Do I have to state that a function is continuous when using the IVT on an FRQ?

Yes. The continuity statement is a required rubric element. Even when continuity seems obvious (for a polynomial, say), you must explicitly write that the function is continuous on the closed interval, or you will lose the justification point.

Can I just plug the value into a limit and write the answer?

Only if direct substitution gives a defined number. If you get an indeterminate form like 0/0, you must show algebraic work — factoring, a conjugate, or a combined fraction — before evaluating. A bare final answer with no supporting manipulation typically earns fewer points.

How is a continuity-with-parameter part graded?

Graders reward the equation you set up between the one-sided limits and the function value, plus the correct solved value of the parameter. Show the equation explicitly rather than jumping straight to the number, since the setup is usually a separate point.

What is the biggest time-saver on Unit 1 FRQs?

Read all parts before starting. Later parts frequently reuse an earlier result, such as a parameter value or a limit, so a clear early answer prevents cascading errors and lets you carry results forward efficiently.

AP-CALCBC-1-FRQ

U1 FRQ Practice

Master AP Calculus BC Unit 1 free-response problems: limits by algebra, continuity with parameters, and IVT root-existence, scored the AP rubric way.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U1 FRQ Practice, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Unit 1 free-response questions rarely test just one skill. A single problem might ask you to evaluate a limit by factoring, then choose a parameter that makes a piecewise function continuous, then justify a root using the Intermediate Value Theorem. This guide shows you how to attack that multi-part format and, just as importantly, how to write answers that earn every rubric point.

The difference between a 3 and a 5 on these questions is usually communication, not computation. You will learn the exact phrases graders look for, how to state hypotheses before conclusions, and how to avoid the classic point-losers like a bare answer with no supporting work. We assume you already know the underlying concepts from Unit 1 — here we focus purely on execution under exam conditions.

How Unit 1 FRQs Are Structured and Scored

A Unit 1 free-response question typically has three or four parts labeled (a), (b), (c), each worth 1 to 3 points. Points are awarded for specific milestones, not for a correct final answer alone. A part worth 2 points might give 1 point for correct setup (the right limit expression or algebraic manipulation) and 1 point for the correct value.

The most important habit: show the intermediate step that the rubric rewards. For a limit solved by factoring, the graders want to see the factored form before you cancel. For continuity, they want to see the equation you set up between one-sided limits and the function value.

Task	What earns the point	Common way to lose it
Evaluate a limit	Show algebraic step, then value	Only writing the final number
Continuity parameter	Set up and solve an equation	Guessing the value with no equation
IVT justification	State continuity, compare signs, conclude	Skipping the continuity statement

Budget roughly 12 to 15 minutes per full FRQ. Read all parts first — later parts often reuse a result from an earlier part, so a mistake early can cascade. Write clearly and box or underline final answers.

Attacking Limit Parts by Algebraic Manipulation

Most Unit 1 limit parts hand you an indeterminate form

\frac{0}{0}

and expect you to resolve it. The graders want evidence of the technique, so always convert the indeterminate expression into a determinate one before substituting.

The three techniques you must recognize instantly are factoring and canceling, rationalizing with a conjugate, and combining complex fractions. For example,

\lim_{x\to 3}\frac{x^2-9}{x-3}

becomes

\lim_{x\to 3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\to 3}(x+3)=6

. Write the factored line — that middle step is often the point.

For a radical like

\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}

, multiply numerator and denominator by the conjugate

\sqrt{x+4}+2

to get

\lim_{x\to 0}\frac{x}{x(\sqrt{x+4}+2)}=\frac{1}{4}

.

A common misconception is that you may plug in the value first; if you do and get

\frac{0}{0}

, that is not an answer — it is a signal to manipulate. Another trap is forgetting one-sided behavior when the problem uses a piecewise definition. When a limit at a point requires a piecewise function, evaluate both

\lim_{x\to a^-}

and

\lim_{x\to a^+}

and state that they must be equal for the two-sided limit to exist.

Continuity With a Parameter

A signature Unit 1 FRQ move gives you a piecewise function containing an unknown constant, often called

k

c

, and asks for the value that makes the function continuous at the seam. The winning strategy is the formal three-part definition of continuity at

x=a

: the function value

f(a)

exists, the limit

\lim_{x\to a}f(x)

exists, and they are equal.

In practice you set the left-hand piece equal to the right-hand piece at the boundary point and solve for the parameter. If

f(x)=x^2+k

for

x\le 2

and

f(x)=3x-1

for

x>2

, continuity at

x=2

requires

2^2+k=3(2)-1

, so

4+k=5

and

k=1

. Show that equation explicitly.

When the boundary piece involves a removable discontinuity, you may need to simplify a fraction first, then evaluate the limit, then match. Be careful about which piece owns the endpoint: only the piece with the

\le

\ge

determines

f(a)

directly, but for continuity the limit must agree regardless. On the exam, state the equation, solve it, and write a brief conclusion like "so

f

is continuous at

x=2

when

k=1

IVT Root-Existence Justifications

The Intermediate Value Theorem is the most heavily rubric-driven part of Unit 1 FRQs because it demands a specific justification template. To earn full credit you must do three things in order: state that the function is continuous on the closed interval, show the function values at the endpoints straddle the target value (usually by opposite signs for a root), and then conclude that IVT guarantees a value

c

in the open interval.

A model sentence: "Because

f

is continuous on

[1,2]

and

f(1)=-3<0<4=f(2)

, by the Intermediate Value Theorem there exists a value

c

(1,2)

such that

f(c)=0

."

