Find the values of $m$ such that the polynomial $P(x) = x^3-mx^2+12x+11$ has no turning points $P(x) = x^3-mx^2+12x+11$
differentiate:  $y' = 3x^2 -2mx +12$
After this point I'm unsure how to progress 
 A: Guide: Two cases
Case $1$:


*

*The derivative is never $0$.


The derivative is a quadratic equation, makes its discriminant negative.
Case $2$:


*

*The derivative is $0$ but at the value that it attains $0$, the second derivative is zero as well, We then examine those values separately. 


$$y'= 3x^2-2mx+12=0$$
$$y''= 6x-2m=0$$
$$x=\frac{m}{3}$$
$$\frac{m^2}{3}-\frac{2m^2}3+12=0$$
$$-\frac{m^2}3+12=0$$
$$(m, x)\in\{ (6,2), (-6, -2)\}$$
If I substitute $m=6$, I get $y''=6(x-2)$, if $x<2$,$y''<0$;  if $x>2$,$y''>0$, hence that is an inflection point.
Similarly for $m=-6$.
Combined with the first part, we get $-6 \le m \le 6$.
A: If we make $y'=0$ ,this means y has some local minimum or maximum.And as there is some maximum and minimum point so there must be a turn in the graph.But in your case this turning is not allowed.so inveresly we can say that y can not have any minimum or maximum point.so $$y'\ne 0$$
this gives, $$x\ne\dfrac{1}{3}\bigg(m-\sqrt{m^2-36}\bigg)~~or~~\dfrac{1}
{3}\bigg(m+\sqrt{m^2-36}\bigg)$$
Let,
$$f(m)=\dfrac{1}{3}\bigg(m-\sqrt{m^2-36}\bigg)$$
$$g(m)=\dfrac{1}{3}\bigg(m+\sqrt{m^2-36}\bigg)$$
So,Domain of these functions are not allowed.
Domain of $f(m)$ and $g(m)$ is=$$\{m \in \mathbb{R}:m\le-6~~or~~m\ge 6\}$$
so,these are not allowed.
Hence, $$-6\lt m \lt 6$$
A: A turning point is a point at which the derivative changes sign.
The derivative of the given function is zero at:
$$P'(x)=3x^2 -2mx +12=0 \Rightarrow x_{1,2} =\frac{m\pm\sqrt{m^2-36}}{3}.$$
For $D>0 \Rightarrow x_1\ne x_2$ and the sign will change (because the parabola intersects the $x$-axis). 
For $D=0 \Rightarrow x_1=x_2$ and the sign does not change (because the parabola only touches the $x$-axis and stays on one side (above in this case). So, the point will not be a turning point (because the sign is not changing).
For $D<0 \Rightarrow \emptyset$. It implies no turning point at all.
So, $D=0$ and $D<0$ are what we need to solve:
$$D\le 0 \Rightarrow m^2-36\le 0 \Rightarrow m\in [-6,6].$$
Note: The points $x=-2$ (when $m=-6$) and $x=2$ (when $m=6$) are called inflection points. 
