What is linear approximation? Basically, it’s a method from calculus used to “straighten out” the graph of a function near a particular point. Scientists often use linear approximation to understand complicated relationships among variables.

In this review article, we’ll explore the methods and applications of linear approximation. We’ll also take a look at plenty of examples along the way to better prepare you for the AP Calculus exams.

## Linear Approximation and Tangent Lines

By definition the **linear approximation** for a function *f*(*x*) at a point *x* = *a* is simply the equation of the *tangent line* to the curve at that point. And that means that *derivatives* are key! (Check out How to Find the Slope of a Line Tangent to a Curve or Is the Derivative of a Function the Tangent Line? for some background material.)

### Formula for the Linear Approximation

Given a point *x* = *a* and a function *f* that is differentiable at *a*, the **linear approximation** *L*(*x*) for *f* at *x* = *a* is:

*L*(*x*) = *f*(*a*) + *f* '(*a*)(*x* – *a*)

The main idea behind linearization is that the function *L*(*x*) does a pretty good job approximating values of *f*(*x*), at least when *x* is near *a*.

In other words, *L*(*x*) ≈ *f*(*x*) whenever *x* ≈ *a*.

### Example 1 — Linearizing a Parabola

Find the linear approximation of the parabola *f*(*x*) = *x*^{2} at the point *x* = 1.

A. *x*^{2} + 1

B. 2*x* + 1

C. 2*x* – 1

D. 2*x* – 2

#### Solution

**C**.

Note that *f* '(*x*) = 2*x* in this case. Using the formula above with *a* = 1, we have:

*L*(*x*) = *f*(1) + *f* '(1)(*x* – 1)

*L*(*x*) = 1^{2} + 2(1)(*x* – 1) = 2*x* – 1

#### Follow-Up: Interpreting the Results

Clearly, the graph of the parabola *f*(*x*) = *x*^{2} is not a straight line. However, near any particular point, say *x* = 1, the tangent line does a pretty good job following the direction of the curve.

How good is this approximation? Well, at *x* = 1, it’s exact! *L*(1) = 2(1) – 1 = 1, which is the same as *f*(1) = 1^{2} = 1.

But the further away you get from 1, the worse the approximation becomes.

x | f(x) = x^{2} | L(x) = 2x - 1 |
---|---|---|

1.1 | 1.21 | 1.2 |

1.2 | 1.44 | 1.4 |

1.5 | 2.25 | 2 |

2 | 4 | 3 |

## Approximating Using Differentials

The formula for linear approximation can also be expressed in terms of **differentials**. Basically, a differential is a quantity that approximates a (small) change in one variable due to a (small) change in another. The differential of *x* is *dx*, and the differential of *y* is *dy*.

Based upon the formula *dy*/*dx* = *f* '(*x*), we may identify:

*dy* = *f* '(*x*) *dx*

The related formula allows one to approximate near a particular fixed point:

*f*(*x* + *dx*) ≈ *y* + *dy*

### Example 2 — Using Differentials With Limited Information

Suppose *g*(5) = 30 and *g* '(5) = -3. Estimate the value of *g*(7).

A. 24

B. 27

C. 28

D. 33

#### Solution

**A**.

In this example, we do not know the expression for the function *g*. Fortunately, we don’t need to know!

First, observe that the change in *x* is *dx* = 7 – 5 = 2.

Next, estimate the change in *y* using the differential formula.

*dy* = *g* '(*x*) *dx* = *g* '(5) · 2 = (-3)(2) = -6.

Finally, put it all together:

*g*(5 + 2) ≈ *y* + *dy* = *g*(5) + (-6) = 30 + (-6) = 24

### Example 3 — Using Differentials to Approximate a Value

Approximate using differentials. Express your answer as a decimal rounded to the nearest hundred-thousandth.

#### Solution

**1.03333**.

Here, we should realize that even though the cube root of 1.1 is not easy to compute without a calculator, the cube root of 1 is trivial. So let’s use *a* = 1 as our basis for estimation.

