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—it's linear!).
This "tangent line as a function" has a special name: the linearization.
| Property | Value |
|---|---|
| Concept | Linear Approximations and Differentials |
| Chapter | 2.9 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
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 (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 $a$.
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.
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.
These are worth memorizing for quick estimates:
| Function | Linearization at $a = 0$ |
|---|---|
| $\sqrt{1 + x}$ | $1 + \frac{1}{2}x$ |
| $(1 + x)^n$ | $1 + nx$ |
| $\sin x$ | $x$ |
| $\cos x$ | $1$ |
| $\tan x$ | $x$ |
| $e^x$ | $1 + x$ |
| $\ln(1 + x)$ | $x$ |
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 $x = 0$.
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) $k(x) = \frac{1}{1 - 2x}$
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.
Looking back:
Looking ahead:
| Previous | Up | Next |
|---|---|---|
| Implicit Differentiation | Skills Index | Differentials |
Last updated: 2026-01-22