Unit 4 · Topic 4.6 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Calculus AB 4.6: Approximating Values of a Function Using Local Linearity and Linearization

Q: Why approximate when I could just evaluate the function?

Often you cannot. The function may be known only from a table or a rate, as with the phone battery, or its value, such as 26, cannot be found exactly by hand. The tangent line needs only f(a) and f'(a).

Q: How close to the point of tangency is close enough?

There is no fixed distance. It depends on how sharply the graph bends. For x4 with the tangent line at 1, the estimate was off by about 0.016 at 1.05 and by about 2.06 at 1.5.

Q: Is a linearization the same thing as a tangent line approximation?

Yes. The tangent line is the graph of the linearization, and its value at an input is the tangent line approximation there.

Q: What if the second derivative is 0 near the point?

If f''(x) = 0 on a whole interval, f is linear there, the tangent line is the graph itself, and the estimate is exact on that interval. If f'' is 0 only at the point and changes sign there, as for x3 at 0, the line can be below the graph on one side and above it on the other.

Q: What reason should I give for over or under?

Use the information you were given. A clear reason says that f'' is positive (or negative) on an interval containing the point of tangency and the input, so the graph lies above (or below) its tangent line there.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 4: 10 to 15% of the multiple-choice section

What you'll be able to do

Use the equation of a tangent line to approximate a value of a function near the point of tangency, and decide whether that approximation is an overestimate or an underestimate.

Key vocabulary

Term	What it means	The exam trap
Tangent line	The line through a point on the graph of $f$ whose slope is the derivative there; at $x = a$ it is $y = f (a) + f^{'} (a) (x - a)$ .	The line touches the graph at the point of tangency, but away from it the two can drift far apart.
Point of tangency	The point $(a, f (a))$ where the tangent line touches the graph.	Name it before writing anything else; the line's point and slope both come from there.
Local linearity	The property that the graph of a function differentiable at a point looks like a line when you zoom in near that point.	It needs a derivative at the point; at a corner, no amount of zooming makes the graph straight.
Locally linear approximation (linearization)	Using the tangent line's value in place of the function's value for inputs near the point of tangency.	It is an estimate, not the function's value, so say "approximately" and write $\approx$ , never $=$ .
Underestimate and overestimate	An estimate below the true value it approximates is an underestimate; one above it is an overestimate.	Compare the estimate with the true value in that order: an estimate that is too big is the overestimate.

Two views of the graph of y = e^x and its tangent line y = 1 + x at the point (0, 1). On the left, for x from -1.5 to 1.5, the curve and the line separate away from (0, 1), with a gap of 0.718 at x = 1. On the right, zoomed in to x from -0.1 to 0.1, they are hard to tell apart, with a gap of 0.005 at x = 0.1.

Part 1 Think first Ten minutes from now

Commit to a guess before reading on. The lesson starts from it.

Predict first

At 2:00 pm your phone's battery reads 80 percent, and the battery screen says it is draining at 0.5 percent per minute right now. You have no formula for the charge. What is a sensible estimate of the charge at 2:10 pm?

In short. A tangent line approximation estimates $f (x)$ for an input $x$ near $a$ by the value of the tangent line there, $y = f (a) + f^{'} (a) (x - a)$ .

Near the point of tangency, the graph of a differentiable function looks almost straight, so its tangent line stays close to it.
The estimate is trusted only near $a$ ; for inputs far from $a$ the line and the graph can be far apart.
How the graph bends near $a$ , read from the sign of $f^{''}$ , can show whether the line's value is too low or too high.

Part 2 Learn it Three ideas, each checked as you go

4a. Up close, a smooth curve looks like a line

Look at the picture at the top. On the left, the graph of $y = e^{x}$ and its tangent line at $(0, 1)$ soon part company: at $x = 1$ the curve is at about 2.718 and the line is at 2, a gap of about 0.718. On the right is the same pair zoomed in to $- 0.1 \leq x \leq 0.1$ . There they are almost one line, and at $x = 0.1$ the gap is about 0.005.

That happens at any point where a function is differentiable. Zoom in far enough and the graph looks straight, and the straight line it looks like is the tangent line. This is called local linearity.

