Calculus · step 3 of 13

Tangent lines & linear approximation

The tangent at x = a is the line through (a, f(a)) with slope f′(a), giving y = f(a) + f′(a)(x − a). Reading that as an approximation formula — f(x) ≈ f(a) + f′(a)(x − a) for x near a — is linearisation, the first-order Taylor polynomial, and the basis of Newton’s method and error propagation.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Compute the point and the slopeEvaluate f(a) for the point and f′(a) for the gradient. Both are needed.
Write point-slope formy − f(a) = f′(a)(x − a). Rearranging to y = mx + c is optional unless asked.
Choose a good base point for approximationPick an a near the target where f(a) and f′(a) are exact — a perfect square or a standard angle.
Estimate the error directionIf f is concave down the tangent lies above the curve, so the approximation overestimates; concave up underestimates.

Worked example

Use linear approximation to estimate √4.1.

Take f(x) = √x and a = 4, where f(4) = 2 exactly.
f′(x) = 1/(2√x), so f′(4) = 1/4.
Linearisation: f(x) ≈ 2 + (1/4)(x − 4).
At x = 4.1: 2 + (1/4)(0.1) = 2.025.

Answer. √4.1 ≈ 2.025. The true value is 2.02485…, so the estimate is a slight overestimate — as expected since √x is concave down.

Where marks get dropped

These are the specific errors that cost credit on tangent lines & linear approximation questions — QED's rubric penalises each of them separately.

Using the slope at the target point instead of at the base point. The whole point is that a is where you know the exact values.
Choosing a base point far from the target. Linearisation is only accurate nearby, with error growing like (x − a)².
Confusing the tangent line with the normal. The normal has slope −1/f′(a).

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Tangent lines & linear approximation — frequently asked questions

How accurate is a linear approximation?

The error is bounded by M(x−a)²/2 where M bounds |f″| between the points. So halving the distance quarters the error.

What is the relationship to Taylor series?

Linearisation is the degree-1 Taylor polynomial about a. Adding the quadratic term gives a better approximation with error of order (x−a)³.

Where is this used?

Newton’s method solves f(x) = 0 by repeatedly following the tangent to the axis, and error propagation in physics uses δy ≈ f′(x)δx.

The rest of Calculus

Limits, derivatives, integrals and their applications. Each subtopic below has its own method, worked example and mark-losing traps.

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