what is the domain and range of f(x)= ln(x) + ln(1-x)?

Answer 1

Domain:(0, 1) since x > 0 and 1-x > 0

f'(x) = 1/x - 1/(1-x) = (1-2x)/[x(1-x)] = 0
x = 1/2
Since f'(x) changes sign from positive to negative when it passes through x = 1/2, we have
fmax = f(1/2) = ln(1/2)+ln(1 - 1/2) = -2ln2
Range: (-infinity, -2ln2]

Answer 2

domain: 0 < x < 1
range: y ≤ ln(1/2)
f(x)= ln(x(1 - x))
f'(x) = (1 - 2x) / (x - x^2 = 0 for max
2x = 1
x = 1/2

Answer 3

domain: (o,1)

range: goes on forever -2ln2