The number-one point-loser is skipping the continuity statement. IVT does not apply to discontinuous functions, so graders require it explicitly, even if continuity seems obvious. The second most common error is failing to actually compute both endpoint values — you must show the numbers, not just claim they have opposite signs. Note also that IVT proves existence, never uniqueness or an exact location. If a part asks "how many" roots, IVT alone cannot answer that; it only guarantees at least one.

Key terms

Indeterminate form.: An expression like $\frac{0}{0}$ that has no defined value on direct substitution and signals the need for algebraic manipulation before evaluating a limit.
One-sided limit.: The value a function approaches from only the left ( $x\to a^-$ ) or only the right ( $x\to a^+$ ); both must exist and be equal for the two-sided limit to exist.
Continuity at a point.: A function $f$ is continuous at $x=a$ when $f(a)$ exists, $\lim_{x\to a}f(x)$ exists, and $\lim_{x\to a}f(x)=f(a)$ .
Parameter.: An unknown constant in a function (often $k$ or $c$ ) whose value is chosen to satisfy a condition such as continuity.
Intermediate Value Theorem.: If $f$ is continuous on $[a,b]$ and $N$ is between $f(a)$ and $f(b)$ , then there exists $c$ in $(a,b)$ with $f(c)=N$ .
Conjugate.: The expression $\sqrt{x}+b$ paired with $\sqrt{x}-b$ ; multiplying by it rationalizes a radical and often clears an indeterminate form.
Removable discontinuity.: A point where a limit exists but does not equal the function value (or the value is undefined), typically producing a hole that cancels algebraically.

Worked example

Let

f(x)=\begin{cases}\dfrac{x^2-4}{x-2}, & x<2\\[4pt] k, & x=2\\[4pt] 3x-b, & x>2\end{cases}

. (a) Find

\lim_{x\to 2^-}f(x)

. (b) Find the value of

b

so that

\lim_{x\to 2}f(x)

exists. (c) Using your value of

b

, determine

k

f

is continuous at

x=2

, and justify that

f(x)=5

has a solution on

[2,4]

Part (a): For

x<2

, factor the numerator:

\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2

for

x\ne 2

. So

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

. Show the factored line to earn the setup point.

Part (b): The right-hand limit is

\lim_{x\to 2^+}(3x-b)=6-b

. For the two-sided limit to exist, the one-sided limits must be equal:

4=6-b

, so

b=2

. State the equation before solving.

Part (c): With

b=2

\lim_{x\to 2}f(x)=4

. For continuity at

x=2

we need

f(2)=k

to equal the limit, so

k=4

.

Now justify the equation

f(x)=5

[2,4]

. For

x\ge 2

f(x)=3x-2

, which is a polynomial and therefore continuous on

[2,4]

. Evaluate endpoints:

f(2)=3(2)-2=4

and

f(4)=3(4)-2=10

. Since

f

is continuous on

[2,4]

and

4<5<10

, by the Intermediate Value Theorem there exists a value

c

(2,4)

such that

f(c)=5

. The three rubric elements — continuity, straddling values, conclusion — are all present.

Practice questions

Evaluate

\lim_{x\to 0}\dfrac{\sqrt{x+9}-3}{x}

$0$
$\dfrac{1}{6}$
$\dfrac{1}{3}$
The limit does not exist

Answer: $\dfrac{1}{6}$

Direct substitution gives

\frac{0}{0}

, so multiply by the conjugate

\sqrt{x+9}+3

. The numerator becomes

(x+9)-9=x

, giving

\frac{x}{x(\sqrt{x+9}+3)}=\frac{1}{\sqrt{x+9}+3}

. Substituting

x=0

yields

\frac{1}{3+3}=\frac{1}{6}

Let

g(x)=\begin{cases}ax+1, & x\le 3\\ x^2-2, & x>3\end{cases}

. Find the value of

a

that makes

g

continuous at

x=3

, and show the equation you used.

Answer: $a=2$

Continuity at

x=3

requires the left piece to equal the right piece at the boundary:

a(3)+1=3^2-2

. This gives

3a+1=7

, so

3a=6

and

a=2

. The graded step is setting up the equation

3a+1=7

; simply stating

a=2

without it would lose the setup point.

A continuous function

h

satisfies

h(1)=-2

and

h(5)=6

. Explain fully why

h(x)=0

must have a solution on

(1,5)

, and state what the Intermediate Value Theorem does not guarantee here.

Answer: By IVT there exists at least one $c$ in $(1,5)$ with $h(c)=0$ ; it does not guarantee uniqueness or the location.

Because

h

is continuous on

[1,5]

and

h(1)=-2<0<6=h(5)

, the value

0

lies between the endpoint values, so the Intermediate Value Theorem guarantees at least one

c

(1,5)

with

h(c)=0

. The theorem is an existence statement only: it does not tell you how many roots exist or exactly where they are, and it cannot be applied without first stating continuity.

FAQ

Do I have to state that a function is continuous when using the IVT on an FRQ?: Yes. The continuity statement is a required rubric element. Even when continuity seems obvious (for a polynomial, say), you must explicitly write that the function is continuous on the closed interval, or you will lose the justification point.
Can I just plug the value into a limit and write the answer?: Only if direct substitution gives a defined number. If you get an indeterminate form like $\frac{0}{0}$ , you must show algebraic work — factoring, a conjugate, or a combined fraction — before evaluating. A bare final answer with no supporting manipulation typically earns fewer points.
How is a continuity-with-parameter part graded?: Graders reward the equation you set up between the one-sided limits and the function value, plus the correct solved value of the parameter. Show the equation explicitly rather than jumping straight to the number, since the setup is usually a separate point.
What is the biggest time-saver on Unit 1 FRQs?: Read all parts before starting. Later parts frequently reuse an earlier result, such as a parameter value or a limit, so a clear early answer prevents cascading errors and lets you carry results forward efficiently.

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