Linearization
Why Approximate with a Line?
Here’s a problem: you need to compute $\sqrt{4.02}$ without a calculator. The exact value is irrational and impossible to write down. But notice that $4.02$ is very close to $4$, where we know $\sqrt{4} = 2$ exactly.
The key insight is that near any point where a function is differentiable, the function looks almost exactly like its tangent line. If you zoom in far enough on a smooth curve, it becomes indistinguishable from a straight line. So instead of computing $f(x)$ (which may be hard), we compute values on the tangent line (which is always easy, since it is linear).
This “tangent line as a function” has a special name: the linearization.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Linear Approximations and Differentials |
| Chapter | 2.9 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
Key Concepts
The Linearization Formula
If $f$ is differentiable at $a$, the linearization of $f$ at $a$ is:
$$\boxed{L(x) = f(a) + f'(a)(x - a)}$$
This is simply the tangent line at $(a, f(a))$, rewritten as a function.
| Symbol | Meaning |
|---|---|
| $a$ | The base point (where we know exact values) |
| $f(a)$ | The function value at the base point |
| $f'(a)$ | The slope of the tangent line at the base point |
| $L(x)$ | The linear function approximating $f$ near $a$ |
The Linear Approximation
The linear approximation (or tangent line approximation) says:
$$f(x) \approx L(x) = f(a) + f'(a)(x - a) \quad \text{when } x \text{ is near } a$$
The approximation improves as $x$ gets closer to $a$.
Geometric Picture
y
│ curve y = f(x)
│ /
│ ●───────── tangent line L(x)
│ /│
│ / │ f(a)
│ / │
│/ │
├───────┼───────── x
aNear $x = a$, the curve and tangent line are nearly identical. Farther away, they diverge.
Why This Works
By the definition of the derivative, as $x \to a$:
$$\frac{f(x) - f(a)}{x - a} \to f'(a)$$
Rearranging: $f(x) - f(a) \approx f'(a)(x - a)$, so $f(x) \approx f(a) + f'(a)(x - a) = L(x)$.
The derivative tells us the best linear approximation to a function near a point.
Standard Linearizations at $a = 0$
These are worth memorizing for quick estimates:
| Function | Linearization at $a = 0$ |
|---|---|
| $\sqrt{1 + x}$ | $1 + \frac{1}{2}x$ |
| $(1 + x)^n$ | $1 + nx$ |
| $\sin x$ | $x$ |
| $\cos x$ | $1$ |
| $\tan x$ | $x$ |
| $e^x$ | $1 + x$ |
| $\ln(1 + x)$ | $x$ |
Practice Problems
For $f(x) = x^2$ at $a = 3$, identify:
(a) The value $f(a)$
(b) The value $f'(a)$
(c) The linearization $L(x)$
Find the linearization of $f(x) = \sqrt[3]{x}$ at $a = 8$.
Use linearization to estimate $\sqrt[3]{8.3}$.
Consider $g(x) = \ln x$ and its linearization at $a = 1$.
(a) Find the linearization $L(x)$.
(b) Use it to estimate $\ln(1.2)$.
(c) Is your estimate an overestimate or underestimate? Explain using the concavity of $g$.
The linearization of $f(x) = e^x$ at $a = 0$ is $L(x) = 1 + x$.
(a) Write an inequality expressing that $L(x)$ approximates $e^x$ to within an error of $0.05$.
(b) Using the fact that $e^x > L(x)$ for $x \neq 0$ (since $e^x$ is concave up), determine the interval of $x$ values where the approximation is accurate to within $0.05$. You may use that $e^{0.3} \approx 1.35$ and $e^{-0.35} \approx 0.70$.
(c) Explain why the interval is not symmetric about $x = 0$.
CCI-Style Conceptual Questions
The graph shows $y = f(x)$ and its tangent line at $x = 2$, where $f(2) = 3$ and the tangent line passes through $(4, 7)$.
Without computing $f'(2)$ algebraically, write down the linearization $L(x)$ of $f$ at $a = 2$.
Three functions have linearizations at $a = 0$ given by $L(x) = 1 + 2x$.
Which of the following could NOT be one of these functions?
(A) $f(x) = e^{2x}$
(B) $g(x) = (1 + x)^2$
(C) $h(x) = 1 + 2x + x^2$
(D) $k(x) = \frac{1}{1 - 2x}$
Common Misconceptions
the linearization $L(x)$ is accurate for all values of $x$, not just those near the base point $a$.
This is the rate-as-fixed-number error applied to linear approximation. The linearization $L(x) = f(a) + f'(a)(x - a)$ captures only the slope at $x = a$; farther from $a$, curvature causes the curve to diverge from the tangent line. For $f(x) = e^x$ at $a = 0$, the approximation $L(x) = 1 + x$ gives $L(2) = 3$ while $e^2 \approx 7.39$, an error of more than 4 -- far too large for practical use.
the linearization value $L(a) = f(a)$ tells the rate at which $f$ is changing near $a$.
This is the height-vs-slope error. The value $f(a)$ is the height of the function at $x = a$; the rate of change near $a$ is the slope $f'(a)$, which is the coefficient of $(x - a)$ in the linearization. For $f(x) = x^2$ at $a = 3$, the function value is $f(3) = 9$ and the slope is $f'(3) = 6$; these are distinct numbers conveying entirely different geometric information.
Mastery Checklist
Mental Model
The “Zoom In” Analogy:
Imagine looking at a curved road on a map. From far away, you see all the bends and turns. But if you zoom in close enough to any smooth section, it looks like a straight line.
Linearization is like using that “zoomed-in straight line” to estimate positions on the actual curved road. The estimate is excellent if you stay close to your zoom point, but becomes unreliable if you wander too far.
Connections
Looking back:
- The tangent line equation $y - f(a) = f'(a)(x - a)$ is exactly the linearization, just rearranged
- This gives geometric meaning to the derivative: it’s the slope of the best linear approximation
Looking ahead:
- Differentials reframe linearization using $dy = f'(x)dx$ notation
- Error estimation applies linearization to measurement uncertainty
- Taylor polynomials (Chapter 11) extend this idea to quadratic, cubic, and higher approximations
| Previous | Up | Next |
|---|---|---|
| Implicit Differentiation | Skills Index | Differentials |
Last updated: 2026-01-22