Linear Approximation Calculator

Q: What is linear approximation?

Replacing a curve near a point a by its tangent line: L(x) = f(a) + f'(a)(x - a). Close to a, f(x) \approx L(x).

Q: Is the estimate too big or too small?

If the curve is concave up near a (f'' > 0) the tangent lies below it, so the estimate is too small. If concave down (like \sqrt{x}), the estimate is too big.

Q: What is the difference between linearization and differentials?

Same idea, different notation. The differential dy = f'(a)\,dx is the change along the tangent line, and \Delta y \approx dy for small dx.

Estimate a hard value with an easy tangent line. Pick the function, a nearby point you know, and the value you want; see the estimate, the true value and the error.

Function f(x)

Reads as

\sqrt{x}

Try:

Known point aEstimate f at x

Estimate

f(4.1) \approx 2.025

true value 2.0248457 · error 0.000154

Step by step

Value at the base point
$f(4) = 2$
Slope at the base point
$f'(x) = \frac{1}{2 \sqrt{x}} \;\Rightarrow\; f'(4) = \frac{1}{4}$
Linearization
L(x) = f(a) + f′(a)(x − a).
$L(x) = 2 + \frac{1}{4}\left(x - 4\right)$
Estimate
$f(4.1) \approx L(4.1) = 2.025$
Check
f″(a) < 0: the curve is concave down, so the tangent sits above it and the estimate is an overestimate.
$\text{true value } 2.0248457,\ \text{error } 0.000154$

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Why a tangent line makes a good estimate

Zoom in far enough on any smooth curve and it looks straight. The straight line it looks like is the tangent, $L(x) = f(a) + f'(a)(x-a)$ . So for $x$ near $a$ , reading off the line instead of the curve costs very little accuracy. The error grows roughly with $(x-a)^2$ , which is why the estimate is excellent at $4.1$ and poor at $9$ .

Adding a squared term, a cubed term and so on gives better and better approximations: that is exactly the Taylor polynomial, of which $L(x)$ is the first-degree case.

Questions students ask

What is linear approximation?

Replacing a curve near a point $a$ by its tangent line: $L(x) = f(a) + f'(a)(x - a)$ . Close to $a$ , $f(x) \approx L(x)$ .

How do I estimate √4.1 or similar by hand?

Pick a nearby value where you know the answer exactly ( $a = 4$ for $\sqrt{x}$ ). Then $f(4) = 2$ , $f'(4) = \frac{1}{4}$ , so $\sqrt{4.1} \approx 2 + \frac14(0.1) = 2.025$ . The true value is 2.02485…

Is the estimate too big or too small?

If the curve is concave up near $a$ ( $f'' > 0$ ) the tangent lies below it, so the estimate is too small. If concave down (like $\sqrt{x}$ ), the estimate is too big.

What is the difference between linearization and differentials?

Same idea, different notation. The differential $dy = f'(a)\,dx$ is the change along the tangent line, and $\Delta y \approx dy$ for small $dx$ .