Linear Approximation Calculator

This linear approximation calculator builds the linearization \( L(x) \) of a function at a point \( a \), uses it to estimate \( f(x) \) at a nearby value, and compares the estimate with the true value so you can see the error. It covers the standard “use a linear approximation to estimate…” problems from Calculus 1.

It’s free, instant and needs no sign-up.

How to use the linear approximation calculator

1. Enter the function \( f(x) \). Use ^ for powers, sqrt(x) for square roots, cbrt(x) for cube roots, sin(x), ln(x), e^(x) and pi. Implicit multiplication like 2x works.
2. Enter the center \( a \): a nearby point where \( f(a) \) is easy to compute, such as a perfect square for sqrt(x) or 1 for ln(x).
3. Enter the value \( x \) you want to estimate.
4. Press Linearize. The output shows \( L(x) \) in slope-intercept form, the estimate \( L(x) \), the actual \( f(x) \) and the absolute error \( |f(x) - L(x)| \).

Formula

The linearization of \( f \) at \( x = a \) is its tangent line:

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

For \( x \) close to \( a \), \( f(x) \approx L(x) \). The change in \( f \) is approximated by \( dy = f'(a)\,dx \) with \( dx = x - a \). A linear approximation calculator differentiates \( f \) exactly. It evaluates \( f(a) \) and \( f'(a) \). Then it simplifies \( L(x) \) to \( mx + b \) form.

Worked example

Example 1. Estimate \( \sqrt[3]{8.3} \).

Take \( f(x) = \sqrt[3]{x} \) and \( a = 8 \), since \( \sqrt[3]{8} = 2 \).

• \( f'(x) = \frac{1}{3x^{2/3}} \), so \( f'(8) = \frac{1}{3\cdot4} = \frac{1}{12} \).
• \( L(x) = 2 + \frac{1}{12}(x - 8) = \frac{x}{12} + \frac{4}{3} \).
• \( L(8.3) = 2 + \frac{0.3}{12} = 2.025 \).

The true value is \( \sqrt[3]{8.3} \approx 2.0246939 \), so the error is about 0.00031. That’s the default example in the calculator above.

Example 2. Estimate \( \ln(0.9) \) using \( a = 1 \). Here \( f(1) = 0 \) and \( f'(1) = 1 \), so \( L(x) = x - 1 \) and \( L(0.9) = -0.1 \). The true value is \( \ln 0.9 \approx -0.1053605 \), an error of about 0.0054. Use a linear approximation calculator to check. Type ln(x), 1 and 0.9.

Over- or underestimate? Reading the error

The output gives the size of the error, not its sign. Concavity tells you the direction:

• If \( f''(x) < 0 \) near \( a \) (concave down), the tangent line lies above the curve, so \( L(x) \) is an overestimate. Both examples are like this: \( \sqrt[3]{x} \) and \( \ln x \) are concave down for \( x > 0 \), and indeed \( 2.025 > 2.0246939 \) and \( -0.1 > -0.1053605 \).
• If \( f''(x) > 0 \) (concave up), \( L(x) \) is an underestimate.

Two more tips:

• Stay close to \( a \). The error grows roughly with \( (x - a)^2 \). Estimating \( \sqrt[3]{9} \) from \( a = 8 \) is fine; estimating \( \sqrt[3]{20} \) from \( a = 8 \) is not. Choose the nearest “nice” center instead (27 would be better there).
• Angles are in radians. For sin(x) or cos(x), enter \( a \) and \( x \) in radians, for example pi/6, not 30.

If you need more accuracy than a straight line gives, add a quadratic term; that’s the idea behind the Taylor series, and the Taylor series calculator builds those higher-degree polynomials. The by-hand method and a table of standard linearizations are in the linear approximation guide.

Worked example: square roots without a calculator

Approximate \( \sqrt{4.1} \) using \( f(x) = \sqrt{x} \) near \( a = 4 \). Since \( f(4) = 2 \) and \( f'(x) = \frac{1}{2\sqrt{x}} \), \( f'(4) = \frac{1}{4} \):

$$\sqrt{4.1} \approx 2 + \frac{1}{4}(0.1) = 2.025$$

The true value is 2.02485, so the error is about 0.00015. The approximation overestimates because \( \sqrt{x} \) is concave down.

The small-angle approximation

Near 0, \( \sin x \approx x \) and \( \cos x \approx 1 \) (in radians). For \( x = 0.1 \), \( \sin 0.1 = 0.09983 \), very close to 0.1. Physicists use this to simplify the pendulum equation for small swings.

