Unit 4Linear Approximation Formula:

L(x)=f(a)+f'(a)(x-a)

Concept 4.178Built for May 2027 AP Exam

Linear Approximation Formula: $L(x)=f(a)+f'(a)(x-a)$

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A differentiable function looks straight when you zoom in far enough, and the straight line it resembles is the tangent line at a nearby point. That line, written as $L(x) = f(a) + f'(a)(x-a)$ , is the linear approximation. The formula is the point-slope form of the tangent line, nothing more, and the value $a$ is the input where everything is easy to compute. Using it means starting at $(a, f(a))$ , moving the small distance $x - a$ , and adding the slope times that distance. The approximation is good near $a$ and poor far from it, and whether it lands above or below the true value is decided by concavity.

On the AP Exam

What you’ll see: A request to approximate a number that has no closed form, such as $\sqrt{4.1}$ or $\ln(1.02)$ , or a request to write the tangent line at a point and use it to estimate a nearby value.
What you’ll do: Choose $a$ as the nearby input where the function is exact, compute $f(a)$ and $f'(a)$ , write $L(x)$ , then substitute the target input and state whether the estimate is above or below the true value.
Memory hook: Anchor, slope, step. Start at the anchor $a$ , add the slope, and multiply by the step $x - a$ .
Use it on the exam: For MCQs: The options usually differ in the sign of the correction term, so a sign error in $f'(a)$ is visible immediately. For FRQs: On 2023 Q3(b) the tangent line at $t = 0$ is used to approximate $M(2)$ , and the guideline scores the line and the value separately. Writing $L$ before substituting protects both points.

Concept Review

The Tangent Line Used as a Stand-In

Figure 7 — A curve and its tangent at the anchor point. Near the anchor the two are almost indistinguishable, and the gap widens with distance, which is why the estimate is good near $a$ and poor far from it.

The tangent touches at one point and runs close on both sides. The distance from the anchor is called the step, and the approximation's error grows roughly with the square of that step. Doubling the distance from the anchor therefore multiplies the error by about four rather than by two.

A concave down curve with its tangent at an anchor point, and a marked vertical gap between the line and the curve further along

The Formula, Decoded

$L(x) = f(a) + f'(a)(x - a)$

Piece	What it is	Where it comes from
$f(a)$	the exact value at the anchor	substitution
$f'(a)$	the slope at the anchor	differentiation, then substitution
$x - a$	the step away from the anchor	the target input minus the anchor
$f'(a)(x-a)$	the correction	slope times step

Every linear approximation question is those four rows in order. A solution that writes the formula with symbols still in it has finished row three and has not substituted the target.

Choosing the Anchor

The anchor is the input near the target where the function and its derivative are both easy. For $\sqrt{4.1}$ the anchor is $4$ ; for $\ln(1.02)$ it is $1$ ; for $\sqrt[3]{8.1}$ it is $8$ . Choosing an anchor far from the target is legal and useless, because the step becomes large and the error with it. The exam always supplies a target that has a comfortable anchor, and the work is to notice which one it is.

Above or Below, and Why

The tangent sits above a curve that is concave down and below a curve that is concave up, which means the sign of $f''$ at the anchor decides whether the estimate overstates or understates:

$\text{if } f''(a) < 0, \text{ then } L(x) > f(x) \text{ near } a$

The same idea appears in a differential-equation setting, where the sign of the second derivative of the solution decides whether a tangent-line estimate is an overestimate or an underestimate, and the reasoning transfers without change.

Differentials and the Same Idea

Writing the step as $dx$ and the correction as $f'(a)\,dx$ gives the differential form $dy = f'(a)\,dx$ , which is the same statement in different notation. A question that asks for the approximate change in a quantity when its input changes slightly is asking for $f'(a)\,dx$ , and the answer has the units of the output rather than of the slope.

Common Misconception Check

Myth 1: "Linear approximation means using the tangent line instead of the function, everywhere." Truth: It means using it near the anchor. Far away the tangent is a different line with a different value, and the word approximation quietly stops applying.
Myth 2: "The estimate is always an underestimate." Truth: It is an underestimate exactly when the curve is concave up at the anchor, and an overestimate when it is concave down.
Myth 3: "The anchor has to be an integer." Truth: It has to be a point where both $f$ and $f'$ are easy. A convenient fraction works just as well when the function allows it.

Topic-Specific Study Method

Anchor, Slope, Step

Borrowed idea: Choose the anchor before choosing anything else. The anchor is the nearby input where the function and its slope are both easy to write down, and everything else on this page follows from it. Anchoring far from the target is legal and useless, because the step grows and the error grows faster. So which input do you anchor at when the target is a logarithm close to one? Write the line first, substitute the target second, and judge the estimate third. The order protects all three points.

