3 ms·
Correction: For some reason I was thinking x^n where n < 1 rather than n^f(x). If f(x) monotonically decreases with x then what I said still applies.
by btrettel 9y ago
Correction: For some reason I was thinking x^n where n < 1 rather than n^f(x). If f(x) monotonically decreases with x then what I said still applies.