TITLE: Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.
QUESTION [6 upvotes]: Prove that $f : [-1, 1] \rightarrow \mathbb{R}$, $x \mapsto x^2 + 3x + 2$ is strictly increasing.

I do not have use derivatives, so I decided to apply the definition of being a strictly increasing function, which should be: 

If we pick 2 numbers $a$ and $b$ from the domain of a function $f$, where $a < b$, then $f(a) < f(b)$.

Now (if that is a correct definition), I have tried to apply it to my case:

Let $a$, $b \in [-1, 1]$ and $a < b$. We want to show that $f(a) < f(b)$ or that $f(b) - f(a) > 0$.
We know that $x^2 + 3x + 2 = (x + 2)(x + 1)$, thus we have that $f(a) = (a + 2)(a + 1)$ and $f(b) = (b + 2)(b + 1)$ , therefore we need to show that:
$$(b + 2)(b + 1) - (a + 2)(a + 1) > 0$$
We can see that $(b + 2)$ and $(a + 2)$ will always be positive, and that $(b + 2) > (a + 2)$ , since $b > a$ (by assumption).
Since $b > a$, we know that $b > -1$ (otherwise $a \geq b$), thus $(b + 1) > 0$ (so we know that $(b + 2)(b + 1) > 0$ .  We also have at most $(a + 1) = 0$, and thus  we have that $(a + 2)(a + 1) \ge 0$.
So, here is the proof that  $f(b) - f(a) > 0$ .  

Am I correct? If yes, what can I improve it? If not, where are the erros and possible solutions?

REPLY [4 votes]: The proof given looks fine to me, modulo my comment (see above).
Here's another way to see it, sans calculus:
Choose $x_1, x_2 \in [-1, 1]$ with $x_1 > x_2$.  Then
$f(x_1) - f(x_2) = x_1^2 + 3x_1 + 2 - x_2^2 - 3x_2 - 2 = x_1^2 - x_2^2 + 3(x_1 - x_2)$
$= (x_1 + x_2)(x_1 - x_2) + 3(x_1 - x_2) = (x_1 + x_2 + 3)(x_1 - x_2); \tag{1}$
note that $x_1 - x_2 > 0$ since $x_1 > x_2$ and that $x_1 + x_2 + 3 > 0$ since $x_1 + x_2 > -2$ since $x_1 > x_2 \ge -1$.  Thus
$f(x_1) -f(x_2) > 0, \tag{2}$
or
$f(x_1) > f(x_2), \tag{3}$
and we're done!
Note that all we really needed was $x_1 > x_2 \ge -1$.
Of course, with calculus we see that $f'(x) = 2x + 3 >0$ for $x \ge -1$, etc. etc. etc.  And since the minimum occurs at $x = -3/2$, Mark Bennet's comment is corroborated.