Question
For the curve f(x)=x ln x+(1-x)ln(1-x), find the value of x in the interval 0<x<1 where f(x) has a minimum.
Answers
Reiny
f ' (x) = x(1/x) + lnx + (1-x)(-1/(1-x)) + -1(ln(1-x)
= 1 + lnx -1 - ln(1-x)
= lnx - ln(1-x)
= ln(x/(1-x))
= 0 for a min
then e^0 = x/(1-x)
1 = x/(1-x)
1-x = x
x = 1/2
Proof:
confirmed by Wolfram
http://www.wolframalpha.com/input/?i=plot+y+%3D+x+ln+x%2B%281-x%29ln%281-x%29
= 1 + lnx -1 - ln(1-x)
= lnx - ln(1-x)
= ln(x/(1-x))
= 0 for a min
then e^0 = x/(1-x)
1 = x/(1-x)
1-x = x
x = 1/2
Proof:
confirmed by Wolfram
http://www.wolframalpha.com/input/?i=plot+y+%3D+x+ln+x%2B%281-x%29ln%281-x%29