Linear Approximation and Local Linearization Explained

What Is Linear Approximation And Local Linearization?

Linear approximation is the estimate of $f(x)$ for $x$ near a point $a$ by the value of the tangent line at $a$, and local linearization is the name for that tangent line written as a function. The tangent line is the straight line through $(a, f(a))$ with slope equal to the derivative $f'(a)$, and as a function it is

$$L(x) = f(a) + f'(a)(x - a), \qquad f(x) \approx L(x) \text{ for } x \text{ near } a.$$

The only hypothesis is that $f$ is differentiable at $a$. Finding the tangent line itself is covered on the tangent line equations page; this article is about what happens when you use that line as a stand-in for the curve: how good the estimate is, which side of the truth it lands on, and when it breaks.

Are Linear Approximation And Linearization The Same Thing?

People on Quora ask exactly this ("What is linear approximation and linearization?"), and the honest answer is that three phrases describe one idea from three angles.

• Linearization (or local linearization) is the function $L(x) = f(a) + f'(a)(x - a)$ itself.

• Linear approximation (also called tangent line approximation) is the act of using $L(x)$ in place of $f(x)$, written $f(x) \approx L(x)$.

• Local linearity is the property that makes it work: a differentiable function looks more and more like a straight line the closer you zoom in on a point.

AP-style courses use "local linearity" for the property and "local linearization" for the estimate; most university texts use "linearization" and "linear approximation". The numbers are identical either way. The naming is a real source of confusion (even experienced teachers swap the terms mid-lesson), so it is worth fixing them once here. The same estimate written in the form $dy = f'(x),dx$ is the language of differentials, which has its own article in this series.

Linearization is also the first member of the family of Taylor polynomials: it is the degree 1 Taylor polynomial of $f$ at $a$.

How Do You Find The Linear Approximation Of A Function?

Another Quora question asks, "What are some ways of finding the linear approximation of a function?" The answer depends on what you are given, but every route ends at the same line.

1. From a formula. Compute $f(a)$ and $f'(a)$, then write $L(x) = f(a) + f'(a)(x - a)$ and substitute the target $x$.

2. From a table or graph. If only data are known, estimate $f'(a)$ from neighbouring values (a symmetric difference quotient works well), then use the same formula. Example 4 below does this.

3. From a known small-$x$ rule. Near 0, results such as $e^x \approx 1 + x$ or $\sqrt{1 + x} \approx 1 + \frac{x}{2}$ can be used directly after rescaling (see the square-root section below).

For the formula route, the procedure is four short steps:

1. Choose $a$ close to the target where $f(a)$ is exact: for $\sqrt{26}$, choose $a = 25$.

2. Compute $f(a)$ and $f'(a)$.

3. Write the step $h = x - a$ with its sign.

4. Evaluate $L(x) = f(a) + f'(a),h$, and decide whether it is an overestimate or an underestimate.

Step 4 is the step most textbooks skip, and it is the one that turns a guess into an answer you can trust.

What Does Local Linearity Mean On A Graph?

Local linearity means that the graph of a differentiable function, viewed through a strong enough zoom around a point, is indistinguishable from its tangent line. The zoom panels at the top of this page show it for $\sqrt{x}$ at $x = 25$: at the widest view the curve clearly bends, and at the narrowest the curve and the line overlap.

The zoom test also shows where linearization fails. The graph of $|x|$ has a corner at 0: zoom in as far as you like and the corner stays a corner, with two different slopes on the two sides. There is no single tangent line, so there is no linearization. That is the geometric content of differentiability of a function: a function is differentiable at a point exactly when it is locally linear there.

Is A Linear Approximation An Overestimate Or An Underestimate?

A linear approximation is an underestimate when the curve is concave up between $a$ and $x$, and an overestimate when the curve is concave down there. A concave-up curve bends upward away from its tangent line, so the line sits below the curve; a concave-down curve bends downward, so the line sits above it. The sign of the second derivative on the interval tells you which.

