Re:A Quirky Function
by d4nte on Saturday June 25, 2016 @09:24AM
(#11976223)
After reading the problem but before reading the solution, I came up with 0^ln(x+1). Did I do my math correctly, and is this a legitimate alternative answer?
[ Parent
| on HN ]
Yes, the results are the same for 0^f(x) where f(x) is any monotonic function with f(0)=0.
f has to be monotonic and increasing, and such as f(x) < 0 if x < 0 , and f(x) > 0 if x > 0 (note the strict inequalities, that makes it a condition slightly stronger than monotonic).