 Exponential sign bracketing

If $f(0)<0$ and $f(x)>0$ for all sufficiently large positive $x$, testing $1,2,4,\dots$ 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 $t\ge 1$, 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.