Question

For the curve f(x)=xln(x)+ (1-x)ln(1-x), find the value of x in the interval 0<x<1 where f(x) has a minimum.

Please help I have no idea how to solve.

Answer 1

y = x ln x + ln(1-x) - x ln (1-x)

= ln x^x + ln [ (1-x)/(1-x)^x

= ln [ x^x (1-x)^1/(1-x)^x ]

= ln [ x^x (1-x)^(1-x) ]

let z = 1-x
dz = - dx

y = ln [ x^x z^z ] = ln x^x + ln z^z

= x ln x + z ln z

y' = x (1/x) + ln x + z(1/z)dz/dx + ln z dz/dx

= 1 + ln x -1 - ln(1-x)

= 0 all the time

Answer 2

http://www.wolframalpha.com/input/?i=plot+xln%28x%29%2B+%281-x%29ln%281-x%29

Answer 3

y = x ln x + (1-x) ln(1-x)

y' = lnx + 1 - (1-x)/(1-x) - ln(1-x)
= lnx - ln(1-x)
= ln(x/(1-x))

y'=0 when x/(1-x) = 1
x = 1-x
x = 1/2