Blog · Calculus guides · 7 November 2026 · 4 min read

By the Asymptote team. Every answer on this page is checked by the same machine that grades the game.

Linear Approximation: Formula, 3 Examples & Practice Problems

Linear approximation estimates a function near a point $a$ using its tangent line there: $f (x) \approx f (a) + f^{'} (a) (x - a)$ . Pick an $a$ close to $x$ where $f (a)$ and $f^{'} (a)$ are easy to work out, then plug in. For example, $4.1 \approx 2 + \frac{1}{4} (0.1) = 2.025$ .

The formula

L (x) = f (a) + f^{'} (a) (x - a) f (x) \approx L (x) for x near a

The tangent line, wearing a lab coat. Up close, every smooth curve looks straight, which is the most useful lie in calculus.

When to use it

Use it when you need a quick estimate of a function at a value that is awkward to compute, like $4.1$ , $3 27.3$ or $sin 3 1^{\circ}$ , but sits right next to one that is easy, like $4$ , $327$ or $sin 3 0^{\circ}$ . It also appears as "find the linearization of $f$ at $a$ ", which just means write down $L (x)$ , and in physics whenever someone says "for small angles". The closer $x$ is to $a$ , the better it works. Far away, the curve bends off and the line carries on regardless. It is also the idea behind Newton's method for solving equations, which follows tangent lines down to a root.

How to do it, step by step

Choose a nice $a$ . Pick the point nearest to $x$ where you know $f (a)$ exactly: a perfect square, a perfect cube, an angle from the unit circle.
Find $f (a)$ and $f^{'} (a)$ . Differentiate $f$ , then evaluate both at $a$ . $f^{'} (a)$ is the slope of the tangent line.
Write the linearization. $L (x) = f (a) + f^{'} (a) (x - a)$ . It is the point-slope form of the tangent line with a new name.
Plug in $x$ . $f (x) \approx L (x)$ . If asked, say whether it is an over- or underestimate: where the curve is concave down the line sits above it, so the estimate is too big.

Worked examples

Example 1: A square root near a perfect square

Easy

Use linear approximation to estimate $4.1$ .

Take $f (x) = x$ and $a = 4$ , since $4 = 2$ is easy.

f^{'} (x) = \frac{1}{2 x} \Rightarrow f^{'} (4) = \frac{1}{4}

L (x) = 2 + \frac{1}{4} (x - 4)

Now put in $x = 4.1$ . The step away from $a$ is $0.1$ , and a quarter of that is $0.025$ .

4.1 \approx 2 + \frac{1}{4} (0.1) = 2.025

A calculator says $2.02485$ , so the line is out by about $0.00015$ . Not bad for no calculator.

Answer

4.1 \approx 2.025

Example 2: A cube root

Medium

Use linear approximation to estimate $3 27.3$ .

Use $f (x) = 3 x = x^{1/3}$ at $a = 27$ , where $f (27) = 3$ . By the power rule,

f^{'} (x) = \frac{1}{3} x^{- 2/3} \Rightarrow f^{'} (27) = \frac{1}{3} \cdot \frac{1}{9} = \frac{1}{27}

To get $2 7^{- 2/3} = \frac{1}{9}$ , take the cube root first and then square: the numbers stay small that way round.

3 27.3 \approx 3 + \frac{1}{27} (0.3) = 3 + \frac{1}{90}

A calculator gives $3.01107$ , so the estimate is good to four decimal places. The cube root curve is concave down for $x > 0$ , so the line should overshoot slightly, and it does: $3.01111 > 3.01107$ .

Answer

3 27.3 \approx 3 + \frac{1}{90} \approx 3.0111

Example 3: An angle in degrees

Harder

Use linear approximation to estimate $sin 3 1^{\circ}$ .

The easy neighbour is $3 0^{\circ}$ . But the derivative of sine is cosine only in radians, so convert first: $3 0^{\circ} = \frac{π}{6}$ , and the extra $1^{\circ}$ is $\frac{π}{180}$ .

