Question

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

Answer 1

Domain: `(0,1)`
Range: `(-\infty, -2log(2))`

Explanation:

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))`