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) = x2 at the point x = 1.
A. x2 + 1
B. 2x + 1
C. 2x – 1
D. 2x – 2
Note that f '(x) = 2x in this case. Using the formula above with a = 1, we have:
L(x) = f(1) + f '(1)(x – 1)
L(x) = 12 + 2(1)(x – 1) = 2x – 1
Follow-Up: Interpreting the Results
Clearly, the graph of the parabola f(x) = x2 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) = 12 = 1.
But the further away you get from 1, the worse the approximation becomes.
|x||f(x) = x2||L(x) = 2x - 1|
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).
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.
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 x1 to x2. Then the exact change in f(x) on that interval is:
Δy = f(x2) – f(x1)
We also use the “Delta” notation for change in x. In fact, Δx and dx typically mean the same thing:
Δx = dx = x2 – x1
However, while Δy 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
Absolute error is the absolute difference between the approximate and exact values, that is, E = | f(a) – L(a) |.
Equivalently, E = | Δy – 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(3x). 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.
Δy = 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.
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 + 3x.
Finally to find the zero, set L(x) = 0 and solve for x
0 = 5.13 + 3x → x = -5.13/3 = -1.71