36 = (x - 6)(x + 6)$ , so $g(x) = x + 6$ for $x \ne 6$ .
So $\lim_{x \to 6} g(x) = 12$ .
Continuity needs $g(6)$ to equal that limit, so $k = 12$ .

Why each choice is right or wrong

A. Incorrect. 0 treats the $\frac{0}{0}$ form as the value 0, but the limit at 6 is found by cancelling first.
B. Incorrect. 6 is the input, not the limit of $g(x)$ as $x$ approaches 6.
C. Correct. For $x \ne 6$ , $g(x) = \frac{(x - 6)(x + 6)}{x - 6} = x + 6$ , which approaches 12. Setting $k = 12$ makes the value equal the limit.
D. Incorrect. 36 comes from $x^2$ at 6, but the top and bottom must be simplified together, giving $x + 6$ , which is 12 at 6.

Explain Explain it

Task verbs: justify, explain

The Intermediate Value Theorem (IVT): if $f$ is continuous on the closed interval $[a, b]$ and $N$ is strictly between $f(a)$ and $f(b)$ , then $f(c) = N$ for at least one $c$ strictly between $a$ and $b$ . It promises that such a $c$ exists. It does not find it, and it promises nothing when $f$ is not continuous on the whole interval.

A full IVT argument names three things: the function is continuous on the closed interval, the target value lies between the two endpoint values (say what they are), and the theorem by name with its conclusion.

Trap No continuity, no IVT

$\frac{1}{x}$ is $-1$ at $x = -1$ and $1$ at $x = 1$ , yet it never equals 0. It is not continuous on $[-1, 1]$ , because it is undefined at 0, so the IVT does not apply. State continuity before you use the theorem.

Worked example Hard

A function $f$ is continuous on $[1, 5]$ , with $f(1) = 4$ , $f(3) = -1$ and $f(5) = 2$ . What is the least number of solutions of $f(x) = 0$ that the Intermediate Value Theorem guarantees on $(1, 5)$ ?

Zero
One
TwoAnswer
More than two

How to solve it

On $[1, 3]$ : $f$ is continuous and 0 is between 4 and -1, so $f(c) = 0$ for some $c$ in $(1, 3)$ .
On $[3, 5]$ : 0 is between -1 and 2, so there is another zero in $(3, 5)$ .
The intervals do not overlap, so at least 2 zeros are guaranteed. There could be more, but the theorem does not promise them.

Why each choice is right or wrong

A. Incorrect. Zero ignores the sign changes: $f$ goes from 4 down to -1 and back up to 2, so it must cross 0 along the way.
B. Incorrect. One counts only the drop from 4 to -1 and misses the rise from -1 to 2 on $[3, 5]$ .
C. Correct. 0 is between 4 and -1, so a zero lies in $(1, 3)$ ; 0 is also between -1 and 2, so another lies in $(3, 5)$ . These intervals do not overlap, so at least two zeros are guaranteed.
D. Incorrect. More than two is not guaranteed: only two sign changes appear in the values, so the theorem promises two zeros and no more.

Justify Your turn

A function $p$ is continuous on $[0, 4]$ , with $p(0) = 7$ and $p(4) = -2$ . Show why $p(x) = 3$ for at least one $x$ between 0 and 4. Name the theorem you use and say why it applies.

Your answer

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A full answer does this

States that $p$ is continuous on $[0, 4]$ as the condition being used
Says that 3 lies between $p(4) = -2$ and $p(0) = 7$
Names the Intermediate Value Theorem as the reason

Model answer

$p$ is continuous on the closed interval $[0, 4]$ , and 3 lies between $p(4) = -2$ and $p(0) = 7$ . So by the Intermediate Value Theorem, $p(c) = 3$ for at least one $c$ with $0 < c < 4$ .

Apply Apply it

Calculator On the exam

On the exam, a graphing calculator is required in Part B of the multiple-choice section (13 of its 42 questions) and in Part A of the free-response section (2 of its 6 questions). It is not allowed in multiple-choice Part A or free-response Part B. So check continuity and work limits by hand, and remember that a calculator table only suggests a limit.

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Continuity and the Intermediate Value Theorem

Know Know it

Worked example Easy

How to solve it

Why each choice is right or wrong

Worked example Medium

How to solve it

Why each choice is right or wrong

Explain Explain it

Trap No continuity, no IVT

Worked example Hard

How to solve it

Why each choice is right or wrong

Justify Your turn

A full answer does this

Model answer

Apply Apply it

Calculator On the exam

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