The last row is the case students meet least and need most. For $\sin x$ at $a = 0$, the linearization is $L(x) = x$, and $f''(0) = 0$.

For $x > 0$ the sine curve is concave down, so $\sin 0.2 = 0.1987$ is below $L(0.2) = 0.2$: an overestimate. For $x < 0$ it is concave up, so $\sin(-0.2) = -0.1987$ is above $-0.2$: an underestimate. The tangent line crosses the curve at an inflection point, and the answer depends on the side.

Notice that the rule reads the second derivative, not the first. Whether $f$ is rising or falling decides the slope of the line, not which side of the curve it lies on.

How Big Is The Error In A Linear Approximation?

The error of a linear approximation is at most $\frac{M}{2}(x - a)^2$, where $M$ is the largest value of $|f''|$ between $a$ and $x$. This follows from Taylor's theorem with the remainder after the linear term:

$$|f(x) - L(x)| \leq \frac{M}{2},(x - a)^2.$$

Two things follow from the formula. The error grows with the square of the step, so halving the distance to $a$ cuts the error by about 4. And the error grows with the curvature $M$, so a nearly straight curve forgives a large step while a sharply bending one does not.

Check the bound on $\sqrt{26}$ from $a = 25$. Here $f''(x) = -\frac{1}{4}x^{-3/2}$, whose size is largest at $x = 25$:

$$M = \frac{1}{4 \cdot 125} = 0.002$$

$$\frac{M}{2}(1)^2 = 0.001$$

The actual error is $5.1 - 5.0990 = 0.00098$, just inside the bound.

The table shows what happens to the same linearization, $L(x) = 5 + \frac{x - 25}{10}$, as the step grows:

At $h = 11$ the "estimate" of $\sqrt{36}$ is off by a tenth, which is absurd for a number you know is exactly 6. Linear approximation is a local tool, and the word "local" is doing real work.

How Can The Linear Approximation Of A Square Root Be Determined?

Quora users ask this one directly, and square roots are the classic use. For $f(x) = \sqrt{x}$ near a perfect square $a = n^2$, the derivative is $f'(x) = \frac{1}{2\sqrt{x}}$, so $f'(n^2) = \frac{1}{2n}$ and

$$\sqrt{n^2 + h} \approx n + \frac{h}{2n}.$$

In words: take the root of the nearest perfect square, then add the leftover divided by twice that root. For $\sqrt{26}$ that is $5 + \frac{1}{10} = 5.1$; for $\sqrt{1.04}$ it is $1 + \frac{0.04}{2} = 1.02$, against a true $1.0198$. Because $\sqrt{x}$ is concave down, every one of these estimates is a slight overestimate.

The square-root rule is one row of a short list of small-$x$ linear approximations that physics and engineering use constantly:

The last row is the first two terms of the binomial series, and it contains the square-root row as the case $k = \frac{1}{2}$.

How Does Linearization Power Newton's Method?

Linearization does not only estimate outputs; it can also find inputs. To solve $f(x) = 0$, replace $f$ by its linearization at a guess $a$ and solve the straight-line equation instead:

$$f(a) + f'(a)(x - a) = 0 \quad\Longrightarrow\quad x = a - \frac{f(a)}{f'(a)}.$$

That single formula is Newton's method. Applied to $f(x) = x^2 - 10$ from $a = 3$, it gives

$$x_1 = 3 - \frac{9 - 10}{6} = 3.1667$$

and one more step gives $3.162281$, against $\sqrt{10} = 3.162278$. Each step is one linear approximation, and repeating it quickly pins down the answer.

