Day 24: AP Calculus Theorems Review

The Intermediate Value Theorem (IVT)

IVT guarantees that a continuous function takes every value between its endpoint values. On the FRQ it is how you prove an equation has a solution, or that some output is achieved, without ever solving.

Core Theorem

If

f

is continuous on the closed interval

[a, b]

and

N

is any number between

f(a)

and

f(b)

, then there is at least one

c

in

(a, b)

with

f(c) = N

.
Hypotheses to cite: (1)

f

continuous on

[a, b]

, (2)

N

strictly between

f(a)

and

f(b)

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

A differentiable function

g

has

g(2) = -1

and

g(7) = 4

. Justify that

g(x) = 0

for some

x

in

(2, 7)

.

How to think about it

State that the function is continuous, show the target value lies between the two endpoint values, then conclude. Skipping the continuity sentence loses the point even if the arithmetic is right.

Key idea: Differentiable implies continuous. Is $0$ between $g(2)$ and $g(7)$ ?

Full solution

1. State the hypotheses

g

is differentiable on

[2, 7]

, so

g

is continuous on

[2, 7]

.

2. Locate the target value

g(2) = -1 < 0 < 4 = g(7)

, so

0

is between

g(2)

and

g(7)

.

3. Invoke the theorem

By the Intermediate Value Theorem, there is a

c

in

(2, 7)

with

g(c) = 0

.

Step-by-Step SOP

1
Check continuity on the closed interval
Say it explicitly — "differentiable, hence continuous" counts.
2
Show the target value is between the endpoint values
Write the inequality $f(a) < N < f(b)$ (or the reverse).
3
Write the conclusion
"By the IVT, there exists $c$ in $(a, b)$ with $f(c) = N$ ."

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

A table gives

h(1) = 3

,

h(3) = 3

,

h(6) = 8

, with

h

continuous. Must there be an

x

in

(1, 6)

with

h(x) = 5

?

Need a hint?

You only need one sub-interval whose endpoint values straddle 5.

Show solution

1. Pick a sub-interval that works

On

[3, 6]

,

h

is continuous and

h(3) = 3 < 5 < 8 = h(6)

.

2. Apply IVT

By the IVT, there is an

x

in

(3, 6) \subset (1, 6)

with

h(x) = 5

. Yes.

Common Pitfalls

⚠
Not stating continuityThe single most common lost point. A justification with no mention of continuity earns nothing, even if the arithmetic is right.
⚠
Using IVT to claim a unique solutionIVT only guarantees at least one $c$ . To claim exactly one, you also need monotonicity (e.g. $f' > 0$ ).

The Extreme Value Theorem (EVT) and the Closed Interval Method

EVT guarantees that a continuous function on a closed interval actually attains an absolute maximum and an absolute minimum. The Closed Interval Method is how you find them.

Core Theorem

If

f

is continuous on the closed interval

[a, b]

, then

f

attains an absolute maximum value and an absolute minimum value on

[a, b]

.
To find them (Closed Interval Method): evaluate

f

at every critical number in

(a, b)

and at both endpoints; the largest output is the absolute max, the smallest is the absolute min.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Find the absolute maximum and minimum of

f(x) = x^3 - 3x

on

[0, 2]

.

How to think about it

On a closed interval the extrema are among a short list: critical numbers inside

(a,b)

plus both endpoints. Evaluate

f

at each and compare — and report the output value, not the input.

Key idea: Find critical numbers from $f'(x) = 0$ , keep only those in $(0, 2)$ , then test them against the endpoints.

Full solution

1. Find critical numbers

f'(x) = 3x^2 - 3 = 0 \implies x = \pm 1

. Only

x = 1

is in

(0, 2)

.

2. Evaluate at the critical number and endpoints

f(0) = 0

,

f(1) = 1 - 3 = -2

,

f(2) = 8 - 6 = 2

.

3. Compare

Absolute maximum

= 2

at

x = 2

; absolute minimum

= -2

at

x = 1

.

Step-by-Step SOP

1
Confirm continuity on a closed interval
EVT needs both — a closed interval and continuity.
2
Find critical numbers in the open interval
Solve $f'(x) = 0$ and find where $f'$ is undefined; discard any outside $(a, b)$ .
3
Evaluate f at those points and at a and b
Build a short list of outputs.
4
Report the largest and smallest
Name the value and where it occurs.

