2.1.2 AP-style problems

Achievable AP Calculus AB

2. Derivative basics

2.1. The derivative

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

AP-style problems

5 min read

Font

Discuss

Feedback

What you’ll learn

Using graphs and tables to estimate and evaluate rates of change using graphs and tables.
Recognizing derivatives from specific limit expressions.

Using tables

The AP exam frequently uses tabular data to test your ability to connect core concepts like functions and derivatives. For example,

A bathtub is being filled with water and the depth of the water is measured over a few minutes. Based on the table below:

Time $t$ (minutes) Depth (inches)

$2$ $3$

$4$ $8$

$6$ $14$

$7$ $16$

$9$ $19$

a) Find the average rate of change over the time interval $4 \leq t \leq 7$ .

b) Estimate the instantaneous rate of change at $t = 3$ .

Time $t$ (minutes)	Depth (inches)
$2$	$3$
$4$	$8$
$6$	$14$
$7$	$16$
$9$	$19$

Solutions

a) Average rate of change over $[4, 7]$

(spoiler)

In AP Calculus, time $t$ is often the input/independent variable. Let water depth be the function of time defined by $D (t)$ .

The average rate of change of the depth of the water over the interval $[4, 7]$ is

$\frac{D ( 7 ) - D ( 4 )}{7 - 4} = \frac{16 - 8}{3} = \frac{8}{3}$

The units are inches per minute because depth (numerator) is measured in inches and time (denominator) is measured in minutes.

So on average, the water depth increased by $\frac{8}{3}$ inches per minute between minutes 4 and 7.

b) Instantaneous rate of change at $t = 3$

(spoiler)

Without an explicit function, the best estimate that can be made for the instantaneous rate of change $D^{'} (3)$ uses the average rate of change over the smallest subinterval that contains $t = 3$ , which is $[2, 4]$ :

$D^{'} (3) \approx \frac{D ( 4 ) - D ( 2 )}{4 - 2} = \frac{8 - 3}{2} = \frac{5}{2} in/min$

This is the best estimate for how quickly the water level was changing at exactly 3 minutes.

Using graphs

Shown below is the graph of function $f$ .

Finding average rate of change
Achievable

If $g$ is the function defined by

$g (x) = f (3 x)$

find the average rate of change of $g (x)$ over the interval $[- 1, \frac{4}{3}]$ .

Solution

(spoiler)

The average rate of change of $g$ over $[- 1, \frac{4}{3}]$ is

$\frac{g ( \frac{4}{3} ) - g ( - 1 )}{\frac{4}{3} - ( - 1 )}$

Since $g (x) = f (3 x)$ ,

$g (\frac{4}{3}) = f (3 \cdot \frac{4}{3}) = f (4) = 1$
$g (- 1) = f (- 3) = 4$

Then the average rate of change of $g$ is

$\frac{g ( \frac{4}{3} ) - g ( - 1 )}{\frac{4}{3} - ( - 1 )} = \frac{1 - 4}{\frac{4}{3} + 1} = \frac{- 3}{\frac{7}{3}} = - \frac{9}{7}$

Recognizing derivatives from the definition

Most derivative problems ask you to find $f^{'} (x)$ given a function $f (x)$ . However, some problems involve identifying the original function $f (x)$ and the point $a$ from a given limit expression.

The key is to recognize which limit definition is being used:

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a} or f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

Examples

Find a function $f (x)$ and a number $a$ such that the following limit represents $f^{'} (a)$ .

$h \to 0 lim \frac{( 3 + h ) ^{2} - 9}{h}$

Solution

(spoiler)

Since the limit notation specifies $h \to 0$ , match it to the 2nd definition

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

By matching the given expression to the template, we can identify $a$ and $f (x)$ :

$(3 + h)^{2}$ matched to $f (a + h)$ gives $a = 3$ and $f (x) = x^{2}$ .

This is confirmed by checking $f (a)$ :

$f (3) = (3)^{2} = 9$

which matches the second constant term in the numerator of the limit expression.

Therefore, the limit represents the derivative of $f (x) = x^{2}$ evaluated at $x = 3$ .

