How do I choose the base point a?

Pick the closest value to your target where both f(a) and f'(a) are easy to compute exactly. For sqrt(4.1) use a=4; for (1.02) use a=1. The nearer a is to the target, the smaller the error.

How do I tell if my linear approximation is too high or too low?

Check the sign of f'' near a. If f''>0 the graph is concave up and the tangent line lies below it, so you underestimate. If f''<0 the graph is concave down and you overestimate. On the AP exam you must state this concavity reasoning to earn justification points.

What is the difference between linearization and differentials?

They are the same tangent-line idea. Linearization L(x)=f(a)+f'(a)(x-a) estimates a function value. The differential dy=f'(x)\,dx estimates the change in the value. Setting dx=x-a makes dy=L(x)-f(a), so both describe the tangent line.

How does this connect to Taylor polynomials in Unit 10?

The linearization is exactly the first-order Taylor polynomial P_1(x) centered at a. Taylor polynomials add higher-degree terms like f''(a)/2!(x-a)^2 to reduce error; linearization just keeps the constant and linear terms.

AP-CALCBC-4.6

U4.6 Linearization (Local Linear Approximation)

Q: What is the difference between linearization and differentials?

They are the same tangent-line idea. Linearization L(x)=f(a)+f'(a)(x-a) estimates a function value. The differential dy=f'(x)\,dx estimates the change in the value. Setting dx=x-a makes dy=L(x)-f(a), so both describe the tangent line.

Q: How does this connect to Taylor polynomials in Unit 10?

The linearization is exactly the first-order Taylor polynomial P_1(x) centered at a. Taylor polynomials add higher-degree terms like f''(a)/2!(x-a)^2 to reduce error; linearization just keeps the constant and linear terms.

Master AP Calculus BC linearization: use the tangent line L(x)=f(a)+f'(a)(x−a) to estimate values near a, connect to differentials and Taylor.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U4.6 Linearization (Local Linear Approximation), then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U4.6 Linearization (Local Linear Approximation) free

What this lesson covers

When a function is hard to evaluate directly — like

\sqrt{4.1}

\ln(1.02)

— you can lean on something easy: the tangent line at a nearby point you already know. Because a smooth curve hugs its tangent line very closely near the point of tangency, the line's height gives a fast, accurate estimate of the function's value.

This lesson shows you how to build the linearization

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

, use it to approximate values, decide whether your estimate is too high or too low, and connect the idea to differentials (

dy=f'(x)\,dx

) and to the first-order Taylor polynomial you'll meet in Unit 10. These threads all describe the same core move: replace a curve by its tangent line near a point.

Building the Linearization

The linearization of

f

x=a

is the equation of the tangent line at that point, written as a function of

x

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

For

x

near

a

f(x)\approx L(x)

. This is called local linear approximation because it is only reliable close to

a

— the further

x

drifts from

a

, the more the curve peels away from its tangent line.

To apply it you need three ingredients: a good base point

a

where

f(a)

is easy, the value

f(a)

, and the slope

f'(a)

. Choosing

a

well is the whole game. To estimate

\sqrt{4.1}

, take

f(x)=\sqrt{x}

and

a=4

because

\sqrt{4}=2

is clean and

4.1

is close by.

Step	What you do
1	Identify $f$ and a convenient base point $a$
2	Compute $f(a)$
3	Compute $f'(x)$ , then $f'(a)$
4	Write $L(x)=f(a)+f'(a)(x-a)$
5	Plug in the target $x$ to estimate $f(x)$

A common exam slip is forgetting the

(x-a)

factor or evaluating

f'

at the target instead of at

a

. Always evaluate both

f

and

f'

at the base point

a

Over- or Underestimate? Concavity Decides

The tangent line lies on one side of the curve depending on concavity, which is governed by

f''

. If

f''>0

near

a

, the graph is concave up and bends above its tangent line, so the tangent line sits below the curve — the linear approximation underestimates

f(x)

. If

f''<0

, the graph is concave down and the tangent line sits above the curve — the approximation overestimates.

