= Exponential sign bracketing
If $f(0)<0$ and $f(x)>0$ for all sufficiently large positive $x$, testing $1,2,4,\ldots$ eventually finds a positive endpoint. Keep the previous nonpositive endpoint to obtain a bracket. If $t$ is the first successful exponent, there are $t+1$ endpoint tests. Subsequent <sign-bracketing binary search> in the last interval takes $t-1$ tests for $t\geq1$, since its width is $2^{t-1}$. No monotonicity is needed, though intermediate sign crossings may be skipped. Fixed-width <integer overflow> and unreliable floating-point signs require explicit handling; eventual real positivity alone does not guarantee representable machine endpoints.
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