The limit expression

$h \to 0 lim \frac{4 ^{3 + h} - 64}{h}$

represents the derivative of a function $f (x)$ at $x = a$ .

a) Write an equivalent limit expression for $f^{'} (a)$ .

b) What is the value of $f (2)$ ?

Solutions

a) Equivalent limit expression

(spoiler)

Since the limit notation specifies $h \to 0$ , this limit mirrors the definition:

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

Matching the expressions:

$4^{3 + h}$ matched to $f (a + h)$ gives $a = 3$ and $f (x) = 4^{x}$ .

Checking $f (a)$ ,

$f (3) = 4^{3} = 64$

which matches the constant in the numerator.

Since $f (x) = 4^{x}$ and $a = 3$ , then the equivalent limit expression in the form

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

is written as

$f^{'} (3) = x \to 3 lim \frac{4 ^{x} - 64}{x - 3}$

b) Find $f (2)$ .

(spoiler)

Since $f (x) = 4^{x}$ , then

$f (2) = 4^{2} = 16$

Working backwards from limit definitions

Two definitions to recognize: $f^{'} (a) = lim_{x \to a} \frac{f ( x ) - f ( a )}{x - a}$ and $f^{'} (a) = lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}$
Identify which definition applies based on whether the limit is $x \to a$ or $h \to 0$
Extract $f (x)$ and $a$ by pattern-matching terms in the numerator, then verify by checking $f (a)$

Using tables for rates of change

Average rate of change over $[a, b]$ : $\frac{f ( b ) - f ( a )}{b - a}$ using table values directly
Instantaneous rate of change at a point: use average rate of change over the smallest available surrounding interval as the best estimate
Always include units (e.g., inches per minute) when interpreting results

Next | 2.2 Tangent lines & slopes

AP-style problems

What you’ll learn

Using graphs and tables to estimate and evaluate rates of change using graphs and tables.
Recognizing derivatives from specific limit expressions.

Using tables

The AP exam frequently uses tabular data to test your ability to connect core concepts like functions and derivatives. For example,

A bathtub is being filled with water and the depth of the water is measured over a few minutes. Based on the table below:

Time $t$ (minutes) Depth (inches)

$2$ $3$

$4$ $8$

$6$ $14$

$7$ $16$

$9$ $19$

a) Find the average rate of change over the time interval $4 \leq t \leq 7$ .

b) Estimate the instantaneous rate of change at $t = 3$ .

Time $t$ (minutes)	Depth (inches)
$2$	$3$
$4$	$8$
$6$	$14$
$7$	$16$
$9$	$19$

Solutions

a) Average rate of change over $[4, 7]$

(spoiler)

In AP Calculus, time $t$ is often the input/independent variable. Let water depth be the function of time defined by $D (t)$ .

The average rate of change of the depth of the water over the interval $[4, 7]$ is

$\frac{D ( 7 ) - D ( 4 )}{7 - 4} = \frac{16 - 8}{3} = \frac{8}{3}$

The units are inches per minute because depth (numerator) is measured in inches and time (denominator) is measured in minutes.

So on average, the water depth increased by $\frac{8}{3}$ inches per minute between minutes 4 and 7.

b) Instantaneous rate of change at $t = 3$

(spoiler)

$D^{'} (3) \approx \frac{D ( 4 ) - D ( 2 )}{4 - 2} = \frac{8 - 3}{2} = \frac{5}{2} in/min$

This is the best estimate for how quickly the water level was changing at exactly 3 minutes.

Using graphs

Shown below is the graph of function $f$ .

If $g$ is the function defined by

$g (x) = f (3 x)$

find the average rate of change of $g (x)$ over the interval $[- 1, \frac{4}{3}]$ .

