Question #b91f5
1 Answer
Jan 29, 2015
The solution is
The argument of the first absolute value (i.e.
The argument of the second absolute value (i.e.
So there are three intervals:
- For
#x>=0# the arguments are both negative, so the absolute value of them are their opposite:
(remember that
- For
#0<x<=1# the first argument is negative and the second is positive.
(not acceptable, out of interval)
- For
#x>1# the arguments are both positive:
#2^(x-1)-2^x=2^x-1+1rArr2^(x-1)=2.2^xrArr2^(x-1)=2^(x+1)rArrx-1=x+1rArr-1=1#
Impossible.