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Is this true?

Newton's method always converges from any starting point

No — it is false.

The counterexample

This one comes with the Notes pack.

See pricing

85 counterexamples are free — including most of the ones Part C leans on hardest.

The kind of mistake this is

Numerical convergence

A scheme assumed to converge, or to converge at its advertised rate.

Drill statements like this
numerical

Others that fail the same way

From Numerical Analysis › Root finding: bisection, Newton–Raphson, fixed point, order of convergence

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