Solution

(spoiler)

The average rate of change of $g$ over $[- 1, \frac{4}{3}]$ is

$\frac{g ( \frac{4}{3} ) - g ( - 1 )}{\frac{4}{3} - ( - 1 )}$

Since $g (x) = f (3 x)$ ,

$g (\frac{4}{3}) = f (3 \cdot \frac{4}{3}) = f (4) = 1$
$g (- 1) = f (- 3) = 4$

Then the average rate of change of $g$ is

$\frac{g ( \frac{4}{3} ) - g ( - 1 )}{\frac{4}{3} - ( - 1 )} = \frac{1 - 4}{\frac{4}{3} + 1} = \frac{- 3}{\frac{7}{3}} = - \frac{9}{7}$

Recognizing derivatives from the definition

The key is to recognize which limit definition is being used:

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a} or f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

Examples

Find a function $f (x)$ and a number $a$ such that the following limit represents $f^{'} (a)$ .

$h \to 0 lim \frac{( 3 + h ) ^{2} - 9}{h}$

Solution

(spoiler)

Since the limit notation specifies $h \to 0$ , match it to the 2nd definition

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

By matching the given expression to the template, we can identify $a$ and $f (x)$ :

$(3 + h)^{2}$ matched to $f (a + h)$ gives $a = 3$ and $f (x) = x^{2}$ .

This is confirmed by checking $f (a)$ :

$f (3) = (3)^{2} = 9$

which matches the second constant term in the numerator of the limit expression.

Therefore, the limit represents the derivative of $f (x) = x^{2}$ evaluated at $x = 3$ .

The limit expression

$h \to 0 lim \frac{4 ^{3 + h} - 64}{h}$

represents the derivative of a function $f (x)$ at $x = a$ .

a) Write an equivalent limit expression for $f^{'} (a)$ .

b) What is the value of $f (2)$ ?

Solutions

a) Equivalent limit expression

(spoiler)

Since the limit notation specifies $h \to 0$ , this limit mirrors the definition:

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

Matching the expressions:

$4^{3 + h}$ matched to $f (a + h)$ gives $a = 3$ and $f (x) = 4^{x}$ .

Checking $f (a)$ ,

$f (3) = 4^{3} = 64$

which matches the constant in the numerator.

Since $f (x) = 4^{x}$ and $a = 3$ , then the equivalent limit expression in the form

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

is written as

$f^{'} (3) = x \to 3 lim \frac{4 ^{x} - 64}{x - 3}$

b) Find $f (2)$ .

(spoiler)

Since $f (x) = 4^{x}$ , then

$f (2) = 4^{2} = 16$

Achievable AP Calculus AB

2. Derivative basics

2.1. The derivative

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.

AP-style problems

5 min read

Font

Discuss

Feedback

What you’ll learn

Using graphs and tables to estimate and evaluate rates of change using graphs and tables.
Recognizing derivatives from specific limit expressions.

Using tables

The AP exam frequently uses tabular data to test your ability to connect core concepts like functions and derivatives. For example,

A bathtub is being filled with water and the depth of the water is measured over a few minutes. Based on the table below:

Time $t$ (minutes) Depth (inches)

$2$ $3$

$4$ $8$

$6$ $14$

$7$ $16$

$9$ $19$

a) Find the average rate of change over the time interval $4 \leq t \leq 7$ .

b) Estimate the instantaneous rate of change at $t = 3$ .

Time $t$ (minutes)	Depth (inches)
$2$	$3$
$4$	$8$
$6$	$14$
$7$	$16$
$9$	$19$

Solutions

a) Average rate of change over $[4, 7]$

(spoiler)

In AP Calculus, time $t$ is often the input/independent variable. Let water depth be the function of time defined by $D (t)$ .

The average rate of change of the depth of the water over the interval $[4, 7]$ is

$\frac{D ( 7 ) - D ( 4 )}{7 - 4} = \frac{16 - 8}{3} = \frac{8}{3}$

The units are inches per minute because depth (numerator) is measured in inches and time (denominator) is measured in minutes.

So on average, the water depth increased by $\frac{8}{3}$ inches per minute between minutes 4 and 7.

b) Instantaneous rate of change at $t = 3$

(spoiler)

$D^{'} (3) \approx \frac{D ( 4 ) - D ( 2 )}{4 - 2} = \frac{8 - 3}{2} = \frac{5}{2} in/min$

This is the best estimate for how quickly the water level was changing at exactly 3 minutes.

Using graphs

Shown below is the graph of function $f$ .