Take $f (x) = sin x$ at $a = \frac{π}{6}$ . Then $f (a) = \frac{1}{2}$ and $f^{'} (a) = cos \frac{π}{6} = \frac{3}{2}$ .

sin 3 1^{\circ} \approx \frac{1}{2} + \frac{3}{2} \cdot \frac{π}{180} = \frac{1}{2} + \frac{3 π}{360}

That is $0.5 + 0.01511 = 0.51511$ , against a true value of $0.51504$ . Sine is concave down here, so the line sits just above the curve and the estimate is slightly too big, as predicted.

Had you used a step of 1 instead of $\frac{π}{180}$ , you would have got $1.37$ , which is a remarkable value for a sine.

Answer

sin 3 1^{\circ} \approx \frac{1}{2} + \frac{3 π}{360} \approx 0.5151

Linear approximation practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Use the linearization of $f (x) = x$ at $a = 9$ to estimate:

9.2

Stuck? Show the answer

Answer

\frac{91}{30}

How to start

$f (9) = 3$ and $f^{'} (x) = \frac{1}{2 x}$ , so $f^{'} (9) = \frac{1}{6}$ . The step from $a$ is $0.2$ .

Problem 2

Medium

Use the linearization of $f (x) = x^{5}$ at $a = 2$ to estimate:

(2.01)^{5}

Stuck? Show the answer

Answer

32.8

How to start

$f (2) = 32$ and $f^{'} (x) = 5 x^{4}$ , so $f^{'} (2) = 80$ . Multiply that by the step, $0.01$ .

Problem 3

Harder

Use the linearization of $f (x) = x^{2} + 9$ at $a = 4$ to estimate:

(4.1)^{2} + 9

Stuck? Show the answer

Answer

5.08

How to start

$f (4) = 25 = 5$ . By the chain rule $f^{'} (x) = \frac{x}{x ^{2} + 9}$ , so $f^{'} (4) = \frac{4}{5}$ .

Want more? Asymptote has 900 related problems, free, with no account.

Common mistakes

Using an $a$ that is not easy. The whole point is that $f (a)$ is known exactly. Linearizing $x$ at $a = 4.1$ to estimate $4.1$ is going round in a circle.
Forgetting the $(x - a)$ . $f (a) + f^{'} (a)$ is a number, not a line. The slope has to multiply the step away from $a$ .
Degrees in a trig derivative. $\frac{d}{d x} sin x = cos x$ only holds in radians. Convert the small step to radians before you multiply by it.
Trusting it far from $a$ . The line hugs the curve near $a$ and drifts away from it. $x$ linearized at $a = 4$ gives $9 \approx 3.25$ , which is off by a quarter.
Writing the estimate as exact. Use $\approx$ , not $=$ . $4.1$ is not $2.025$ ; it is very close to it, and knowing the difference is part of the question.

Practice it now

Five minutes, no account

The game has no linear approximation problems as such. The closest thing is in Sprint: finding $f^{'} (a)$ , and the slope of the tangent line at a point, which is the half of every linear approximation that takes any calculus.

Play a Sprint See it on the formula sheet

FAQ

Is linear approximation the same as the tangent line?: Yes. The linearization $L (x)$ is the equation of the tangent line at $x = a$ , and linear approximation means using that line to estimate values of $f$ nearby.
What is the formula for linear approximation?: $f (x) \approx f (a) + f^{'} (a) (x - a)$ , for $x$ close to $a$ .
How do you tell if it is an overestimate or an underestimate?: Look at the concavity at $a$ . Concave up ( $f^{''} (a) > 0$ ): the tangent line lies below the curve, so it underestimates. Concave down: it overestimates. See concavity and inflection points.
Is linearization the same as differentials?: Same idea, different bookkeeping. The differential $d y = f^{'} (x) d x$ estimates the change in $y$ , and $L (x)$ is $f (a)$ plus that change.
Why does linear approximation work?: Because a differentiable function is one that looks more and more like a straight line as you zoom in on it, and the tangent line is that straight line. The error shrinks roughly like the square of the step, so halving the distance from $a$ makes the estimate about four times better.
Where can I find linear approximation practice problems?: Right on this page: the practice problems section has three linear approximation problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 900 related problems after that, free and with no account: try a five-minute Sprint.

