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How can a tangent line be used to estimate the value of a function near a known point?

Topic 4.6 Approximating Values of a Function Using Local Linearity and Linearization: use the tangent line to approximate function values near a point.

A focused answer to AP Calculus AB Topic 4.6, using local linearity and the tangent line to approximate function values near a point, building the linearization formula, and determining whether the estimate is an over- or under-estimate using concavity, with worked examples.

Generated by Claude Opus 4.810 min answer

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  1. What this topic is asking
  2. The linearization formula
  3. Why the tangent line approximates
  4. A reliable procedure
  5. Over- or under-estimate from concavity
  6. Using differentials and reading the error

What this topic is asking

The College Board (Topic 4.6) uses the idea of local linearity - that a differentiable function looks like a straight line when you zoom in - to approximate function values near a known point. The approximating line is the tangent line at that point, and using it as a stand-in for the function is called linearization or linear approximation.

The linearization formula

This is just the point-slope equation of the tangent line, with the function value f(a)f(a) as the base point and the derivative f(a)f'(a) as the slope.

Why the tangent line approximates

A reliable procedure

Build the tangent line at a convenient nearby point where ff and ff' are easy, then evaluate it at the target input.

Over- or under-estimate from concavity

A signature AP question asks whether your linear approximation is too big or too small, and the answer comes from the second derivative. The tangent line is straight while the curve bends, so the curve sits on one side of its tangent. If f>0f''> 0 near aa (concave up, the curve cups upward), the tangent line lies below the curve, and the approximation under-estimates the true value. If f<0f'' < 0 (concave down), the tangent line lies above the curve, and the approximation over-estimates. To justify on the exam, compute ff'', state its sign on the relevant interval, name the concavity, and conclude the direction of the error. This concavity argument is worth a scored point and is the part students most often leave incomplete by stating the conclusion without the ff'' evidence.

Using differentials and reading the error

The same idea appears in differential language: writing dy=f(a)dxdy = f'(a)\,dx, the differential dydy approximates the actual change Δy=f(a+Δx)f(a)\Delta y = f(a + \Delta x) - f(a) for a small step dx=Δxdx = \Delta x. This is the marginal-change idea from Topic 4.1 in disguise, and it is why estimating "the next unit" of cost and approximating 9.2\sqrt{9.2} are the same mathematics. The approximation is most accurate when (a) the target is close to the base point and (b) the curve is nearly straight there (small f|f''|). Choosing a base point where the function is easy to evaluate exactly - a perfect square, a perfect cube, a multiple of π\pi - is the practical art of these problems. On the no-calculator section, linearization is the standard way to produce a decimal estimate for an awkward root or trig value without a calculator, so recognizing "approximate this value near a nice point" as a tangent-line problem is the key cue.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice, no calculator). The tangent line to ff at x=4x = 4 is used to approximate f(4.1)f(4.1). If f(4)=2f(4) = 2 and f(4)=3f'(4) = 3, the approximation is (A) 2.12.1 (B) 2.32.3 (C) 55 (D) 2.032.03
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The correct answer is (B), 2.32.3.

The linear approximation is L(x)=f(4)+f(4)(x4)=2+3(x4)L(x) = f(4) + f'(4)(x - 4) = 2 + 3(x - 4). At x=4.1x = 4.1: L(4.1)=2+3(0.1)=2.3L(4.1) = 2 + 3(0.1) = 2.3. The change is f(4)Δx=3×0.1=0.3f'(4)\,\Delta x = 3 \times 0.1 = 0.3 added to f(4)=2f(4) = 2.

AP 2023 (style)4 marksSection II (free response, no calculator). Let f(x)=xf(x) = \sqrt{x}. (a) Write the linearization of ff at x=9x = 9. (b) Use it to approximate 9.2\sqrt{9.2}. (c) Is the approximation an over- or under-estimate? Justify using concavity.
Show worked answer →

A 4-point linearization question.

(a) (2 points) f(9)=3f(9) = 3, f(x)=12xf'(x) = \frac{1}{2\sqrt{x}} so f(9)=16f'(9) = \frac{1}{6}. Thus L(x)=3+16(x9)L(x) = 3 + \frac{1}{6}(x - 9).
(b) (1 point) L(9.2)=3+16(0.2)=3+0.263.0333L(9.2) = 3 + \frac{1}{6}(0.2) = 3 + \frac{0.2}{6} \approx 3.0333.
(c) (1 point) f(x)=14x3/2<0f''(x) = -\frac{1}{4}x^{-3/2} < 0, so ff is concave down and the tangent line lies above the curve: the approximation is an over-estimate.

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