So near the point of tangency, the tangent line is a locally linear approximation of the function: you can read an approximate value of $f$ off the line instead of off the curve. The word locally matters. The approximation is trusted only for inputs close to the point of tangency. Far away, as at $x = 1$ here, the line and the curve can be well apart.

Why approximate at all? Because often you cannot evaluate $f$ exactly. There may be no formula, as with the phone battery, or the value, such as $\sqrt{26}$ , may not be one you can work out by hand.

Quick check · Fill the statement

No calculator. Fill each blank from the list.

Zoom in on the graph of $f$ near a point where $f$ is differentiable, and the graph looks almost like a . So the tangent line at that point is a locally approximation of $f$ , and it is trusted only for inputs the point of tangency.

4b. Writing the line and using it

To use the tangent line you need its equation. At $x = a$ the line passes through $(a, f (a))$ with slope $f^{'} (a)$ , so in point-slope form it is

y = f (a) + f^{'} (a) (x - a) .

To approximate $f$ at an input $x$ near $a$ , put that $x$ into the line. Keep three things straight:

$f (a)$ and $f^{'} (a)$ are both taken at the point of tangency $a$ , not at the input you are estimating.
The slope multiplies $x - a$ , the distance from $a$ , not $x$ itself.
The point of tangency should be an input where you know $f$ and $f^{'}$ exactly, close to the input you want.

Example: approximate $g (1.05)$ for $g (x) = x^{4}$ . Use $a = 1$ . Then $g (1) = 1$ and $g^{'} (x) = 4 x^{3}$ , so $g^{'} (1) = 4$ . The tangent line is $y = 1 + 4 (x - 1)$ , and at $x = 1.05$ it gives $1 + 4 (0.05) = 1.2$ . The true value is ${1.05}^{4} = 1.21550625$ , so the estimate is off by about 0.016.

At $x = 1.5$ the same line gives 3, while $g (1.5) = 5.0625$ . Farther from the point of tangency, the estimate is much worse.

Quick check · Calculate

No calculator. Let $f (x) = x^{3}$ . Use the line tangent to the graph of $f$ at $x = 2$ to approximate $f (2.1)$ . Give an exact decimal.

Your answer

4c. Overestimate or underestimate?

A tangent line value is an estimate, so ask which side of the true value it lands on. The answer comes from how the graph bends near the point of tangency, and the second derivative from Unit 3 measures that.

$f^{''}$ is the rate of change of $f^{'}$ . Suppose $f^{''} (x) > 0$ on an interval around $a$ , so the slopes of $f$ grow as you move right. To the right of $a$ its slopes are greater than the line's, so it rises faster (or falls more slowly) and ends up above. To the left of $a$ its slopes are smaller than the line's, so walking back from $a$ it loses less height than the line (or gains more) and again ends up above. A graph that bends up like this is concave up, a term Unit 5 builds on. The tangent line lies below the graph near $a$ , and its value is an underestimate.

If $f^{''} (x) < 0$ near $a$ , everything flips: the graph is concave down, the line lies above it, and its value is an overestimate.

The picture at the top shows the first case: $e^{x}$ has second derivative $e^{x} > 0$ , and the curve sits above its tangent line on both sides of 0.

This is a local statement: it holds for inputs in an interval around $a$ where the sign of $f^{''}$ does not change. If $f^{''}$ changes sign between $a$ and your input, its sign at $a$ alone does not settle it.

Quick check · Decide and justify

No calculator. Let $f (x) = \ln x$ . Its tangent line at $x = 1$ is $y = x - 1$ , which gives 0.5 as an approximation of $\ln 1.5$ . A classmate says 0.5 is an overestimate of $\ln 1.5$ . Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Using the derivative at the input being estimated, $f^{'} (x)$ , as the slope of the line instead of $f^{'} (a)$ .	The tangent line at $a$ has one slope, the derivative at the point of tangency.	Evaluate both $f (a)$ and $f^{'} (a)$ at $a$ , and put the new input only into $x - a$ .
Multiplying the slope by $x$ instead of by $x - a$ .	Point-slope form measures the run from $a$ . Using $x$ adds the slope times the whole input.	Write $y = f (a) + f^{'} (a) (x - a)$ and work out $x - a$ , with its sign, first.
Deciding overestimate or underestimate from the sign of $f^{'}$ .	An increasing function can lie above or below its tangent line. What decides it is how the graph bends, the sign of $f^{''}$ .	Find the sign of $f^{''}$ on an interval containing both $a$ and the input, and say what it means for the graph and the line.

