TITLE: Increasing or decreasing functions proof without derivation
QUESTION [0 upvotes]: We know that by definition we get that:
If $x_1 < x_2$ and if $f(x_1) < f(x_2)$, so increasing. 
Otherwise, if $f(x_1) > f(x_2)$ is decreasing.
We get that I can use that to prove if $\sqrt[3]{2x+1}$ either increasing or decreasing. And see which inequality works
Let's suppose $f(x) < f(x+1)$ 
so I get that:  $\sqrt[3]{2x -1} < \sqrt[3]{2x + 1}$
$-1 < 1$
That tell us is an increasing function because the inequality works.
Right? And then if I have a different function:
$g(x)= 3x^2+1$ how can I proceed? divided on cases? using $f(x+k)$?
 We do know that that function is increasing in the  interval (0,infinite) and decreasing in (infinite, 0) how can I prove that?

REPLY [0 votes]: If $$0\leq x<y$$ then  $$ x^2<y^2$$ then  $$3x^2<3y^2$$  and finally $$ 3x^2+1<3y^2+1.$$ Conclusion: From $0\leq x<y$ follows $g(x)<g(y).$ The function is strictly increasing in $[0,\infty).$
Similarly, if $ x<y< 0$ we get $x^2>y^2\dots$ The function is strictly decreasing in $(-\infty,0).$ 
You can freely decide whether to enclose $0$ in the first or second interval.
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