Target	Anchor $a$	$f(a)$	$f'(a)$	$L$ at the target
$\sqrt{4.1}$	$4$	$2$	$\frac{1}{4}$	$2.025$
$\ln(1.02)$	$1$	$0$	$1$	$0.02$
$\sqrt[3]{8.1}$	$8$	$2$	$\frac{1}{12}$	$2 + \frac{0.1}{12}$
$(1.03)^{10}$	$1$	$1$	$10$	$1.3$
$\sin(0.1)$	$0$	$0$	$1$	$0.1$

① For each row, compute the true value with a calculator and subtract the estimate. The sign of that difference is the concavity test in numbers, and seeing it once makes the rule memorable.

② Take a target twice as far from the anchor and compare the two errors. The error roughly quadruples, which is the quadratic behaviour the figure describes.

③ Write the tangent line as a function $L(x)$ before substituting the target. A solution that substitutes first has no line to show, and the line is what the guideline scores.

Training & Error Diagnosis

Core Archetype (Simplest Form): Use a linear approximation to estimate $\ln(1.2)$ .

Standard Execution:

Anchor. $a = 1$ , where $\ln 1 = 0$ and the derivative is $1$ .

Slope. $f'(x) = \frac{1}{x}$ , so $f'(1) = 1$ .

Line. $L(x) = 0 + 1(x-1) = x - 1$ .

Substitute. $L(1.2) = 0.2$ , so $\ln(1.2) \approx 0.2$ .

Judge. Since $f''(x) = -\frac{1}{x^2} < 0$ , the curve is concave down and the estimate is an overestimate. The true value is about $0.182$ .

Exam Variation (Context / Parameter Shift): A quantity is modeled by $M(t)$ with $M(0) = 5$ and $M'(0) = -0.6$ . Use the tangent line at $t = 0$ to estimate $M(2)$ , and determine whether the estimate is too large or too small given that $M''(t) > 0$ for all $t$ .

Solution: The tangent line is $L(t) = 5 - 0.6t$ , so $L(2) = 5 - 1.2 = 3.8$ . Since $M'' > 0$ the graph is concave up, so the tangent line lies below the curve and the estimate of $3.8$ is too small.

Common Traps & Failure Signals:

The unsubstituted line (Signal: an answer still containing $a$ or $x$ ): the line was written and the target was never put into it.
The wrong anchor (Signal: a step larger than half the anchor): a convenient point was not found, and the approximation is being asked to do work it cannot do.
The missing verdict (Signal: a number with no statement about over or under): concavity was never checked, so half the question went unanswered.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Anchor Lost: You substituted before choosing an anchor.
Slope Sign: You computed $f'(a)$ correctly and then moved in the wrong direction, which happens when the step is written as $a - x$ instead of $x - a$ and is invisible on a target larger than the anchor.
Concavity Skipped: You produced the estimate and never asked whether the line sits above or below the curve, so the answer has a number and no verdict, and the verdict is where the second point lives.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three released BC sets use a tangent line to produce a nearby value, and two of them do it inside a differential-equation question where the line is the first step of a longer argument
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; Counted where the tangent line is used to approximate a value, not merely written down as a slope
FRQ Years Appeared: 2017 (Q4a), 2023 (Q3b), 2026 (Q3C)

Practice Examples

MCQ 1

A Square Root

Using the tangent line to the graph of $f(x) = \sqrt{x}$ at $x = 9$ , what is the linear approximation of $\sqrt{9.2}$ ?

MCQ 2

Above or Below

The function $f$ satisfies $f(2) = 7$ , $f'(2) = -3$ , and $f''(x) < 0$ for all $x$ near $2$ . A linear approximation at $a = 2$ is used to estimate $f(2.1)$ . Which statement is correct?

MCQ 3

A Power

Using a linear approximation at $x = 1$ , what is the approximate value of $(1.03)^{10}$ ?

FRQ 1 • Estimating a Logarithm

Let $f(x) = \ln x$ for $x > 0$ .

(a) Write the linear approximation $L(x)$ of $f$ at $a = 1$ .

(b) Use $L$ to estimate $\ln(1.2)$ , and determine whether the estimate is an overestimate or an underestimate. Give a reason.

(c) Use the same line to estimate $\ln(0.9)$ .

(d) Explain why the estimate in part (b) is closer to the true value than the estimate in part (c).

Cheatsheet