Part 3 Watch it done A square root by hand

Open each step in turn, then do the last one yourself.

Use a tangent line to $f (x) = \sqrt{x}$ to approximate $\sqrt{26}$ , and say whether the approximation is an overestimate or an underestimate.

Step 1 · Pick the point of tangency
Use $a = 25$ . It is close to 26, and $f (25) = \sqrt{25} = 5$ is exact.
Why this step?
Why not use $a = 26$ , the input you actually want?
The line at $a = 26$ needs $f (26) = \sqrt{26}$ , the very number you are trying to find. The point of tangency has to be an input where $f$ and $f^{'}$ are known exactly, and 25 is the nearest such input.
Step 2 · Find the slope
$f^{'} (x) = \frac{1}{2 \sqrt{x}}$ , so $f^{'} (25) = \frac{1}{2 \cdot 5} = \frac{1}{10}$ .
Step 3 · Write the tangent line
$y = 5 + \frac{1}{10} (x - 25)$ .
Step 4 · Evaluate at 26
$y = 5 + \frac{1}{10} (26 - 25) = 5.1$ , so $\sqrt{26} \approx 5.1$ .
Step 5 · Decide over or under
$f^{''} (x) = - \frac{1}{4} x^{- 3 / 2}$ , which is negative for every $x > 0$ . The graph of $\sqrt{x}$ is concave down, so it lies below its tangent line, and 5.1 is an overestimate of $\sqrt{26}$ .
Your step · Calculate

No calculator. Finish it. Use the same tangent line, $y = 5 + \frac{1}{10} (x - 25)$ , to approximate $\sqrt{24.5}$ . Give an exact decimal.

Your answer

Part 4 Try it A draining tank and a cube root

Try it · Calculate

No calculator. Water drains from a tank. The tank holds $W (t)$ gallons $t$ minutes after noon, where $W$ is differentiable, with $W (20) = 340$ and $W^{'} (20) = - 6$ . Use the line tangent to the graph of $W$ at $t = 20$ to estimate the time $t$ , in minutes after noon, at which the tank holds 322 gallons.

Your answer

Try it · Spot the error

No calculator. A student uses a tangent line to approximate $\sqrt[3]{8.3}$ , with $f (x) = \sqrt[3]{x} = x^{1 / 3}$ and $a = 8$ . Tap the first line with a mistake.

Try it · Multiple choice

No calculator. You want to approximate $\sin 0.05$ by hand with a line tangent to the graph of $f (x) = \sin x$ . Which tangent line should you use?

Part 5 Explain it Write it as on the exam

Free response on the exam is written by hand, so work this on paper. Nothing you type here leaves the page. Then open the model answer and score yourself against each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. A bowl of soup is served and begins to cool. Its temperature, in degrees Fahrenheit, $t$ minutes after it is served is $H (t)$ , where $H$ is twice differentiable. It is known that $H (10) = 150$ , $H^{'} (10) = - 3.5$ , and $H^{''} (t) > 0$ for $0 < t < 30$ .

(a) Write an equation for the line tangent to the graph of $H$ at $t = 10$ .

(b) Use the tangent line from part (a) to approximate $H (12)$ . Show the work that leads to your answer, and give units.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $y = 150 - 3.5 (t - 10)$ .

(b) $H (12) \approx 150 - 3.5 (12 - 10) = 150 - 7 = 143$ degrees Fahrenheit.

(c) An underestimate. $H^{''} (t) > 0$ for $0 < t < 30$ , an interval containing both 10 and 12, so the graph of $H$ is concave up there and lies above its tangent line at $t = 10$ . So $H (12) > 143$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes the tangent line $y = 150 - 3.5 (t - 10)$ , with $H (10)$ as the point and $H^{'} (10)$ as the slope.
Evaluates the line at $t = 12$ and gets 143 degrees Fahrenheit, with units.
Says the approximation is an underestimate because $H^{''} (t) > 0$ on an interval containing 10 and 12, so the graph is concave up and lies above its tangent line.

Part 6 Check yourself A hike up a ridge

Both case questions use the table and the fact above it.

