Linearization and Tangent Line Approximation

If you zoom in far enough on a smooth curve, it starts to look like a straight line: its tangent line. That idea, called local linearity, lets you estimate hard-to-compute values like $\sqrt{4.1}$ or $\ln 1.05$ using nothing more than a line. On the AP exam, tangent line approximations appear often, usually with a follow-up question: is your estimate too big or too small?

Key ideas

Local linearity

If $f$ is differentiable at $x = a$, then near $a$ the graph of $f$ is very close to its tangent line. The closer $x$ is to $a$, the better the match.

The linearization

The tangent line at $x = a$ passes through $(a, f(a))$ with slope $f'(a)$. Written as a function, it is the linearization of $f$ at $a$:

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

For $x$ near $a$:

$$f(x) \approx L(x)$$

This is the same line as the point-slope form $y - f(a) = f'(a)(x - a)$, just solved for $y$.

Choosing a

Pick $a$ so that $f(a)$ and $f'(a)$ are easy to find exactly, and $a$ is close to the value you want. To estimate $\sqrt{4.1}$, use $a = 4$. To estimate $e^{0.1}$, use $a = 0$.

Overestimate or underestimate?

The concavity of $f$ between $a$ and $x$ tells you which side of the curve the tangent line is on:

Concavity	Tangent line is	Approximation $L(x)$ is
Concave up, $f''(x) \gt 0$	below the curve	an underestimate
Concave down, $f''(x) \lt 0$	above the curve	an overestimate

On the AP exam, justify with the second derivative: ”$f''(x) \lt 0$ on the interval, so the graph is concave down there, and the tangent line lies above it. The approximation is an overestimate.”

Near $x = 4$, the tangent line is almost the same as $y = \sqrt{x}$. The curve is concave down, so the line sits above it.

In context

In a word problem, the linearization reads: “new value $\approx$ value now $+$ rate now $\times$ time elapsed”. If a tank holds $500$ L at $t = 10$ min and is filling at $12$ L/min, then about $0.5$ min later it holds about $500 + 12(0.5) = 506$ L.

Worked examples

Example 1: Estimating a square root

Use a tangent line approximation to estimate $\sqrt{4.1}$. Is your estimate too big or too small?

Solution. Let $f(x) = \sqrt{x}$ and $a = 4$. Then $f(4) = 2$ and

$$f'(x) = \frac{1}{2\sqrt{x}} \quad\Rightarrow\quad f'(4) = \frac{1}{4}$$ $$L(x) = 2 + \frac{1}{4}(x - 4) \quad\Rightarrow\quad \sqrt{4.1} \approx L(4.1) = 2 + \frac{1}{4}(0.1) = 2.025$$

Check the concavity: $f''(x) = -\dfrac{1}{4}x^{-3/2} \lt 0$ for $x \gt 0$, so $f$ is concave down and the tangent line lies above the curve. The estimate $2.025$ is an overestimate. (Check: a calculator gives $\sqrt{4.1} \approx 2.024846$, a little less.)

Example 2: Linearizing ln x

Find the linearization of $f(x) = \ln x$ at $x = 1$, and use it to estimate $\ln 1.05$.

Solution. $f(1) = \ln 1 = 0$ and $f'(x) = \dfrac{1}{x}$, so $f'(1) = 1$:

$$L(x) = 0 + 1(x - 1) = x - 1$$ $$\ln 1.05 \approx L(1.05) = 0.05$$

Since $f''(x) = -\dfrac{1}{x^2} \lt 0$, the graph is concave down, so $0.05$ is an overestimate. (The true value is about $0.048790$.)

Example 3: Working from given values

A function $f$ has $f(3) = 7$ and $f'(3) = -2$, and $f''(x) \gt 0$ for all $x$. Estimate $f(3.2)$ and say whether the estimate is too big or too small.

Solution.

$$L(x) = 7 - 2(x - 3) \quad\Rightarrow\quad f(3.2) \approx 7 - 2(0.2) = 6.6$$

$f''(x) \gt 0$ means $f$ is concave up, so the tangent line lies below the graph. The estimate $6.6$ is an underestimate.

Example 4: A tank filling

$V(t)$ is the volume of water in a tank, in litres, $t$ minutes after a pump is turned on. $V(10) = 500$ and $V'(10) = 12$. Use the tangent line at $t = 10$ to estimate $V(10.5)$. If $V''(t) \lt 0$ for $10 \le t \le 10.5$, is the estimate too big or too small?

Solution.

$$V(10.5) \approx V(10) + V'(10)(10.5 - 10) = 500 + 12(0.5) = 506$$

About $506$ litres. Since $V'' \lt 0$, the graph is concave down: the rate of filling is slowing, so the tangent line (which assumes the rate stays at $12$ L/min) lies above the graph. The estimate is an overestimate.

Common mistakes

Using $f(x)$ in place of $f(a)$ in the formula. The point of the linearization is that you only need values at $a$. Write $L(x) = f(a) + f'(a)(x - a)$ and plug in numbers for $f(a)$ and $f'(a)$.

Choosing an $a$ that’s far away or awkward. For $\sqrt{4.1}$, $a = 4$ is ideal. Using $a = 1$ would work in principle but gives a poor estimate.

Getting the concavity rule backwards. Concave up means the curve bends upward away from the tangent line, so the line is below: underestimate. Picture $y = x^2$ and its tangent at the origin.

