Answer:
All real values of x where x < −1
Step-by-step explanation:
Solve the function as an inequality:
(x-3)(x+1)>0
There are two cases in this:
Both factors are greater than 0.
x-3>0 and x+1>0
x>3 and x>-1
The intersection is x >3
OR
Both factors less than 0.
x-3<0 and x+1<0
x< 3 and x<-1
The intersection is x< -1
We have obtained that the function is positive for the intervals x < -1 and x > 3. But in one of those intervals the function is decreasing and in the other is increasing.
The answer is: all real values of x where x < −1 ....
