Linearization & Local Approximation

You’ll be able to

Write the equation of the tangent line as a local linear approximation
Use linearization to estimate function values near a known point
Determine whether an estimate is an overestimate or underestimate from concavity

The tangent line as a stand-in

Near a point of tangency, a smooth curve and its tangent line are nearly indistinguishable. Linearization exploits this: the tangent line at x = a, L(x) = f(a) + f′(a)(x − a), gives a quick estimate of f(x) for x close to a. The idea is that a hard-to-compute value like √4.1 can be approximated with the easy tangent line to √x at the convenient point x = 4.

Over- or underestimate? Ask concavity

Whether the tangent line lies above or below the curve is decided by concavity. Where f is concave up (f″ > 0), the curve bends above its tangent, so the linear approximation is an underestimate. Where f is concave down (f″ < 0), the curve bends below its tangent, so the approximation is an overestimate. Checking the sign of f″ turns a guess into a justified statement.

Linear approximation

L(x) = f(a) + f′(a)·(x − a)

The tangent line at x = a. Use it to estimate f(x) for x near a. Concave up ⇒ underestimate; concave down ⇒ overestimate.

Worked example

Use linearization at a = 4 to estimate √4.1, and state whether it is an over- or underestimate.

1.Let f(x) = √x = x^(1/2). Then f(4) = 2 and f′(x) = 1/(2√x), so f′(4) = 1/4.
2.Build the tangent line: L(x) = 2 + (1/4)(x − 4).
3.Evaluate at x = 4.1: L(4.1) = 2 + (1/4)(0.1) = 2 + 0.025 = 2.025.
4.Check concavity: f″(x) = −1/(4x^(3/2)) < 0, so f is concave down at x = 4.

Answer: √4.1 ≈ 2.025. Because √x is concave down there, the tangent line lies above the curve, so 2.025 is an overestimate (the true value is ≈ 2.0248).

Tip

Choose the base point a to be the nearest value where f(a) and f′(a) are easy — a perfect square, a nice angle, a whole number. The closer a is to your target, the more accurate the linear estimate.

Checkpoint

If f(3) = 10 and f′(3) = 2, what does the linear approximation estimate for f(3.5)?

Checkpoint

A function is concave up on an interval. A tangent-line approximation of f there will be:

On the exam

To justify over- versus underestimate, cite concavity explicitly: "f″ < 0 on the interval, so the graph is concave down and the tangent line overestimates." A bare "overestimate" with no reason to concavity does not earn the justification point.

Answer the 2 checkpoints as you read.