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