Concavity at $a$	Sign of $f''$	Tangent line vs. curve	$L(x)$ vs. $f(x)$
Concave up	$f''>0$	line below curve	underestimate
Concave down	$f''<0$	line above curve	overestimate

For

\sqrt{x}

f''(x)=-\tfrac14 x^{-3/2}<0

, so the curve is concave down and any tangent-line estimate of a square root is an overestimate. AP free-response questions frequently ask you to state whether the approximation is greater or less than the true value and to justify with a statement about concavity or the sign of the second derivative. Simply saying "because

f''<0

the graph is concave down, so the tangent line lies above the curve" earns the justification point.

Differentials and Estimating Change

Linearization and differentials are the same idea dressed differently. Define

dy=f'(x)\,dx

, where

dx

is a small change in

x

and

dy

is the corresponding change predicted along the tangent line. The true change is

\Delta y=f(x+dx)-f(x)

, and for small

dx

\Delta y\approx dy

.

This form is ideal for error-propagation and "how much does the output change" problems. If the radius of a sphere is measured as 5 cm with a possible error of

0.05

cm, the volume

V=\tfrac43\pi r^3

has

dV=4\pi r^2\,dr=4\pi(25)(0.05)

^3

, estimating the resulting volume error without recomputing

V

twice.

The link to linearization is direct: since

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

, we have

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

, which is exactly

dy=f'(a)\,dx

with

dx=x-a

. So the differential measures the tangent-line change while linearization gives the tangent-line value. Watch units and keep

dx

small — a differential estimate degrades just like a linearization when

dx

grows.

Connection to the First-Order Taylor Polynomial

Linearization is the first-order Taylor polynomial centered at

a

. The Taylor polynomial of degree

n

about

a

begins

P_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n.

Keeping only the first two terms gives

P_1(x)=f(a)+f'(a)(x-a)=L(x)

. So everything you learn here is the seed of the Taylor series work in Unit 10, where adding higher-degree terms sharpens the approximation.

Thinking of

L

P_1

also explains the error behavior. The concavity term

\tfrac{f''(a)}{2}(x-a)^2

is the leading piece of what

L

throws away. Its sign matches

f''(a)

, confirming the over/underestimate rule: a positive second-derivative term means the true value exceeds

L

, so

L

underestimates.

Because the dropped term contains

(x-a)^2

, the error shrinks quadratically as

x\to a

. Halving the distance from

a

roughly quarters the approximation error. This is why local linear approximation is excellent very near

a

and why choosing the closest convenient base point matters so much.

Key terms

Linearization $L(x)$ .: The tangent-line function $L(x)=f(a)+f'(a)(x-a)$ used to approximate $f(x)$ for $x$ near $a$ .
Local linear approximation.: Using the tangent line at $a$ to estimate nearby function values, accurate only close to $a$ .
Base point $a$ .: The chosen center where $f(a)$ and $f'(a)$ are easy to compute; the point of tangency.
Differential.: The tangent-line change $dy=f'(x)\,dx$ , approximating the true change $\Delta y$ for small $dx$ .
Concavity.: Curving direction set by $f''$ ; $f''>0$ is concave up (line below), $f''<0$ is concave down (line above).
First-order Taylor polynomial.: $P_1(x)=f(a)+f'(a)(x-a)$ , identical to the linearization; higher terms improve accuracy.
Approximation error.: The gap $f(x)-L(x)$ ; its leading part is $\frac{f''(a)}{2}(x-a)^2$ , shrinking quadratically as $x\to a$ .

Worked example

Use a linearization to estimate

\sqrt[3]{8.06}

, and state whether the estimate is an overestimate or an underestimate.

Let

f(x)=x^{1/3}

and choose the base point

a=8

because

\sqrt[3]{8}=2

is exact and

8.06

is close.

Compute

f(8)=2

. The derivative is

f'(x)=\tfrac13 x^{-2/3}

, so

f'(8)=\tfrac13\cdot 8^{-2/3}=\tfrac13\cdot\tfrac1{4}=\tfrac1{12}

.

