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

Domain: we need to make sure that both logarithms exist, and every logarithm has the same request: its argument must be strictly positive. So, the first term, #ln(x)#, asks for #x>0#, while the second one asks for #1-x>0#, which easily leads to #x<1#.

Again, both conditions need to be satisfied, so we must consider only the numbers between #0# and #1#: if we go before #0# the first logarithm doesn't exist, and if we go beyond #1# the second doesn't exist. So, #D = (0,1)#

Range: As for the range, it's useful to see what happens at the borders of the domain. We have

#lim_{x\to 0^+} f(x) = ln(0^+)+ln(1) = -\infty+0 = -\infty#

Similarly, we have

#lim_{x\to 1^-} f(x) = ln(1)+ln(0^+) = 0-\infty = -\infty#

So, our function looks like an upside-down "U". The range is thus unbounded towards negative infinity. As for the maximum, we simply need to derive:

#f'(x) = 1/x-1/(1-x) = (1-x-x)/(x(1-x))=(1-2x)/(x(1-x))#

and we have #f'(x)=0 \iff 1-2x=0\iff x=1/2#

Since #f(1/2) = ln(1/2)+ln(1-1/2) = ln(1/2)+ln(1/2) =2ln(1/2) = -2log(2)#, the range is #(-\infty, -2log(2))#