Common Pitfalls

⚠
Testing only critical numbersThe endpoints are candidates too. Absolute extrema on a closed interval very often occur at $a$ or $b$ .
⚠
Confusing the x-value with the extreme valueThe absolute maximum is the output $f(c)$ , not the input $c$ . Answer with the value; also state where it happens.

The Mean Value Theorem (MVT) and Rolle's Theorem

MVT guarantees a point where the instantaneous rate of change equals the average rate of change over the interval. It is how the FRQ makes you prove that some derivative value must occur.

Core Theorem

If

f

is continuous on

[a, b]

and differentiable on

(a, b)

, then there is a

c

in

(a, b)

with

f'(c) = \frac{f(b) - f(a)}{b - a}.

Rolle's Theorem is the special case

f(a) = f(b)

, which forces

f'(c) = 0

.
Hypotheses to cite: (1) continuous on

[a, b]

, (2) differentiable on

(a, b)

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

f

is differentiable,

f(1) = 4

, and

f(5) = 16

. Justify that

f'(c) = 3

for some

c

in

(1, 5)

.

How to think about it

On the review, the credit comes from stating the hypotheses, not the arithmetic. Say that

f

is continuous on the closed interval and differentiable on the open one, compute the average rate of change, then name the MVT as the reason some

c

hits it.

Key idea: Compute the average rate of change and cite the two hypotheses.

Full solution

1. State the hypotheses

f

is differentiable on

[1, 5]

, so it is continuous on

[1, 5]

and differentiable on

(1, 5)

.

2. Compute the average rate of change

\dfrac{f(5) - f(1)}{5 - 1} = \dfrac{16 - 4}{4} = 3

.

3. Invoke MVT

By the Mean Value Theorem, there is a

c

in

(1, 5)

with

f'(c) = 3

.

Step-by-Step SOP

1
State both hypotheses
Continuous on the closed interval AND differentiable on the open interval.
2
Compute the secant slope
$\frac{f(b) - f(a)}{b - a}$ .
3
Write the conclusion
"By the MVT, there exists $c$ in $(a, b)$ with $f'(c) =$ [that slope]."

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

A car's position is

s(t)

, continuous and differentiable, with

s(0) = 0

and

s(2) = 120

(miles, hours). Explain why the car's speed was exactly 60 mph at some instant.

Need a hint?

Average velocity over

[0, 2]

is the secant slope. Speed = magnitude of velocity.

Show solution

1. Average velocity

\dfrac{s(2) - s(0)}{2 - 0} = \dfrac{120}{2} = 60

mph.

2. Apply MVT

s

is continuous on

[0, 2]

and differentiable on

(0, 2)

, so by the MVT there is a time

c

in

(0, 2)

with

s'(c) = 60

— the car's velocity, hence its speed, was 60 mph at that instant.

Common Pitfalls

⚠
Dropping the differentiability hypothesisMVT needs differentiability on the open interval. A function with a corner in $(a, b)$ can fail the conclusion.
⚠
Using MVT when only a table is givenIf the problem says "differentiable" you may apply MVT. If it only says "continuous," you get IVT, not MVT.

The Mean Value Theorem for Integrals

This is the integral analogue of MVT: a continuous function attains its own average value somewhere on the interval. It links the average-value formula to an actual point on the graph.

Core Theorem

If

f

is continuous on

[a, b]

, then there is a

c

in

(a, b)

with

f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx.

The right-hand side is the average value of

f

on

[a, b]

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

For

f(x) = x^2

on

[0, 3]

, find the value

c

guaranteed by the Mean Value Theorem for Integrals.

How to think about it

This is the average-value theorem: some

c

in the interval attains the function's average value. Compute

f_{\text{avg}} = \frac{1}{b - a}\int_a^b f\,dx

first, then solve

f(c) = f_{\text{avg}}

and keep only the solution inside the interval.

Key idea: Compute the average value first, then solve $f(c) =$ average value with $c$ in $(0, 3)$ .

Full solution

1. Average value

\dfrac{1}{3}\int_0^3 x^2\,dx = \dfrac{1}{3}\cdot 9 = 3

.