When to use it

Use it when you need a quick estimate of a function at a value that is awkward to compute, like

4.1

3 27.3

sin 3 1^{\circ}

, but sits right next to one that is easy, like

4

327

sin 3 0^{\circ}

. It also appears as "find the linearization of

f

a

", which just means write down

L (x)

, and in physics whenever someone says "for small angles". The closer

x

is to

a

, the better it works. Far away, the curve bends off and the line carries on regardless. It is also the idea behind Newton's method for solving equations, which follows tangent lines down to a root.

How to do it, step by step

Choose a nice $a$ . Pick the point nearest to

x

where you know

f (a)

exactly: a perfect square, a perfect cube, an angle from the unit circle.

Find $f (a)$ and $f^{'} (a)$ . Differentiate

f

, then evaluate both at

a

f^{'} (a)

is the slope of the tangent line.

Write the linearization.

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

. It is the point-slope form of the tangent line with a new name.

Plug in $x$ .

f (x) \approx L (x)

. If asked, say whether it is an over- or underestimate: where the curve is concave down the line sits above it, so the estimate is too big.

Worked examples

Example 1: A square root near a perfect square

Easy

Use linear approximation to estimate $4.1$ .

Take $f (x) = x$ and $a = 4$ , since $4 = 2$ is easy.

f^{'} (x) = \frac{1}{2 x} \Rightarrow f^{'} (4) = \frac{1}{4}

L (x) = 2 + \frac{1}{4} (x - 4)

Now put in $x = 4.1$ . The step away from $a$ is $0.1$ , and a quarter of that is $0.025$ .

4.1 \approx 2 + \frac{1}{4} (0.1) = 2.025

A calculator says $2.02485$ , so the line is out by about $0.00015$ . Not bad for no calculator.

Answer

4.1 \approx 2.025

Example 2: A cube root

Medium

Use linear approximation to estimate $3 27.3$ .

Use $f (x) = 3 x = x^{1/3}$ at $a = 27$ , where $f (27) = 3$ . By the power rule,

f^{'} (x) = \frac{1}{3} x^{- 2/3} \Rightarrow f^{'} (27) = \frac{1}{3} \cdot \frac{1}{9} = \frac{1}{27}

To get $2 7^{- 2/3} = \frac{1}{9}$ , take the cube root first and then square: the numbers stay small that way round.

3 27.3 \approx 3 + \frac{1}{27} (0.3) = 3 + \frac{1}{90}

Answer

3 27.3 \approx 3 + \frac{1}{90} \approx 3.0111

Example 3: An angle in degrees

Harder

Use linear approximation to estimate $sin 3 1^{\circ}$ .

The easy neighbour is $3 0^{\circ}$ . But the derivative of sine is cosine only in radians, so convert first: $3 0^{\circ} = \frac{π}{6}$ , and the extra $1^{\circ}$ is $\frac{π}{180}$ .

Take $f (x) = sin x$ at $a = \frac{π}{6}$ . Then $f (a) = \frac{1}{2}$ and $f^{'} (a) = cos \frac{π}{6} = \frac{3}{2}$ .

sin 3 1^{\circ} \approx \frac{1}{2} + \frac{3}{2} \cdot \frac{π}{180} = \frac{1}{2} + \frac{3 π}{360}

That is $0.5 + 0.01511 = 0.51511$ , against a true value of $0.51504$ . Sine is concave down here, so the line sits just above the curve and the estimate is slightly too big, as predicted.

Had you used a step of 1 instead of $\frac{π}{180}$ , you would have got $1.37$ , which is a remarkable value for a sine.

