8/(2-x) - 1/(1+x)
Why would this curve have no stationary points?
The differential of this expression is : 8/(2-x)^2 + (1+x)^-2
There is a shorter way if you are a bit more observant. Since the square of any real number must be non-negative, we can notice that the denominators of both fractions of the derivative must be positive, meaning both fractions must be positive. So we are adding two positive numbers together, which can't possibly equal zero.