Finding average rate of change
Achievable

If $g$ is the function defined by

$g (x) = f (3 x)$

find the average rate of change of $g (x)$ over the interval $[- 1, \frac{4}{3}]$ .

Solution

(spoiler)

The average rate of change of $g$ over $[- 1, \frac{4}{3}]$ is

$\frac{g ( \frac{4}{3} ) - g ( - 1 )}{\frac{4}{3} - ( - 1 )}$

Since $g (x) = f (3 x)$ ,

$g (\frac{4}{3}) = f (3 \cdot \frac{4}{3}) = f (4) = 1$
$g (- 1) = f (- 3) = 4$

Then the average rate of change of $g$ is

$\frac{g ( \frac{4}{3} ) - g ( - 1 )}{\frac{4}{3} - ( - 1 )} = \frac{1 - 4}{\frac{4}{3} + 1} = \frac{- 3}{\frac{7}{3}} = - \frac{9}{7}$

Recognizing derivatives from the definition

The key is to recognize which limit definition is being used:

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a} or f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

Examples

Find a function $f (x)$ and a number $a$ such that the following limit represents $f^{'} (a)$ .

$h \to 0 lim \frac{( 3 + h ) ^{2} - 9}{h}$

Solution

(spoiler)

Since the limit notation specifies $h \to 0$ , match it to the 2nd definition

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

By matching the given expression to the template, we can identify $a$ and $f (x)$ :

$(3 + h)^{2}$ matched to $f (a + h)$ gives $a = 3$ and $f (x) = x^{2}$ .

This is confirmed by checking $f (a)$ :

$f (3) = (3)^{2} = 9$

which matches the second constant term in the numerator of the limit expression.

Therefore, the limit represents the derivative of $f (x) = x^{2}$ evaluated at $x = 3$ .

The limit expression

$h \to 0 lim \frac{4 ^{3 + h} - 64}{h}$

represents the derivative of a function $f (x)$ at $x = a$ .

a) Write an equivalent limit expression for $f^{'} (a)$ .

b) What is the value of $f (2)$ ?

Solutions

a) Equivalent limit expression

(spoiler)

Since the limit notation specifies $h \to 0$ , this limit mirrors the definition:

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

Matching the expressions:

$4^{3 + h}$ matched to $f (a + h)$ gives $a = 3$ and $f (x) = 4^{x}$ .

Checking $f (a)$ ,

$f (3) = 4^{3} = 64$

which matches the constant in the numerator.

Since $f (x) = 4^{x}$ and $a = 3$ , then the equivalent limit expression in the form

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

is written as

$f^{'} (3) = x \to 3 lim \frac{4 ^{x} - 64}{x - 3}$

b) Find $f (2)$ .

(spoiler)

Since $f (x) = 4^{x}$ , then

$f (2) = 4^{2} = 16$

Working backwards from limit definitions

Two definitions to recognize: $f^{'} (a) = lim_{x \to a} \frac{f ( x ) - f ( a )}{x - a}$ and $f^{'} (a) = lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}$
Identify which definition applies based on whether the limit is $x \to a$ or $h \to 0$
Extract $f (x)$ and $a$ by pattern-matching terms in the numerator, then verify by checking $f (a)$

Using tables for rates of change

Average rate of change over $[a, b]$ : $\frac{f ( b ) - f ( a )}{b - a}$ using table values directly
Instantaneous rate of change at a point: use average rate of change over the smallest available surrounding interval as the best estimate
Always include units (e.g., inches per minute) when interpreting results

Next | 2.2 Tangent lines & slopes

AP-style problems

What you’ll learn

Using graphs and tables to estimate and evaluate rates of change using graphs and tables.
Recognizing derivatives from specific limit expressions.

Using tables

The AP exam frequently uses tabular data to test your ability to connect core concepts like functions and derivatives. For example,

A bathtub is being filled with water and the depth of the water is measured over a few minutes. Based on the table below:

Time $t$ (minutes) Depth (inches)

$2$ $3$

$4$ $8$

$6$ $14$

$7$ $16$

$9$ $19$

a) Find the average rate of change over the time interval $4 \leq t \leq 7$ .

b) Estimate the instantaneous rate of change at $t = 3$ .

