The question:
So, I have done almost everything. I am in the last part of the question. This is how I did it:
I differentiated the curve $y = x^3 + px^2 + px$ and got:
$dy/dx = 3x^2 + 2px + p$
I am not sure how to proceed from here. I would like to know how I would solve from here. Any help would be very highly appreciated. Final answer:
$0 < p < 3$
On
Basic approach. You now have for the derivative a quadratic expression of the form $ax^2+bx+c$. The original curve has no stationary points if and only if the derivative has no zeros; that is, when $ax^2+bx+c = 0$ has no solutions. Using the quadratic formula
$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$
as a guide, identify the values of $p$ for which that derivative has no zeros. (Hint: When is the quadratic formula undefined in the reals?)
No stationary points means $$3x^2+2px+p\neq 0$$
Use the discriminant of this quadratic equation $D=(2p)^2-4\cdot 3\cdot p$. In order for a quadratic equation to have no real roots the discriminant has to be negative. Hence, $4p^2-12p<0$ or $4p(p-3)<0$.
Can you complete it from here?
EDIT: A product of two real numbers can only be negative if only one of them is negative. As $p>0$ from the beginning you can conclude $p-3<0$ or $p<3$. Combined with $p>0$ we get the result $0<p<3$.