Answer

sin 3 1^{\circ} \approx \frac{1}{2} + \frac{3 π}{360} \approx 0.5151

Linear approximation practice problems

Three to try, easy to harder. Type an answer and it is checked instantly by the same checker that grades the game; any correct form counts. Stuck? The answer and the first step are under each one.

Problem 1

Easy

Use the linearization of $f (x) = x$ at $a = 9$ to estimate:

9.2

Stuck? Show the answer

Answer

\frac{91}{30}

How to start

$f (9) = 3$ and $f^{'} (x) = \frac{1}{2 x}$ , so $f^{'} (9) = \frac{1}{6}$ . The step from $a$ is $0.2$ .

Problem 2

Medium

Use the linearization of $f (x) = x^{5}$ at $a = 2$ to estimate:

(2.01)^{5}

Stuck? Show the answer

Answer

32.8

How to start

$f (2) = 32$ and $f^{'} (x) = 5 x^{4}$ , so $f^{'} (2) = 80$ . Multiply that by the step, $0.01$ .

Problem 3

Harder

Use the linearization of $f (x) = x^{2} + 9$ at $a = 4$ to estimate:

(4.1)^{2} + 9

Stuck? Show the answer

Answer

5.08

How to start

$f (4) = 25 = 5$ . By the chain rule $f^{'} (x) = \frac{x}{x ^{2} + 9}$ , so $f^{'} (4) = \frac{4}{5}$ .

Want more? Asymptote has 900 related problems, free, with no account.

Common mistakes

Using an $a$ that is not easy. The whole point is that

f (a)

is known exactly. Linearizing

x

a = 4.1

to estimate

4.1

is going round in a circle.

Forgetting the $(x - a)$ .

f (a) + f^{'} (a)

is a number, not a line. The slope has to multiply the step away from

a

Degrees in a trig derivative.

\frac{d}{d x} sin x = cos x

only holds in radians. Convert the small step to radians before you multiply by it.

Trusting it far from $a$ . The line hugs the curve near

a

and drifts away from it.

x

linearized at

a = 4

gives

9 \approx 3.25

, which is off by a quarter.

Writing the estimate as exact. Use

\approx

, not

=

4.1

is not

2.025

; it is very close to it, and knowing the difference is part of the question.

Practice it now

Five minutes, no account

Play a Sprint See it on the formula sheet

FAQ

Is linear approximation the same as the tangent line?

Yes. The linearization

L (x)

is the equation of the tangent line at

x = a

, and linear approximation means using that line to estimate values of

f

nearby.

What is the formula for linear approximation?

f (x) \approx f (a) + f^{'} (a) (x - a)

, for

x

close to

a

How do you tell if it is an overestimate or an underestimate?

Look at the concavity at

a

. Concave up (

f^{''} (a) > 0

): the tangent line lies below the curve, so it underestimates. Concave down: it overestimates. See concavity and inflection points.

Is linearization the same as differentials?

Same idea, different bookkeeping. The differential

d y = f^{'} (x) d x

estimates the change in

y

, and

L (x)

f (a)

plus that change.

Why does linear approximation work?

Because a differentiable function is one that looks more and more like a straight line as you zoom in on it, and the tangent line is that straight line. The error shrinks roughly like the square of the step, so halving the distance from

a

makes the estimate about four times better.

Where can I find linear approximation practice problems?

Right on this page: the practice problems section has three linear approximation problems, easy, medium and harder, with the answer and the first step under each one, and a box that checks what you type with the same checker that grades Asymptote. The game has 900 related problems after that, free and with no account: try a five-minute Sprint.

Linear Approximation: Formula, 3 Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A square root near a perfect square

Example 2: A cube root

Example 3: An angle in degrees

Linear approximation practice problems

Common mistakes

FAQ

Keep going

Linear Approximation: Formula, 3 Examples & Practice Problems

The formula

When to use it

How to do it, step by step

Worked examples

Example 1: A square root near a perfect square

Example 2: A cube root

Example 3: An angle in degrees

Linear approximation practice problems

Common mistakes

FAQ

Keep going