Time $t$ (minutes)	Depth (inches)
$2$	$3$
$4$	$8$
$6$	$14$
$7$	$16$
$9$	$19$

Solutions

a) Average rate of change over $[4, 7]$

(spoiler)

In AP Calculus, time $t$ is often the input/independent variable. Let water depth be the function of time defined by $D (t)$ .

The average rate of change of the depth of the water over the interval $[4, 7]$ is

$\frac{D ( 7 ) - D ( 4 )}{7 - 4} = \frac{16 - 8}{3} = \frac{8}{3}$

The units are inches per minute because depth (numerator) is measured in inches and time (denominator) is measured in minutes.

So on average, the water depth increased by $\frac{8}{3}$ inches per minute between minutes 4 and 7.

b) Instantaneous rate of change at $t = 3$

(spoiler)

$D^{'} (3) \approx \frac{D ( 4 ) - D ( 2 )}{4 - 2} = \frac{8 - 3}{2} = \frac{5}{2} in/min$

This is the best estimate for how quickly the water level was changing at exactly 3 minutes.

Using graphs

Shown below is the graph of function $f$ .

If $g$ is the function defined by

$g (x) = f (3 x)$

find the average rate of change of $g (x)$ over the interval $[- 1, \frac{4}{3}]$ .

Solution

(spoiler)

The average rate of change of $g$ over $[- 1, \frac{4}{3}]$ is

$\frac{g ( \frac{4}{3} ) - g ( - 1 )}{\frac{4}{3} - ( - 1 )}$

Since $g (x) = f (3 x)$ ,

$g (\frac{4}{3}) = f (3 \cdot \frac{4}{3}) = f (4) = 1$
$g (- 1) = f (- 3) = 4$

Then the average rate of change of $g$ is

$\frac{g ( \frac{4}{3} ) - g ( - 1 )}{\frac{4}{3} - ( - 1 )} = \frac{1 - 4}{\frac{4}{3} + 1} = \frac{- 3}{\frac{7}{3}} = - \frac{9}{7}$

Recognizing derivatives from the definition

The key is to recognize which limit definition is being used:

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a} or f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

Examples

Find a function $f (x)$ and a number $a$ such that the following limit represents $f^{'} (a)$ .

$h \to 0 lim \frac{( 3 + h ) ^{2} - 9}{h}$

Solution

(spoiler)

Since the limit notation specifies $h \to 0$ , match it to the 2nd definition

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

By matching the given expression to the template, we can identify $a$ and $f (x)$ :

$(3 + h)^{2}$ matched to $f (a + h)$ gives $a = 3$ and $f (x) = x^{2}$ .

This is confirmed by checking $f (a)$ :

$f (3) = (3)^{2} = 9$

which matches the second constant term in the numerator of the limit expression.

Therefore, the limit represents the derivative of $f (x) = x^{2}$ evaluated at $x = 3$ .

The limit expression

$h \to 0 lim \frac{4 ^{3 + h} - 64}{h}$

represents the derivative of a function $f (x)$ at $x = a$ .

a) Write an equivalent limit expression for $f^{'} (a)$ .

b) What is the value of $f (2)$ ?

Solutions

a) Equivalent limit expression

(spoiler)

Since the limit notation specifies $h \to 0$ , this limit mirrors the definition:

$f^{'} (a) = h \to 0 lim \frac{f ( a + h ) - f ( a )}{h}$

Matching the expressions:

$4^{3 + h}$ matched to $f (a + h)$ gives $a = 3$ and $f (x) = 4^{x}$ .

Checking $f (a)$ ,

$f (3) = 4^{3} = 64$

which matches the constant in the numerator.

Since $f (x) = 4^{x}$ and $a = 3$ , then the equivalent limit expression in the form

$f^{'} (a) = x \to a lim \frac{f ( x ) - f ( a )}{x - a}$

is written as

$f^{'} (3) = x \to 3 lim \frac{4 ^{x} - 64}{x - 3}$

b) Find $f (2)$ .

(spoiler)

Since $f (x) = 4^{x}$ , then

$f (2) = 4^{2} = 16$