Can't solve it? Slide down the tangent.

Most equations have no formula for their roots. Newton–Raphson finds them anyway, one tangent at a time, and each step roughly doubles the number of correct digits.

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$x^3 - 2x - 5 = 0$ has a root between $2$ and $3$: $f(2) = -1$ and $f(3) = 16$, so the curve crosses the axis in between. There is no neat way to find it exactly. Start at $x_0 = 3$.

The tangent at $x_n$ has gradient $f'(x_n)$, so it meets the axis a horizontal distance $\tfrac{f(x_n)}{f'(x_n)}$ further on. Go up to the curve there and repeat:

From $3$ the steps go $2.36$, $2.1271\ldots$, $2.0951\ldots$, $2.0945516\ldots$ Already $x_4$ shares its first seven figures with the root, $2.0945514\ldots$ Each step roughly doubles the correct digits.

If $f'(x_n) = 0$ the tangent is horizontal and never meets the axis. Here that happens at the bottom of the dip, $x = \sqrt{\tfrac{2}{3}}$. A start near such a point is almost as bad: a nearly flat tangent throws the next value far away.

Tangent, axis, curve, repeat.

Locating roots by a change of sign, iteration with staircase and cobweb diagrams, the trapezium rule, and twelve graded exam questions with full worked solutions are waiting.

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