Newton's Method Explained: Finding Roots with Tangent Lines

Newton's method finds where a function equals zero by repeatedly sliding down tangent lines. From a guess $x_n$, draw the tangent and see where it hits the x-axis; that point is the next, better guess. The formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.

Why it works

Near a root, a smooth curve looks almost like a straight line: its tangent. The tangent's x-intercept is easy to find, and it lands close to the true root. Repeating the step usually closes in very fast.

The formula

Worked example 1: $\sqrt{2}$

Solve $x^2 - 2 = 0$, so $f(x) = x^2 - 2$ and $f'(x) = 2x$. Start at $x_0 = 1$.

After four steps the answer is correct to 12 digits.

Worked example 2: $\cos x = x$

Rewrite as $f(x) = \cos x - x = 0$, with $f'(x) = -\sin x - 1$. Starting from $x_0 = 1$ the method settles on $x \approx 0.739085$ in a few steps. No algebra trick solves this equation exactly, which is exactly when numerical methods shine.

When it fails

• Flat tangent: if $f'(x_n) = 0$ you divide by zero.
• Bad start: far from a root, a tangent can send the next guess far away.
• Cycling: occasionally guesses bounce between values.

Try a different starting guess, ideally read from a graph.

Common mistakes

• Forgetting to differentiate correctly.
• Using degrees in trig functions (calculus needs radians).
• Stopping too early: keep going until the guesses stop changing.

Practice questions

1. Do one step of Newton's method for $x^2 - 5$ from $x_0 = 2$.
2. What is $f'(x)$ for $f(x) = x^3 - 2x - 5$?
3. Why does $x_0 = 0$ fail for $f(x) = x^2 - 2$?

Answers: 1) $x_1 = 2.25$ 2) $3x^2 - 2$ 3) $f'(0) = 0$, so the tangent is flat

The Newton's method calculator lists every iteration and animates the tangents. To find all roots in a range at once, use the equation solver. Derivatives are covered in derivatives for beginners.

Frequently asked questions

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