If I know that $x\in[0,2]$, how can I find the set of values that $2x^2-2x$ can take? I know that:
$$x\in[0,2]$$
and I have to find the set of values that
$$2x^2-2x$$
can take. I tried factoring the above expression into:
$$2x(x-1)$$
in the attempt to form inequalities from the given information (the fact that $0\le x \le 2$) and then multiply the inequalities, but I get inequalities that I cannot multiply, they would give the wrong answer. What approach should I use?

$2x^2 - 2x = 2(x-1)x$.
The hardest part is when $x-1 < 0$ and $x > 0$ i.e. when $0 < x < 1$
But note:  If $0 < x < 1$ then $0 < 1-x$ and $0 < x$ and by AM-GM then $(1-x)x \le ....$ Oh, I can't do that in my head....
$\frac {x + (1-x)}2 \le \sqrt{x(1-x)}$ so 
$\frac 12 \le \sqrt {x(1-x)}$ so 
$\frac 14 \le x(1-x)$ and $x(1-x) = \frac 14$ when $x =1-x$ or in other words $x=\frac 12$.
So for $0< x < 1$ we have $2(x-1)x \ge -\frac 12$ with $2(x-1)x = -\frac 12$ when $x = \frac 12$.
For $1< x \le 2$ we have $x,x-1 > 0$ so  $0=2*0*1 < 2(x-1)x\le 2*1*2=4$.
And at $x = 0$ we have $2x^2 - 2x = 0$ and at $x=2$ we have $2x^2 -2x = 4$.
And as $2x^2 -2x$ is continuous we know it doesn't "jump around" so it hits all points.
So the range of values is from $[-\frac 12, 4]$.
=====
Oh, d'oh.....
A basic study of parabolas 
If $y= ax^2 + bx + c$ then the graph is a parabola.  The $x$ intercepts are at $x = \frac {-b \pm {\sqrt b^2 -4ac}}2a; y=0$ and the axis of symmetry is at $x= \frac {-b}{2a}$ And the minimum/max value (if $a$ is positive/negative) is $c-\frac {b^2}{4a}$
So minimum value here where $a=2; b=-2; c=0$ is at $x =\frac{-2}{2*2}=\frac 12$ and is  $y=0 - \frac {4}{8}=-\frac 12$.
And maximum is at the one of the endpoints... in this case at $x = 2; y=4$.
