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Linear Approximation and Its Error

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The tangent line is the function, locally

Near x = a, the graph of a differentiable function is nearly indistinguishable from its tangent line. That gives the linearization L(x) = f(a) + f′(a)(x − a), used to estimate f at a nearby x when f itself is awkward to evaluate. The approximation is only as good as "nearby" is small: the error grows roughly with the square of the distance from a, so an estimate at a + 0.1 is about a hundred times better than one at a + 1.

Concavity decides the direction of the error

A tangent line lies below a concave-up graph and above a concave-down graph. So if f″ > 0 near a, the linear approximation underestimates f; if f″ < 0, it overestimates. This is a full justification on the exam and takes one sentence: state the sign of f″ on the interval, name the concavity, and conclude the direction. Students who answer "overestimate" without citing f″ get the direction right and the point wrong.

Units make the interpretation

If R(t) is measured in liters per hour and t in hours, then R′(t) is in liters per hour per hour — the rate at which the rate is changing. A derivative always carries the units of the output divided by the units of the input, and an integral of a rate carries the units of the product. On free response, "R′(3) = −2" earns nothing without "the flow rate is decreasing at 2 liters per hour per hour at time t = 3 hours."

Linearization at x = a
L(x) = f(a) + f′(a)(x − a) · f(x) ≈ L(x) for x near a
Concave up on the interval ⇒ L underestimates. Concave down ⇒ L overestimates.
Worked example

Let f be a function with f(4) = 6 and f′(4) = −0.5, and suppose f″(x) > 0 for all x. Estimate f(4.2), and state whether the estimate is too big or too small.

  1. 1.Build the linearization at a = 4: L(x) = 6 + (−0.5)(x − 4).
  2. 2.Evaluate at x = 4.2: L(4.2) = 6 + (−0.5)(0.2) = 6 − 0.1 = 5.9.
  3. 3.So f(4.2) ≈ 5.9.
  4. 4.Since f″ > 0, the graph of f is concave up, and a concave-up graph lies above its tangent line.
  5. 5.The tangent-line value 5.9 is therefore below the true value: the estimate is an underestimate.
Answer: f(4.2) ≈ 5.9, an underestimate, because f″ > 0 makes f concave up and a concave-up graph lies above its tangent line.
Watch out

The direction of the error depends on f″, not on whether f′ is positive or negative. A decreasing function can be concave up — as here — and its tangent line still underestimates.

Checkpoint

g(1) = 3 and g′(1) = 4. What does the tangent-line approximation predict for g(1.1)?

Checkpoint

A tangent-line estimate of h(2.1) is made at x = 2, where h″ < 0 on the surrounding interval. The estimate is:

On the exam

A complete over/under justification names three things: the sign of f″, the concavity that follows, and the position of the tangent line relative to the curve. Write all three.

Answer the 2 checkpoints as you read.

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