Interpretation of derivatives Consider the function $f : \mathbb{R} \to \mathbb{R}$,
$$f(x) = 2x^3 − 4x^2 + 2x.$$


*

*Determine for which parts of the domain $f$ is positive/negative/zero. 


Do I take the first derivative? $f'(x)= 6x^2-8x+2$. If the first derivative is greater than zero then it is increasing. How do I determine for what parts? 


*Determine for which parts of the domain $f$ is increasing/decreasing.


Do I take each part of the first derivative and check if it is greater than zero? How would I do this and how does this differ from question 1?


*Determine for which parts of the domain f is concave/convex. 


I take the second derivative $f''(x)=12x-8=0$,
$x=2/3$ which is greater than zero, therefore the function is convex. How do I determine for each part? Or is $2/3$ already for each part?
Help appreciated! Thank you!
 A: As regards part 1) note thet your function is a polynomial which can be easily factorized:
$$f(x)=2x^3 − 4x^2 + 2x=2x(x^2-2x+1)=2x(x-1)^2.$$
The domain of $f$ is $\mathbb{R}$ and


*

*$f$ is positive for $x>0$ and $x\neq 1$, 

*$f$ is negative for $x<0$, 

*$f$ is zero at $x=0$ and $x=1$.


For part 2) and 3), the work to do is similar since 
$$f'(x)=6x^2-8x+2=2(3x-1)(x-1)\quad\mbox{and}\quad f''(x)=12x-8=4(3x-2)$$ are polynomials too.
P.S. Note that $a\cdot b>0$ iff ($a>0$ and $b>0$) or ($a<0$ and $b<0$), so to establish the sign of a product should not be a difficult job.
A: for 1) we have to show that $$6x^2-8x+2>0$$ this is equivalent to $$x^2-\frac{4}{9}x+\frac{4}{9}+\frac{1}{3}-\frac{4}{9}>0$$
this is equivalent to $$\left(x-\frac{2}{9}\right)^2-\frac{1}{9}>0$$
can you proceed?
further we have $$f''(x)=12x-8$$ and you have to solve the inequality $$12x-8>0$$ or $$12x-8<0$$
A: Are these right solutions?
2.
f´is positive for x>1 and x< 1/3
f´is negative for x>1/3 and x<1
f´is zero at x=1 and x=1/3
3.
f"is concave for x<2/3 
f" is convex for x>2/3 
f" is zero when x=2/3
Would be this be the right way to answer the third question? Because f" is just a straight line that is positive.