Justifying with the first derivative. Whether $f$ is increasing or decreasing has nothing to do with over- or underestimating. Use $f''$ (concavity).

Forgetting what the tangent line means in context. In Example 4, the estimate assumes the rate stays at $12$ L/min. If the rate is really slowing down, the estimate is too big.

Practice

1. (Warm-up) Find the linearization of $f(x) = x^2$ at $a = 3$ and use it to estimate $3.1^2$. Is the estimate too big or too small?

Solution

$f(3) = 9$ and $f'(x) = 2x$, so $f'(3) = 6$.

$$L(x) = 9 + 6(x - 3) \quad\Rightarrow\quad 3.1^2 \approx 9 + 6(0.1) = 9.6$$

$f''(x) = 2 \gt 0$, so the graph is concave up and $9.6$ is an underestimate. (Indeed $3.1^2 = 9.61$.)

2. (Warm-up) $f(2) = 5$ and $f'(2) = 3$. Estimate $f(2.1)$.

Solution

$$f(2.1) \approx 5 + 3(0.1) = 5.3$$

3. (Warm-up) Use a tangent line approximation to estimate $\sqrt[3]{8.12}$.

Solution

Let $f(x) = x^{1/3}$ and $a = 8$. Then $f(8) = 2$ and

$$f'(x) = \frac{1}{3}x^{-2/3} \quad\Rightarrow\quad f'(8) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$$$$\sqrt[3]{8.12} \approx 2 + \frac{1}{12}(0.12) = 2.01$$

4. (Core) Use the tangent line to $f(x) = e^x$ at $x = 0$ to estimate $e^{0.1}$. Is your estimate an overestimate or an underestimate? Justify.

Solution

$f(0) = 1$ and $f'(0) = e^0 = 1$, so $L(x) = 1 + x$ and $e^{0.1} \approx 1.1$.

$f''(x) = e^x \gt 0$ for all $x$, so the graph is concave up and the tangent line lies below it. The estimate is an underestimate. (The true value is about $1.105171$.)

5. (Core) Let $f(x) = x^3 - 2x$. Use the linearization at $a = 2$ to estimate $f(1.95)$, and decide whether it’s too big or too small.

Solution

$f(2) = 8 - 4 = 4$ and $f'(x) = 3x^2 - 2$, so $f'(2) = 10$.

$$f(1.95) \approx 4 + 10(1.95 - 2) = 4 - 0.5 = 3.5$$

$f''(x) = 6x \gt 0$ near $x = 2$, so $f$ is concave up there and $3.5$ is an underestimate. (The true value is $3.514875$.)

6. (Core) Find the linearization of $f(x) = \tan x$ at $a = \dfrac{\pi}{4}$ (radians), and use it to estimate $\tan 0.8$ to 3 decimal places.

Solution

$f\left(\dfrac{\pi}{4}\right) = 1$ and $f'(x) = \sec^2 x$, so $f'\left(\dfrac{\pi}{4}\right) = (\sqrt{2})^2 = 2$.

$$L(x) = 1 + 2\left(x - \frac{\pi}{4}\right)$$$$\tan 0.8 \approx 1 + 2\left(0.8 - \frac{\pi}{4}\right) \approx 1 + 2(0.014602) \approx 1.029$$

(A calculator gives $\tan 0.8 \approx 1.030$; the estimate is a slight underestimate, since $\tan x$ is concave up on $\left(0, \dfrac{\pi}{2}\right)$.)

7. (Core) $D(t)$ is the depth of snow on the ground, in centimetres, $t$ hours after midnight. $D(4) = 30$ and $D'(4) = 2.5$.

• (a) Estimate the depth at 4:24 a.m.
• (b) If $D''(t) \lt 0$ for $4 \le t \le 5$, is your estimate too big or too small? Explain.

Solution

(a) 4:24 a.m. is $t = 4.4$ hours.

$$D(4.4) \approx 30 + 2.5(0.4) = 31$$

About $31$ cm.

(b) $D'' \lt 0$, so the graph of $D$ is concave down and the tangent line lies above it. The estimate is too big (an overestimate).

8. (Challenge) A function $f$ has $f(3) = 2$ and $f'(x) = \sqrt{x^2 + 7}$. Estimate $f(3.1)$ and decide whether it’s an overestimate or an underestimate.

Solution

$f'(3) = \sqrt{9 + 7} = 4$, so

$$f(3.1) \approx 2 + 4(0.1) = 2.4$$

Differentiate $f'$ to find the concavity:

$$f''(x) = \frac{x}{\sqrt{x^2 + 7}}$$

which is positive for $x \gt 0$. So $f$ is concave up near $x = 3$, and $2.4$ is an underestimate.

9. (Challenge) The curve $x^2 + xy + y^2 = 7$ passes through $(1, 2)$. Use the tangent line at that point to estimate the $y$-value on the curve near $(1, 2)$ when $x = 1.1$.

Solution

Differentiate implicitly:

$$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{2x + y}{x + 2y}$$

At $(1, 2)$: $\dfrac{dy}{dx} = -\dfrac{4}{5}$.

$$y \approx 2 - \frac{4}{5}(1.1 - 1) = 2 - 0.08 = 1.92$$

(Check: solving $y^2 + 1.1y - 5.79 = 0$ gives $y \approx 1.918$, very close.)