Find the open intervals where $x(x-6)^3$ is concave upwards and concave downwards. Find the open intervals where $f(x)=x(x-6)^3$ is concave upwards and concave downwards.

Solution: We want to find the concavity of $f(x)=x(x-6)^3$, so we have to take the second derivative, find the critical points (where it equals zero and DNE), then split up the number line by these points and test for the sign of $f''(x)$ in each of them. First lets get to the second derivative by using the product rule and the chain rule where appropriate:
$f'(x)=(x)'(x-6)^3+x((x-6)^3)'$
$f'(x)=(1)(x-6)^3+x(3(x-6)^2)(x-6)'$
$f'(x)=(x-6)^3+x(3(x-6)^2)(1)$
$f'(x)=(x-6)^3+3x(x-6)^2$
Okay, now let's do it again
$f''(x)=((x-6)^3)'+(3x)'(x-6)^2+(3x)((x-6)^2)'$
$f''(x)=3(x-6)^2(x-6)'+3(x-6)^2+(3x)(2(x-6))(x-6)'$
$f''(x)=3(x-6)^2(1)+3(x-6)^2+6x(x-6))(1)$
$f''(x)=3(x-6)^2+3(x-6)^2+6x(x-6))$
$f''(x)=6(x-6)^2+6x(x-6))$
Foiling out and collecting like terms yields:
$f''(x)=12x^2-108x+216$
$f''(x)=12(x^2-9x+18)$
$f''(x)=12(x-6)(x-3)$
We want to know when this equals zero...
$f''(x)=12(x-6)(x-3)=0$
$\rightarrow x=6,x=3$
So let's split up our number line by $x=3,6$ and test each region:
$f''(0)=(12)(-6)(-3)=12*18>0$
$f''(5)=12(-1)(2)<0$
$f''(7)=12(1)(4)>0$
So our function is concave upward on $(-\infty,0)$ and $(6,\infty)$, and concave downwards on $(3,6)$

Set $z=x-6$;
This amounts to a translation of the coordinate system and does not change the characteristics of the function.
Then $Y=(z+6)z^3=z^4+6z^3;$
$Y'=4z^3+18z^2$; 
$Y''=12z^2+36z=12z(z+3)$;
This is a parabola.
$Y'' \gt 0$: Convex;
$12z(z+3)\gt 0$;
1) $z>0$; 2) $z <-3$; 
$Y'' <0$: Concave:  $-3 <z <0$.
Reset to $x: x=z+6$;
