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
If there was a stationary point, then the derivative would be zero at that point, i.e.,
$\displaystyle \frac{8}{(2-x)^2} + \frac{1}{(1+x)^2} = 0$
Multiplying by both denominators gives:
$\displaystyle 8(1+x)^2 + (2-x)^2 = 0$
$\displaystyle 8+16x+8x^2 + 4-4x+x^2=0$
$\displaystyle 9x^2+12x+12=0$
$\displaystyle 3x^2+4x+4=0$
Does this quadratic equation have any real solutions?
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.