Build the linearization:

L(x)=2+\tfrac1{12}(x-8).

Evaluate at

x=8.06

L(8.06)=2+\tfrac1{12}(0.06)=2+0.005=2.005.

\sqrt[3]{8.06}\approx 2.005

.

Now check concavity.

f''(x)=-\tfrac29 x^{-5/3}

, which is negative for

x>0

, so the graph is concave down near

a=8

. A concave-down curve lies below its tangent line, so

L(x)

lies above the curve and the estimate is an overestimate. (Indeed

\sqrt[3]{8.06}=2.004992\ldots

, just under

2.005

Practice questions

Let

f(x)=\ln x

. Using the linearization of

f

a=1

, which value best estimates

\ln(1.1)

$0.10$
$0.05$
$1.10$
$0.11$

Answer: $0.10$

a=1

f(1)=\ln 1=0

and

f'(x)=\tfrac1x

f'(1)=1

. Then

L(x)=0+1\cdot(x-1)=x-1

, giving

L(1.1)=0.1

. The true value

\ln(1.1)\approx0.0953

, and since

f''(x)=-\tfrac1{x^2}<0

the curve is concave down, so

0.10

is a slight overestimate — consistent with the tangent-line answer.

A quantity is modeled by

f(x)=\sqrt{x}

. Write the linearization at

a=25

and use it to estimate

\sqrt{25.3}

. State with justification whether your estimate is greater or less than the true value.

Answer: $L(x)=5+\tfrac1{10}(x-25)$ , so $\sqrt{25.3}\approx 5.03$ ; this is an overestimate.

With

f(x)=x^{1/2}

f(25)=5

and

f'(x)=\tfrac1{2\sqrt{x}}

gives

f'(25)=\tfrac1{10}

. Thus

L(x)=5+\tfrac1{10}(x-25)

and

L(25.3)=5+\tfrac1{10}(0.3)=5.03

. Since

f''(x)=-\tfrac14 x^{-3/2}<0

, the graph is concave down, so the tangent line lies above the curve and

5.03

exceeds the true

\sqrt{25.3}\approx5.02991

. The justification must cite the sign of

f''

or concavity, not just assert the direction.

The radius of a circle is measured as 10 cm with a maximum error of

0.1

cm. Use a differential to estimate the maximum error in the computed area

A=\pi r^2

Answer: About $2\pi\approx6.28$ cm $^2$ .

Differentiate:

dA=2\pi r\,dr

. With

r=10

and

dr=0.1

dA=2\pi(10)(0.1)=2\pi\approx6.28

^2

. The differential replaces the true area change by the tangent-line change, which is accurate because

dr

is small relative to

r

. This is linearization applied to change rather than to value.

FAQ

How do I choose the base point $a$ ?: Pick the closest value to your target where both $f(a)$ and $f'(a)$ are easy to compute exactly. For $\sqrt{4.1}$ use $a=4$ ; for $\ln(1.02)$ use $a=1$ . The nearer $a$ is to the target, the smaller the error.
How do I tell if my linear approximation is too high or too low?: Check the sign of $f''$ near $a$ . If $f''>0$ the graph is concave up and the tangent line lies below it, so you underestimate. If $f''<0$ the graph is concave down and you overestimate. On the AP exam you must state this concavity reasoning to earn justification points.
What is the difference between linearization and differentials?: They are the same tangent-line idea. Linearization $L(x)=f(a)+f'(a)(x-a)$ estimates a function value. The differential $dy=f'(x)\,dx$ estimates the change in the value. Setting $dx=x-a$ makes $dy=L(x)-f(a)$ , so both describe the tangent line.
How does this connect to Taylor polynomials in Unit 10?: The linearization is exactly the first-order Taylor polynomial $P_1(x)$ centered at $a$ . Taylor polynomials add higher-degree terms like $\frac{f''(a)}{2!}(x-a)^2$ to reduce error; linearization just keeps the constant and linear terms.

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