mycroft holmes
Last Activity: 10 Years ago
If x>0, then consider the function, f(x) = x1/x. This is monotonic in the intervals (0, e) and (e, infty)
Thus only two solutions exist to x1/x =2 1/2 i.e. x =2 and x=4
If x<0, then f(x) = 2x – x2 monotonically increasing and continuous. Also, f(0)>0 and f(-1)<0 and so one more solution may be found in this interval.
Thus three distinct solutions exist
