How do you find the domain of the following functions #f(x)= ln(x-x^2)#?

Question

Anna Allen · Accepted Answer

The domain is the open interval #{x in RR:0 < x < 1}=(0,1)# Since the domain of #ln(x)# is #{x in RR:x>0}#, it follows that we require #x-x^2>0\leftrightarrow x(1-x)>0\leftrightarrow 0 < x < 1#. Hence the domain of #f(x)=ln(x-x^2)# is #{x in RR:0 < x < 1}=(0,1)#.

Xavier Alonso · Answer

undefined To find the domain of the function $ f(x) = \ln(x-x^2) $, you need to determine the values of $ x $ for which the function is defined. Since the natural logarithm function $ \ln(x) $ is defined only for positive values of $ x $, we need to ensure that $ x - x^2 > 0 $.

1. Solve the inequality $ x - x^2 > 0 $ for $ x $:
   $ x - x^2 > 0 $
   $ x(1 - x) > 0 $

2. Find the critical points by setting each factor equal to zero:
   $ x = 0 $ and $ 1 - x = 0 $
   Solving $ 1 - x = 0 $ gives $ x = 1 $.

3. Create a sign chart to determine the intervals where $ x(1 - x) > 0 $:
   
   $$
   \begin{array}{|c|c|c|c|}
   \hline
   \text{Interval} & x & 1 - x & x(1 - x) > 0 \
   \hline
   (-\infty, 0) & \text{negative} & \text{positive} & \text{negative} \
   (0, 1) & \text{positive} & \text{positive} & \text{positive} \
   (1, \infty) & \text{positive} & \text{negative} & \text{negative} \
   \hline
   \end{array}
   $$

4. Determine the domain:
   Since $ x(1 - x) > 0 $ for $ x \in (0, 1) $, the domain of $ f(x) = \ln(x-x^2) $ is $ \boxed{(0, 1)} $.