Examples Of Linear Approximation And Local Linearization

Example 1

Use local linearization at $a = 25$ to estimate $\sqrt{26}$, and say whether the estimate is too high or too low.

$$f(25) = 5, \qquad f'(x) = \frac{1}{2\sqrt{x}} \Longrightarrow f'(25) = \frac{1}{10}$$

$$L(x) = 5 + \frac{1}{10}(x - 25)$$

$$L(26) = 5 + 0.1 = 5.1$$

Since $f''(x) = -\frac{1}{4}x^{-3/2} < 0$, the curve is concave down, so $5.1$ is an overestimate. The true value is $5.0990$. The step most often skipped here is that last sentence: the estimate gets computed, and the question of which side of the truth it falls on never gets asked.

Final answer: $\sqrt{26} \approx 5.1$, an overestimate.

Example 2

Estimate $e^{-0.1}$ with the linearization of $e^x$ at $a = 0$.

$$f(0) = 1, \qquad f'(0) = e^0 = 1$$

$$L(x) = 1 + x$$

$$L(-0.1) = 0.9$$

Because $e^x$ is concave up everywhere ($f''(x) = e^x > 0$), the tangent line lies below the curve and $0.9$ is an underestimate. The true value is $0.9048$. (The step is negative here, and that changes nothing about the concavity argument.)

Final answer: $e^{-0.1} \approx 0.9$, an underestimate.

Example 3

A function satisfies $f(2) = 3$, $f'(2) = -0.5$ and $f''(x) > 0$ for all $x$. Estimate $f(2.2)$ and decide whether the estimate is too high or too low.

The estimate itself is quick:

$$L(2.2) = 3 + (-0.5)(0.2) = 2.9$$

A tempting line of reasoning for the second part goes like this: the function is decreasing, so the curve drops away and the line, which drops too, overshoots it; the estimate must be too high. Test that claim on a concrete function with the same data, $f(x) = 3 - 0.5(x - 2) + (x - 2)^2$:

$$f(2.2) = 3 - 0.1 + 0.04 = 2.94$$

The true value $2.94$ is above $2.9$, so the estimate is too low, the opposite of the claim. The flaw is that the sign of $f'$ only decides whether the line slopes up or down. The side of the curve is decided by $f''$: here $f'' > 0$, the curve is concave up, the tangent line lies below it, and the linear approximation is an underestimate whatever the slope.

Final answer: $f(2.2) \approx 2.9$, an underestimate.

Example 4

A cup of coffee is cooling. Its temperature (in degrees Celsius) is recorded as $T(8) = 72$, $T(10) = 68$ and $T(12) = 65$, with time in minutes. Estimate $T(11)$ with a local linearization at $t = 10$.

There is no formula, so estimate the slope from the data with a symmetric difference:

$$T'(10) \approx \frac{65 - 72}{12 - 8} = -1.75 \text{ degrees per minute}$$

$$L(t) = 68 - 1.75(t - 10)$$

$$L(11) = 68 - 1.75 = 66.25$$

The drops (4 degrees, then 3 degrees per 2 minutes) are shrinking, which suggests the curve is concave up, so the estimate is a little low. Wait, that overstates it: three data points hint at the concavity but cannot prove it, so the honest statement is that the estimate is probably slightly low.

Final answer: $T(11) \approx 66.25$ degrees Celsius.

Example 5

A deposit of 1,000 grows for 10 years at an annual rate $r$, so $A(r) = 1000(1 + r)^{10}$. Use the linearization at $r = 0.05$ to estimate the balance at $r = 0.052$.

$$A(0.05) = 1000(1.05)^{10} \approx 1628.89$$

$$A'(r) = 10000(1 + r)^9 \Longrightarrow A'(0.05) \approx 15513.28$$

$$L(0.052) = 1628.89 + 15513.28(0.002)$$

$$L(0.052) \approx 1659.92$$

$A''(r) = 90000(1 + r)^8 > 0$, so the estimate is low; the exact balance is $1660.19$. The slope $A'(0.05)$ is the useful number in its own right: each extra 0.1 percentage point of interest adds about $15.51$ to the 10-year balance.

Final answer: about $1659.92$, an underestimate by $0.27$.