How big is the error?

If \( |f''(x)| \le M \) between \( a \) and \( x \), the error of the linear approximation is at most

$$|E| \le \frac{M}{2}(x - a)^2$$

The error grows with the square of the distance from \( a \). Thus, the approximation is excellent close to \( a \), but poor far away. Adding the next term gives a Taylor polynomial for a linear approximation calculator.

Differentials: estimating measurement error

A sphere’s radius is measured as 10 cm with a possible error of 0.1 cm. The volume is \( V = \frac{4}{3}\pi r^3 \approx 4{,}189 \) cm³, and the error in the volume is about

$$dV = 4\pi r^2\,dr = 4\pi(100)(0.1) \approx 125.7 \text{ cm}^3$$

That is 3% of the volume from a 1% error in the radius, because the radius is cubed.

Step-by-step method

1. Choose a nearby point \( a \) where \( f(a) \) is easy to compute.
2. Find \( f'(a) \).
3. Write \( L(x) = f(a) + f'(a)(x - a) \).
4. Evaluate \( L \) at the point you want, and use the sign of \( f'' \) to tell whether it over- or underestimates.

Linear approximation and tangent lines

\( L(x) \) is simply the tangent line at \( x = a \), used to estimate nearby values.

Practice problems

1. Approximate \( \sqrt[3]{8.2} \) using \( a = 8 \).
2. Approximate \( \ln(1.05) \) using \( a = 1 \).

Answers: 1. \( 2 + \frac{1}{12}(0.2) \approx 2.0167 \). 2. \( 0 + 1 \times 0.05 = 0.05 \) (true value 0.04879).

Tips for Choosing 'a' and 'x'

The accuracy of a linear approximation depends heavily on your choice of 'a'. Select 'a' as a point where the function's value and its derivative are easy to calculate. This point 'a' should also be very close to 'x', the value you wish to estimate.

Consider the function's behavior. If the function changes rapidly, even a small distance between 'a' and 'x' can lead to a larger error. For instance, for trigonometric functions, choose 'a' as a standard angle like pi/4 or pi/6 in a linear approximation calculator.

When estimating roots, pick 'a' as the nearest perfect square, cube, or other power. For example, to estimate sqrt(15), choose a = 16. For cbrt(28), choose a = 27. This ensures f(a) is a whole number.

Another Full Example: Exponential Function

Let's estimate e^0.02. This value is hard to compute exactly without a calculator. We will use a linear approximation.

First, define our function. Let f(x) = e^x. Next, choose a point 'a' close to 0.02 where f(a) is easy to find. The simplest choice is a = 0. So, f(0) = e^0 = 1.

Now, find the derivative of f(x). The derivative of e^x is f'(x) = e^x. Evaluate the derivative at 'a'. Therefore, f'(0) = e^0 = 1.

Construct the linearization L(x) = f(a) + f'(a)(x – a). Plugging in our values, L(x) = 1 + 1(x – 0). This simplifies to L(x) = 1 + x.

Finally, estimate f(0.02) using L(x). So, L(0.02) = 1 + 0.02 = 1.02. The actual value of e^0.02 is approximately 1.020201. Our estimate is very close.

The error is about 0.000201. The function e^x is concave up everywhere. Thus, the linear approximation provides an underestimate, which matches our result (1.02 < 1.020201).

Related guides

• Linear Approximation (Linearization)
• How to Find the Equation of a Tangent Line
• Derivative of ln(x)
• Taylor and Maclaurin Series

Further reading

• Linear Approximations (Paul’s Online Math Notes) — the tangent line as an estimate, and why accuracy drops away from \( a \).
• Differentials (Paul’s Online Math Notes) — the \( dy = f'(x)\,dx \) notation and estimating measurement error.

FAQ

What is linear approximation used for?

It estimates function values that are hard to compute directly, such as roots and logarithms of numbers near “nice” values. It also underlies differentials, error propagation and Newton’s method.

Is linearization the same as the tangent line?

Yes. The linearization \( L(x) \) is the tangent line equation at \( x = a \). It approximates \( f(x) \). A linear approximation calculator gives this same line. It also provides the normal line.

How accurate is a linear approximation?

It is very accurate close to \( a \) and gets worse as you move away. The calculator shows the exact error for your values so you can judge.

Why is my estimate bigger than the real value?

Your function is concave down near \( a \), so the tangent line lies above the graph. For concave up functions the estimate comes out too small.