2. Solve f(c) = average value

c^2 = 3 \implies c = \sqrt{3} \approx 1.732

(reject

-\sqrt{3}

, not in the interval).

Step-by-Step SOP

1
Confirm f is continuous on [a, b]
That is the only hypothesis.
2
Compute the average value
$\frac{1}{b - a}\int_a^b f\,dx$ .
3
Solve f(c) = average value
Keep only solutions in $(a, b)$ .

Common Pitfalls

⚠
Confusing it with MVT for derivativesMVT for Integrals is about the average value of $f$ ; MVT for derivatives is about the average rate of change (a slope). Read which one the interval and hypotheses point to.

The Fundamental Theorem of Calculus (Part 1 & Part 2)

The FTC ties integration and differentiation together. Part 1 evaluates a definite integral with an antiderivative; Part 2 differentiates a function defined by an integral. Both show up on nearly every FRQ.

Core Theorem

Part 1: if

f

is continuous on

[a, b]

and

F' = f

, then

\int_a^b f(x)\,dx = F(b) - F(a)

.
Part 2: if

g(x) = \int_a^{u(x)} f(t)\,dt

with

f

continuous, then

g'(x) = f(u(x))\cdot u'(x)

.
Accumulation form:

F(b) = F(a) + \int_a^b F'(x)\,dx

— "final = initial + accumulated change."

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Water flows into a tank at rate

R(t)

gallons/hour. The tank holds 50 gallons at

t = 0

. Write an expression for the amount at

t = 8

and explain it.

How to think about it

Use the accumulation form: final amount = initial amount +

\int

of the rate. The integral of a rate is a net change, not a running total — so it gets added to the starting 50 gallons, never reported on its own.

Key idea: Use the accumulation form: final amount = initial amount + net change.

Full solution

1. Set up the accumulation expression

Amount

(8) = 50 + \int_0^8 R(t)\,dt

.

2. Explain each piece

50

is the starting amount;

\int_0^8 R(t)\,dt

is the total gallons that flowed in over the 8 hours (net change), by the FTC.

Step-by-Step SOP

1
For a value: use Part 1
Find an antiderivative, evaluate at the endpoints, subtract.
2
For a derivative of an integral: use Part 2
Plug the upper limit into the integrand, then multiply by the derivative of the upper limit.
3
For "how much / final amount": use the accumulation form
Initial value $+ \int$ (rate) over the interval.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

If

G(x) = \int_1^{x^2} \sin(t)\,dt

, find

G'(x)

.

Need a hint?

FTC Part 2 with the chain rule: the upper limit is

u(x) = x^2

.

Show solution

1. Apply the chain-rule form

G'(x) = \sin(x^2)\cdot \dfrac{d}{dx}(x^2) = 2x\sin(x^2)

.

Common Pitfalls

⚠
Forgetting the chain-rule factor in Part 2When the upper limit is $x^2$ (not just $x$ ), you must multiply by $2x$ .
⚠
Losing the initial condition in an accumulation problem $\int_a^b F'(x)\,dx$ is only the change. The question usually wants $F(a) + \int_a^b F'(x)\,dx$ .

Worked Examples

1. State the hypotheses

2. Locate the target value

3. Invoke the theorem

Check continuity on the closed interval

Show the target value is between the endpoint values

Write the conclusion

Practice — Your Turn

1. Pick a sub-interval that works

2. Apply IVT

Worked Examples

1. Find critical numbers

2. Evaluate at the critical number and endpoints

3. Compare

Confirm continuity on a closed interval

Find critical numbers in the open interval

Evaluate f at those points and at a and b

Report the largest and smallest

Worked Examples

1. State the hypotheses

2. Compute the average rate of change

3. Invoke MVT

State both hypotheses

Compute the secant slope

Write the conclusion

Practice — Your Turn

1. Average velocity

2. Apply MVT

Worked Examples

1. Average value

2. Solve f(c) = average value

Confirm f is continuous on [a, b]

Compute the average value

Solve f(c) = average value

Worked Examples

1. Set up the accumulation expression

2. Explain each piece

For a value: use Part 1

For a derivative of an integral: use Part 2

For "how much / final amount": use the accumulation form

Practice — Your Turn

1. Apply the chain-rule form

Gary Chang