A hiker climbs a ridge. Her elevation is $E (t)$ meters above sea level $t$ hours after she starts, where $E$ is twice differentiable. Some values of $E$ and $E^{'}$ are in the table, and $E^{''} (t) < 0$ for $0 < t < 3$ .

The hiker's elevation and its rate of change
$t$ (hours)	$E (t)$ (meters)	$E^{'} (t)$ (meters per hour)
0	1200	300
1	1450	200
2	1620	120
3	1700	40

Case question 1 · Calculate

No calculator. Use the line tangent to the graph of $E$ at $t = 2$ to approximate the hiker's elevation, in meters, at $t = 2.5$ hours.

Your answer

Case question 2 · Multiple choice

No calculator. Which statement about the approximation of $E (2.5)$ from the line tangent to the graph of $E$ at $t = 2$ is true?

The rest stand on their own and do not use the case.

Check yourself · Sort it

No calculator. In each case, the line tangent to the graph of the function at the given point is used to approximate the function at the given input. Sort each approximation.

Underestimate

Overestimate

Check yourself · Decide and justify

No calculator. The line tangent to the graph of $h (x) = x^{2}$ at $x = 1$ is $y = 1 + 2 (x - 1)$ . A classmate says this line approximates $h (1.2)$ and $h (3)$ equally well. Is the classmate right?

Which reason supports your answer?

Part 7 Keep it Pull back two earlier ideas

This lesson leaned on the second derivative from Unit 3 and on the derivative of $e^{x}$ from Unit 2. Recalling them now is what keeps them.

Bring it back · Calculate

No calculator. Let $f (x) = \frac{1}{x^{2}}$ . Find $f^{''} (2)$ . Give an exact decimal.

Your answer

Bring it back · Multiple choice

No calculator. What is $\lim_{h \to 0} \frac{e^{2 + h} - e^{2}}{h}$ ?

In three lines

The tangent line at $x = a$ , $y = f (a) + f^{'} (a) (x - a)$ , approximates $f$ for inputs near $a$ .
Choose a point of tangency where $f (a)$ and $f^{'} (a)$ are known exactly, close to the input you want.
If $f^{''} > 0$ between the point and the input, the line gives an underestimate; if $f^{''} < 0$ , an overestimate.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

Write the tangent line in point-slope form with the point named, $y = f (a) + f^{'} (a) (x - a)$ , before you substitute; there is no need to simplify it, and leaving it in that form shows where each number came from. When a question asks whether an estimate is too high or too low, state the sign of $f^{''}$ on an interval containing both inputs and what that means for the graph and its tangent line, and only then give your answer.

Questions students ask

Why approximate when I could just evaluate the function?

Often you cannot. The function may be known only from a table or a rate, as with the phone battery, or its value, such as $\sqrt{26}$ , cannot be found exactly by hand. The tangent line needs only $f (a)$ and $f^{'} (a)$ .

How close to the point of tangency is close enough?

There is no fixed distance. It depends on how sharply the graph bends. For $x^{4}$ with the tangent line at 1, the estimate was off by about 0.016 at 1.05 and by about 2.06 at 1.5.

Is a linearization the same thing as a tangent line approximation?

Yes. The tangent line is the graph of the linearization, and its value at an input is the tangent line approximation there.

What if the second derivative is 0 near the point?

If $f^{''} (x) = 0$ on a whole interval, $f$ is linear there, the tangent line is the graph itself, and the estimate is exact on that interval. If $f^{''}$ is 0 only at the point and changes sign there, as for $x^{3}$ at 0, the line can be below the graph on one side and above it on the other.

What reason should I give for over or under?

Use the information you were given. A clear reason says that $f^{''}$ is positive (or negative) on an interval containing the point of tangency and the input, so the graph lies above (or below) its tangent line there.

Course alignment, for teachers

AP Calculus AB topic 4.6, Unit 4: Contextual Applications of Differentiation.

Skill 1.F: Explain how an approximated value relates to the actual value.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first Ten minutes from now

Part 2 Learn it Three ideas, each checked as you go

4a. Up close, a smooth curve looks like a line

4b. Writing the line and using it

4c. Overestimate or underestimate?

Part 3 Watch it done A square root by hand

Part 4 Try it A draining tank and a cube root

Part 5 Explain it Write it as on the exam

Part 6 Check yourself A hike up a ridge

Part 7 Keep it Pull back two earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic