Local Linearity and Linearization — AP® Calculus AB Notes

Zoom in on the graph of any differentiable function, near any point, and it starts to look like a straight line: its tangent line. This is local linearity. It turns the tangent line into a tool for estimating values that are hard to compute, such as $\sqrt{26}$ or $f(2.1)$ when you only know $f$ at 2. The exam then asks one more question every time: is the estimate too big or too small?

The tangent line as an approximation

The tangent line at $x = a$ passes through $(a, f(a))$ with slope $f'(a)$ . In point-slope form it is the linearization of $f$ at $a$ :

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

For $x$ close to $a$ , $f(x) \approx L(x)$ . Read the formula as a story: start at the known height $f(a)$ , then add the rate $f'(a)$ times the step $x - a$ . It is the "rate times change" idea from the derivative in context, written as a line.

Worked example 1. Estimate $\sqrt{26}$ .

Choose a nearby point where $\sqrt{x}$ is easy: $a = 25$ , $f(25) = 5$ .

$f'(x) = \frac{1}{2\sqrt{x}}$ , so $f'(25) = \frac{1}{10}$ .

$L(26)$ $= 5 + \frac{1}{10}(26 - 25)$ $= 5.1$ . A calculator gives $\sqrt{26} \approx 5.0990$ , so the estimate is close, and slightly too large.

Worked example 2. A table gives $f(5) = 1$ and $f'(5) = 2$ . Estimate $f(4.8)$ .

$L(x) = 1 + 2(x - 5)$ , so $f(4.8)$ $\approx 1 + 2(-0.2)$ $= 0.6$ .

The step $x - a = -0.2$ is negative because 4.8 is to the left of 5. Dropping that sign is a common slip.

Too big or too small? Concavity decides

A graph is concave up where it bends upward, like a cup, and concave down where it bends downward, like a cap. The second derivative tells which: $f'' > 0$ means $f'$ is increasing, so the slopes keep growing and the graph bends up; $f'' < 0$ means the graph bends down. (Unit 5 studies concavity in full; here we need just this much.)

A concave up graph lies above its tangent lines, so the line gives values that are too small: an underestimate. A concave down graph lies below its tangent lines, so the line gives an overestimate.

The graph of y = √x (crimson) and its tangent line at (4, 2), L(x) = 2 + (x − 4)/4 (blue). At x = 7 the line gives 2.75 while √7 ≈ 2.646; a zoomed panel shows the line above the concave down curve, so the estimate is an overestimate. — **Figure 1.** $\sqrt{x}$ is concave down, so its tangent line at $x = 4$ lies above it: $L(7) = 2.75$ overestimates $\sqrt{7} \approx 2.646$ . LevlPrep original diagram.

On the interval between $a$ and $x$	Shape	Tangent line is	$L(x)$ is
$f'' > 0$	concave up	below the graph	an underestimate
$f'' < 0$	concave down	above the graph	an overestimate

Worked example 3. Use the tangent line to $\ln x$ at $x = 1$ to estimate $\ln(1.2)$ , and decide whether the estimate is too large or too small.

$\ln 1 = 0$ and the slope is $\frac{1}{1} = 1$ , so $L(x) = x - 1$ and $\ln(1.2) \approx 0.2$ .

The second derivative of $\ln x$ is $-\frac{1}{x^2}$ , which is negative for every $x > 0$ . The graph is concave down, so the tangent line lies above it, and $0.2$ is an overestimate. (Indeed $\ln 1.2 \approx 0.182$ .)

The justification must name concavity. "Overestimate because $f$ is increasing" does not earn the point: whether $f$ is increasing decides which way the line tilts, not which side of the curve it sits on. The concavity must hold on the whole interval between $a$ and the point you estimate.

How good is the estimate?

The gap between the curve and the line is tiny near $a$ and grows as you move away. Using the tangent line at $x = 1$ , $\ln 1.2$ is estimated as $0.2$ (true value about $0.182$ ), but $\ln 3$ is estimated as $2$ (true value about $1.099$ ). Always build the line at the known point closest to where you estimate.

Slips that cost points

Swapping the height and the slope in $L(x)$ $= f(a) + f'(a)(x - a)$ .
Losing the sign of $x - a$ when estimating to the left of $a$ .
Justifying over or under with the sign of $f'$ , or with no reason.
Using a tangent line built at a point far from the estimate.