Math Core

Zoom in far enough on the graph of a differentiable function and it starts to look like a straight line: its tangent line. That observation, called local linearity, lets you approximate hard-to-compute values of a function using nothing more than a point and a slope.

The tangent line as an approximation

If $f$ is differentiable at $x = a$ , the tangent line at $a$ passes through $(a, f(a))$ with slope $f'(a)$ . In point-slope form,

y = f(a) + f'(a)(x - a).

Near $x = a$ , the curve and this line are very close together, so values on the line are good estimates of values on the curve.

Definition

Linearization

The linearization (or local linear approximation) of $f$ at $x = a$ is

L(x) = f(a) + f'(a)(x - a).

For $x$ near $a$ , $f(x) \approx L(x)$ .

The formula has a natural reading in context: new value ≈ old value + rate × change in input. If a tank holds 200 gallons and is losing 12 gallons per minute, then half a minute later it holds about $200 + (-12)(0.5) = 194$ gallons. That is exactly $L(3.5)$ with $a = 3$ .

y = √x and its tangent line at x = 16. Near x = 16 the two are almost indistinguishable.Open in grapher →

Approximating a change

Sometimes you care about how much a quantity changes, not its new value. Subtract $f(a)$ from both sides of $f(x) \approx L(x)$ and write $\Delta x = x - a$ :

\Delta f \approx f'(a)\,\Delta x.

In words, the change in output is about the rate times the change in input. If a balloon's volume is increasing at 30 cubic inches per second, then over the next 0.2 second it gains about $30(0.2) = 6$ cubic inches. This version is handy on multiple-choice questions that ask "approximately how much will the quantity increase?"

Keep the size of $\Delta x$ small. Linear approximation assumes the rate stays roughly constant, and that assumption breaks down as you move farther from $a$ . An estimate one-tenth of a unit away is usually excellent; an estimate ten units away may be useless.

How good is the approximation?

The approximation is best close to $a$ and gets worse as you move away. On the AP exam, you are often asked whether $L(x)$ is an overestimate or an underestimate. The answer depends on which side of the tangent line the curve lies, which is controlled by concavity.

Concavity decides over or under

Near $x = a$ :

If $f''(x) > 0$ (concave up), the curve lies above its tangent line, so $L(x)$ is an underestimate.
If $f''(x) < 0$ (concave down), the curve lies below its tangent line, so $L(x)$ is an overestimate.

A picture helps you remember: a concave-up curve is shaped like a cup, and a tangent line touches the bottom of the cup from below.

Common mistake

Justify over/underestimate claims with the sign of $f''$ on the interval between $a$ and $x$ , not just at a single point. Also, don't confuse the linearization with the function itself: $L(x)$ is a line that you evaluate, and the result is only an approximation, so write $f(4.1) \approx 3.7$ , not $f(4.1) = 3.7$ .

Worked examples

Worked example: Approximating a square root

Use a linear approximation of $f(x) = \sqrt{x}$ at $x = 16$ to estimate $\sqrt{16.5}$ . Is the estimate too large or too small?

Solution. $f(16) = 4$ and $f'(x) = \dfrac{1}{2\sqrt{x}}$ , so $f'(16) = \dfrac{1}{8}$ . Then

L(x) = 4 + \frac{1}{8}(x - 16), \qquad L(16.5) = 4 + \frac{1}{8}(0.5) = 4.0625.

Since $f''(x) = -\dfrac{1}{4}x^{-3/2} < 0$ for $x > 0$ , the graph is concave down, so the estimate is an overestimate. (Indeed, $\sqrt{16.5} \approx 4.0620$ .)

Worked example: From given values only

A function $g$ satisfies $g(2) = 5$ and $g'(2) = -3$ . Estimate $g(2.1)$ .

Solution. You don't need a formula for $g$ :

g(2.1) \approx g(2) + g'(2)(2.1 - 2) = 5 + (-3)(0.1) = 4.7.

Worked example: Linearization in context

The depth of snow on a mountain, $D(t)$ inches, is measured $t$ hours after midnight. At 6 a.m. the depth is 30 inches and $D'(6) = 1.2$ . Also, $D''(t) < 0$ for $5 \le t \le 7$ . Estimate the depth at 6:30 a.m., and say whether your estimate is too high or too low.

Solution. 6:30 a.m. is $t = 6.5$ :

D(6.5) \approx 30 + 1.2(0.5) = 30.6 \text{ inches}.

Because $D'' < 0$ on the interval, $D$ is concave down there and the tangent line lies above the graph, so 30.6 inches is an overestimate.

Worked example: A linearization with an exponential

Use the tangent line to $y = e^x$ at $x = 0$ to approximate $e^{0.1}$ .

Solution. At $x = 0$ , $y = 1$ and $y' = e^0 = 1$ , so $L(x) = 1 + x$ and $e^{0.1} \approx 1.1$ . Since $(e^x)'' = e^x > 0$ , the curve is concave up and $1.1$ is an underestimate. (The true value is about 1.10517.)

Linearizing implicit curves

Local linearity works for curves defined implicitly, too. Find $\dfrac{dy}{dx}$ at the known point by implicit differentiation, then use $y \approx y_0 + \dfrac{dy}{dx}(x - x_0)$ . The same idea shows up again in Unit 7, where tangent lines to solutions of differential equations give approximate values.

Tip

A quick check: your approximation should be close to $f(a)$ . If $f(2) = 5$ and you estimated $f(2.1) = 35$ , something went wrong, probably a slope multiplied by $x$ instead of by $x - a$ .

Practice

Practice 1

A differentiable function $f$ has $f(3) = 7$ and $f'(3) = 2$ . Use a linear approximation to estimate $f(3.2)$ .