abs(x^3-3x^2+2) = {(x^3-3x^2+2," if ", x^3-3x^2+2 > 0),(-(x^3-3x^2+2)," if ", x^3-3x^2+2 > 0):}
We need to solve the inequalities.
First we solve:
x^3-3x^2+2 = 0
Using the rational zeros theorem or obsrving that the sum of the coefficients is 0, we learn that 1 is a solution.
Therefore x-1 is a factor. Use division or trial and error to factor and get
(x-1)(x^2-2x-2) = 0
The quadratic can be solved using the formula or completing the square. We get:
x= 1-sqrt3, " " 1, " " 1+sqrt3
Analyzing the sign, we get
{: (bb"Interval:",(-oo,1-sqrt3),(1-sqrt3,1),(1,1+sqrt3),(1+sqrt3,oo)),
(darrbb"Factors"darr,"========","======","=====","======"),
(x-2, bb" -",bb" -",bb" +",bb" +"),
(x^2-2x-2,bb" +",bb" -",bb" -",bb" +"),
("==========","========","======","=====","======"),
(x^3-3x^2+2,bb" -",bb" +",bb" -",bb" +")
:}
So we can write
f(x) = abs(x^3-3x^2+2) = {(x^3-3x^2+2," if ", x < 1-sqrt3),(-(x^3-3x^2+2)," if ", 1-sqrt3 < x < 1),(x^3-3x^2+2," if ",1 < x < 1+sqrt3),(-(x^3-3x^2+2)," if ",1+sqrt3 < x):}
Differentiating each piece yields
f'(x) = {(3x^2-6x," if ", x < 1-sqrt3),(-3x^2+6x," if ", 1-sqrt3 < x < 1),(3x^2-6x," if ",1 < x < 1+sqrt3),(-3x^2+6x," if ",1+sqrt3 < x):}
We can quickly see that f'(x) = 0 at 0 and at 2.
It takes a moment to realize that f is not differentiable at the three cut points. x= 1-sqrt3, " " 1, " " 1+sqrt3.
The derivative on the left and right of the cuts have opposite signs because they are not 0 at the cuts.
(We can actually calculate if you like, the left derivative at 1-sqrt3 is 6 and the right derivative is -6. So the (two-sided) derivative does not exist at this point.
The critical numbers are
x= 1-sqrt3, " " 1, " " 1+sqrt3 " " (where f' does not exist)
and
x=0, " " 2 " " (where f' is 0)