Consider the function . Find its derivative (we’ll need it for the approximation formula).

Then, using the differential, , we can estimate the required quantity.

## Exact Change Versus Approximate Change

Sometimes we are interested in the *exact* change of a function’s values over some interval. Suppose *x* changes from *x*_{1} to *x*_{2}. Then the **exact change** in *f*(*x*) on that interval is:

Δ*y* = *f*(*x*_{2}) – *f*(*x*_{1})

We also use the “Delta” notation for change in *x*. In fact, Δ*x* and *dx* typically mean the same thing:

Δ*x* = *dx* = *x*_{2} – *x*_{1}

However, while Δ*y* measures the exact change in the function’s value, *dy* only estimates the change based on a derivative value.

### Example 4 — Comparing Exact and Approximate Values

Let *f*(*x*) = cos(3x), and let *L*(*x*) be the linear approximation to *f* at *x* = π/6. Which expression represents the absolute error in using *L* to approximate *f* at *x* = π/12?

A. π/6 – √2/2

B. π/4 – √2/2

C. √2/2 – π/6

D. √2/2 – π/4

#### Solution

**B.**

**Absolute error** is the absolute difference between the approximate and exact values, that is, *E* = | *f*(*a*) – *L*(*a*) |.

Equivalently, *E* = | Δ*y* – *dy* |.

Let’s compute *dy* ( = *f* '(*x*) *dx* ). Here, the change in *x* is negative. *dx* = π/12 – π/6 = -π/12. Note that by the Chain Rule, we obtain: *f* '(*x*) = -3 sin(3*x*). Putting it all together,

*dy* = -3 sin(3 π/6 ) (-π/12) = -3 sin(π/2) (-π/12) = 3π/12 = π/4

Ok, next we compute the exact change.

Δ*y* = *f*(π/12) – *f*(π/6) = cos(π/4) – cos(π/2) = √2/2

Lastly, we take the absolute difference to compute the error,

*E* = | √2/2 – π/4 | = π/4 – √2/2.

## Application — Finding Zeros

Linear approximations also serve to find zeros of functions. In fact *Newton’s Method* (see AP Calculus Review: Newton’s Method for details) is nothing more than repeated linear approximations to target on to the nearest root of the function.

The method is simple. Given a function *f*, suppose that a zero for *f* is located near *x* = *a*. Just linearize *f* at *x* = *a*, producing a linear function *L*(*x*). Then the solution to *L*(*x*) = 0 should be fairly close to the true zero of the original function *f*.

### Example 5 — Estimating Zeros

Estimate the zero of the function using a tangent line approximation at *x* = -1.

A. -1.48

B. -1.53

C. -1.62

D. -1.71

#### Solution

**D**.

Remember, The purpose of this question is to *estimate* the zero. First of all, the tangent line approximation is nothing more than a linearization. We’ll need to know the derivative:

Then find the expression for *L*(*x*). Note that *g*(-1) ≈ 2.13 and *g* '(-1) = 3.

*L*(*x*) = 2.13 + 3(*x* – (-1)) = 5.13 + 3*x*.

Finally to find the zero, set *L*(*x*) = 0 and solve for *x*

0 = 5.13 + 3*x* → *x* = -5.13/3 = -1.71

#### Improve your SAT or ACT score, guaranteed. Start your 1 Week Free Trial of Magoosh SAT Prep or your 1 Week Free Trial of Magoosh ACT Prep today!

### More from Magoosh

##### About Shaun Ault

Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!

### Leave a Reply

**Magoosh blog comment policy:** To create the best experience for our readers, we will approve and respond to comments that are relevant to the article, general enough to be helpful to other students, concise, and well-written! :) If your comment was not approved, it likely did not adhere to these guidelines. If you are a Premium Magoosh student and would like more personalized service, you can use the Help tab on the Magoosh dashboard. Thanks!