Answer to Question #86643 in Calculus for ish

Question

Find the critical numbers of
f(x)=|x3−3x2+2|
|Enter the points in increasing order and enter N into any blank you don't need to use.

Answer 1

$$f(x)=|x^3-3x^2+2|$$ $$f'(x)=(x^3-3x^2+2)*(x^3-3x^2+2)'/|x^3-3x+2|$$

$$f'(x)=(x^3-3x^2+2)*3x*(x-2)/|x^3-3x^2+2|$$

Critical numbers are number where f'(x)=0 or not exist

$$(x^3-3x^2+2)*3x*(x-2)/|x^3-3x^2+2|=0$$

$$x=0$$ $$x=2$$

f'(x) not exist at:

$$x^3-3x^2+2=0$$

$$x^3-3x^2+2=(x-1)(x^2-2x-2)=0$$

$$x=1$$

$$x^2-2x-2=0$$

$$D=4-4*(-2)=12$$

$$x=(2\pm\sqrt{12})/2=1\pm\sqrt{3}$$ $$1+\sqrt{3}\approx2.7$$

$$1-\sqrt{3}\approx-0.7$$

So, the numbers in increasing order:

$$1-\sqrt{3}; 0; 1; 2; 1+\sqrt{3}$$