Example 6

Use the linearization of $f(x) = x^3 + 2x$ at $a = 1$ to estimate the solution of $f(x) = 3.2$.

This time the output is known and the input is wanted.

$$f(1) = 3, \qquad f'(x) = 3x^2 + 2 \Longrightarrow f'(1) = 5$$

$$3 + 5(x - 1) = 3.2$$

$$x - 1 = 0.04$$

$$x = 1.04$$

The exact solution, found numerically, is $1.0391$, so the linear estimate is off by less than $0.001$.

Final answer: $x \approx 1.04$

Why Did People Approximate Curves With Lines At All?

The short reason is that "lines are the only functions anyone can evaluate easily." Every other function (roots, exponentials, sines, logarithms) needs a table, a series or a machine, and before machines that cost hours.

• Ancient square roots were tangent-line estimates. Heron of Alexandria (about AD 10 to 75, Alexandria) approximated $\sqrt{720}$ by starting at 27, dividing 720 by 27, and averaging: $\frac{1}{2}\left(27 + \frac{720}{27}\right) = 26\tfrac{5}{6}$. Algebraically, $\frac{1}{2}\left(a + \frac{S}{a}\right) = a + \frac{S - a^2}{2a}$, which is exactly the linearization of $\sqrt{x}$ at $a^2$, evaluated at $S$. Both give $26.8333$, against a true $26.8328$.

• Repeating the step gives the modern algorithm. Heron noted that the improved value can be fed back in. That repetition is Newton's method for $x^2 - S = 0$, still used inside software for square roots.

• Engineering still runs on linearizations. Physics formulas such as the pendulum period use $\sin\theta \approx \theta$; circuit and control engineers linearize nonlinear devices around an operating point, because straight-line models can be solved, stacked and predicted.

The destination is worth naming early: almost every model in science is a linearization somewhere, and the second skill after building one is knowing how far it can be trusted.

Where Is Linear Approximation Used In Real Life?

Linear approximation is used wherever a quick, good-enough estimate near a known value beats an exact but expensive calculation.

• Physics: the small-angle rule $\sin\theta \approx \theta$ turns the pendulum equation into simple harmonic motion, and $(1 + x)^k \approx 1 + kx$ simplifies countless formulas when one quantity is much smaller than another.

• Engineering: sensors, amplifiers and control systems are designed around a linearization at their operating point.

• Economics and finance: marginal cost and the slope $A'(r)$ in Example 5 answer "how much does the output change per unit of input?" near current conditions.

• Computing: Newton's method, built from repeated linearizations, solves equations inside calculators and scientific software.

• Mental arithmetic: estimates like $\sqrt{50} \approx 7 + \frac{1}{14}$ or $1.02^{10} \approx 1.2$ come straight from the small-step rules.

The Mistakes Students Make Most Often

Each of these three errors produces a confident wrong number, which is what makes them costly.

Mistake 1: Evaluating The Derivative At The Target Instead Of At The Base Point

Where it slips in: Building $L(x)$ when the target value, such as $26$, looks like the natural input for the derivative.

Don't do this: Use $f'(26)$ in the formula. The whole point is that $f'(26)$ is as hard to compute as $f(26)$.

The correct way: Every number in $L(x)$ comes from the base point: $f(a)$ and $f'(a)$. Only the step $x - a$ involves the target.

The memorizer who learned "plug $x$ into the derivative" from earlier chapters falls into this one; the formula looks familiar, so the subscript $a$ gets read as $x$. Saying the rule out loud ("height and slope at the point I know") is the habit that fixes it.

Mistake 2: Losing The Sign Of The Step

Where it slips in: Estimating below the base point, such as $\sqrt{24}$ from $a = 25$.

Don't do this: Write $h = 1$ because 24 is "one away". That gives $\sqrt{24} \approx 5.1$, larger than $\sqrt{25} = 5$ even though $\sqrt{x}$ is increasing.

