Calculus I · Unit 2B · lesson

Using Linearization

Concept

Learning objectives

Select useful base points and estimate function values.

Choose a Nearby Base Point

Explanation

Before the formulas

The approximation in Using Linearization replaces a nonlinear function with its best local line. The known value supplies the starting height, and the derivative supplies the rate of change. The estimate is useful because lines are easy to calculate with, not because the original curve has become exactly linear.

Always identify the center $a$ and the target $x$ . Write $\Delta x=x-a$ , calculate the predicted change $f'(a)\Delta x$ , and then add it to $f(a)$ . Use concavity or a numerical comparison to understand the direction and size of the error.

Explanation

Choose a base point where both the function and derivative are easy

Linearization works best when the center $a$ is close to the target and $f(a)$ is known exactly. Perfect squares, perfect cubes, familiar angles, and simple exponential values are natural anchors.

Write the line $L(x)=f(a)+f'(a)(x-a)$ before substituting the target. This keeps the derivative calculation, line construction, and numerical estimate as separate checkable steps.

A linearization is most effective when the base point is close to the target and the function value and derivative are easy to compute there. The method trades exactness for speed while keeping the approximation anchored to real local behavior.

Do not choose a base point merely because it is an integer. Choose one that makes the function genuinely easy and keeps the displacement small. A nearby perfect square is useful for square roots; a nearby familiar angle is useful for trigonometric functions.

A good base point $a$ is close to the target and makes both $f(a)$ and $f'(a)$ easy.

Guided walkthrough

Estimate a cube root

Estimate $\sqrt[3]{26}$ .

Answer reveal

Worked solution

Your setup or answer

Write a real attempt before opening the supplied answer.

Worked example

Estimate a trigonometric value

Near $a=0$ , $\sin0=0$ and $\cos0=1$ , so

\sin x\approx x.

Thus $\sin(0.03)\approx0.03$ when the angle is measured in radians.

Common mistake

A linearization is local. Using the tangent at $x=4$ to estimate $\sqrt{100}$ is not a calculus technique; it is an elaborate way to be wrong.

Interactive checklinearization-01

Use the tangent to $\sqrt x$ at $x=9$ to estimate $\sqrt{9.15}$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Use L(x)=f(9)+f'(9)(x-9).

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

Modeling lab

Estimate a square root without a calculator

Use $f(x)=\sqrt{x}$ near $a=100$ . Since

f(100)=10, \qquad f'(100)=\frac1{20},

the linearization is

L(x)=10+\frac{x-100}{20}.

Therefore

\sqrt{102}\approx10+\frac2{20}=10.1.

The actual value is slightly below $10.1$ because $\sqrt{x}$ is concave down, so its tangent line lies above the curve.

After the explanation

Use the section idea

Reading lens

A differentiable curve behaves like its tangent line over a small enough neighborhood, with concavity explaining the direction of error.

Mental model

Linearization, differentials, and Newton's method reuse one local line for estimation, uncertainty, or a better root guess.

Decision

Choose a nearby easy input, record the local slope, and state why the requested change is small enough for the model.

Common trap

Presenting a tangent estimate as exact or running Newton iterations without checking residuals and failure modes.

Check yourself

Can you give the estimate, error direction, units, and a reason the local line is trustworthy here?

Source & rights

Original instruction with traceable references.

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