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?
 A: This is a "bottom down" parabola (I don't know the official term in English) (because of the $+2$ in front of the $x^2$) and its zeroes are obvious from your rewrite: $x=0$ and $x=1$. So the minimum point is at $x=\frac{1}{2}$ at the halfway point between the zeroes. There it assumes the value $2\cdot \frac12 \cdot (-\frac12)=-\frac12$ and then goes upwards to $(0,0)$ on the left point of the domain, and $(2, 4)$ at the right. So the $y$-values are $[-\frac12, 4]$.
A: A graph never hurts to make, so let's start there. If nothing else, it'll confirm your observations and help your intuition, even if not a substitute for a proof:

So, we already can likely assume $f(x)=2x^2-2x$ is a continuous function. Moreover, it's on a finite interval, so you can argue through the intermediate value theorem that, between the maximum and minimum of $f$ on the interval, all values of $f$ are achieved. 
That is, if we let $\min f(x) = m, \max f(x) = M$, then $m \le f(x) \le M$ for all $x \in [0,2]$. Moreover, by continuity and the IVT, for any particular $y \in [m,M]$, there exists $c \in [0,2]$ such that $f(c)=y$. Then, by these, the range of $f$ is given by $[m,M]$.
So, how do we find the minimum and maximum? 
That's simple.


*

*Find the extrema in the interval. Find the derivative $f'(x)$, and find the values of $x$ for which $f'(x) = 0$. Record the values of $f$ at these values.

*Examine the endpoints of the interval. Here, your endpoints are $x=0$ and $x=2$, so just find $f(0),f(2)$.

*The smallest $f(x)$ value will be your minimum $m$, and the largest will be your maximum $M$.

For clarity, let's go through an example.

Find the range of the function $f(x) = x^3 - 3x^2$ on the interval $[1,4]$.

$f$ is a polynomial, so we have continuity. The interval $[1,4]$ is finite in length as well. Therefore, between the minimum and maximum achieved on the interval are all of the values $f$ achieves.
We take the derivative and find $f'(x) = 3x^2 - 6x$. Solving $f'(x) = 0$ gives us
$$3x^2 - 6x = 3x(x - 2) = 0$$
which clearly means $x=0,x=2$ are where $f'(x)=0$. However, $x=0$ is not in the interval of concern, so the only potential (relevant) extremum is $x=2$.
We also have to test the endpoints, $x=1,x=4$. We find the values at each:


*

*$f(2) = -4$

*$f(1) = -2$

*$f(4) = 16$
The biggest of these is $f(4) = 16$ and the smallest is $f(2) = -4$. Therefore, the range of $f$ on the interval $[1,4]$ is $[-4,16]$. And look, our graph confirms this:

A: Here it is another way to tackle the problem. The function $f(x) = 2x^{2} - 2x$ can be rewritten as
\begin{align*}
2x^{2} - 2x = 2(x^{2} - x) & = 2\left(x^{2} - x + \frac{1}{4} - \frac{1}{4}\right)\\
& = 2\left(x-\frac{1}{2}\right)^{2} - \frac{1}{2} \geq -\frac{1}{2}
\end{align*}
which attains its minimum at $x = 1/2$. Since $f'(x) = 4x - 2 < 0$ for $x < 1/2$ and $f'(x) > 0$ for $x > 1/2$, it suffices to check its values on the points $x = 0$ and $x = 2$.
Since $f(0) = 0$ and $f(2) = 4$, we conclude that $f([0,2]) = [-1/2,4]$. Hopefully this helps.
A: $2x^2 - 2x = 2(x-1)x$.
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 $x = 2; y=4$.
