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.
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
http://www.wolframalpha.com/input/?i=plot+xln%28x%29%2B+%281-x%29ln%281-x%29
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
To find the value of x in the interval 0 < x < 1 where f(x) has a minimum, you can use calculus.
First, we need to find the derivative of f(x) with respect to x. Let's call this derivative f'(x).
To find f'(x), you can use the product rule of differentiation. The product rule states that if you have a function u(x) multiplied by another function v(x), then the derivative of the product is given by u'(x)v(x) + u(x)v'(x).
So applying the product rule to f(x) = xln(x) + (1 - x)ln(1 - x), we get:
f'(x) = [ln(x) + 1] + x(1/x) - [ln(1 - x) - 1](1/(1 - x))
Simplifying further, we have:
f'(x) = ln(x) + 1 + 1 - ln(1 - x) + 1/(1 - x)
Combining like terms, we have:
f'(x) = ln(x) - ln(1 - x) + 2/(1 - x)
Now, to find the x-value where f(x) has a minimum, you need to find the critical point by setting f'(x) equal to zero and solving for x:
ln(x) - ln(1 - x) + 2/(1 - x) = 0
To solve this equation, we can simplify it by multiplying through by (1 - x) to get rid of the denominators:
(1 - x)ln(x) - xln(1 - x) + 2 = 0
Now, this equation does not have a simple algebraic solution. However, you can use numerical methods or graphing software to find the approximate value of x where the equation is satisfied.
Once you find the solution(s) for x, you can then substitute these values back into the original function f(x) = xln(x) + (1 - x)ln(1 - x) to determine the corresponding minimum value of f(x) within the specified interval.