The correct way: Compute $h = x - a = 24 - 25 = -1$, so $\sqrt{24} \approx 5 - 0.1 = 4.9$ (true value $4.8990$). A quick check is direction: an increasing function must give a smaller value to the left of $a$.

Mistake 3: Using The Line Far From The Base Point

Where it slips in: Problems where the nearest convenient base point is not close, such as estimating $e = e^1$ from the linearization $1 + x$ at 0, which gives $2$ against a true $2.7183$.

Don't do this: Report a linear estimate without checking the step against the curvature.

The correct way: Check the bound $\frac{M}{2}(x - a)^2$ first. For $e^x$ on $[0, 1]$, $M = e \approx 2.7183$, so the error could be as large as $1.36$; a step of 1 is far too long for this curve, and a closer base point (or a different method) is needed.

The rusher tends to make this mistake: the formula is applied the moment a base point appears, without asking whether the step is small relative to how sharply the curve bends. The real-world version is the pendulum. The textbook period formula rests on the linear approximation $\sin\theta \approx \theta$, and HyperPhysics at Georgia State University notes that the error in the period stays under one percent only for amplitudes below about 22 degrees. Swing a pendulum clock wider than that and the simple formula predicts the wrong period, because the line has been stretched too far from its base point.

Practice Problems On Linear Approximation And Local Linearization

1. Use the linearization of $\sqrt{x}$ at $a = 100$ to estimate $\sqrt{98}$. Is it an overestimate or an underestimate?

2. Use the linearization of $\frac{1}{x}$ at $a = 1$ to estimate $\frac{1}{0.98}$.

3. Estimate $e^{0.02}$ with the linearization of $e^x$ at 0.

4. Estimate $\ln 0.9$ with the linearization of $\ln x$ at $a = 1$.

5. Estimate $\sqrt[3]{127}$ with the linearization of $\sqrt[3]{x}$ at $a = 125$.

6. Estimate $\tan^{-1} 1.1$ with the linearization of $\tan^{-1} x$ at $a = 1$.

Answer to Question 1: $L(x) = 10 + \frac{x - 100}{20}$, so $\sqrt{98} \approx 9.9$; overestimate (true value $9.8995$).

Answer to Question 2: $L(x) = 1 - (x - 1)$, so $\frac{1}{0.98} \approx 1.02$; underestimate (true value $1.0204$).

Answer to Question 3: $e^{0.02} \approx 1.02$; underestimate (true value $1.0202$).

Answer to Question 4: $\ln 0.9 \approx -0.1$; overestimate (true value $-0.1054$).

Answer to Question 5: $L(x) = 5 + \frac{x - 125}{75}$, so $\sqrt[3]{127} \approx 5.0267$; overestimate (true value $5.0265$).

Answer to Question 6: $L(x) = \frac{\pi}{4} + \frac{x - 1}{2}$, so $\tan^{-1} 1.1 \approx 0.8354$; overestimate (true value $0.8330$).

The Short Version

• Linear approximation and local linearization replace $f(x)$ near $a$ by the tangent line $L(x) = f(a) + f'(a)(x - a)$.

• Concave up between $a$ and $x$ means an underestimate, and concave down means an overestimate; the sign of $f''$ decides, not the sign of $f'$.

• The error is at most $\frac{M}{2}(x - a)^2$, so it grows with the square of the step and with the curvature.

• Linearization needs differentiability, so it fails at corners such as $|x|$ at 0.

• Solving the linearization for zero, again and again, is Newton's method.

A Practical Next Step

Work through practice problems 1 to 6 and label each answer as over or under before checking it; if a label comes out wrong, return to the over or under section, and if a number comes out wrong, compare your step sign with Mistake 2.

Book a free demo class

Read More

• Applications of derivatives in estimation and optimization

• Rate of change and what the slope measures

• The mean value theorem behind the error bound

• Maclaurin series for functions near zero