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?

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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.

  • I see, thank you! – Joy Rocha 2 days ago

An easy way to do that is to prove that a composition of increasing functions is increasing and then note that both $g(x) = x^{1/3}$ and $h(x) = 2x+1$ are increasing, with $f(x) = g(h(x))$.

For a direct proof, let $f(x) = (2x+1)^{1/3}$ and assume $x<y$. Then, $$ f(y)\ ?\ f(x) \\ (2y+1)^{1/3}\ ?\ (2x+1)^{1/3} \\ 2y + 1\ ?\ 2x+1 \\ 2(y-x)\ ?\ 0 $$ which means $?$ is really $>$, so $f(y) > f(x)$ iff $y > x$...

UPDATE

As for your other one, it is not true. If $g(x) = 3x^2+1$ then note that $g(-1) = g(1) = 4$ and $g(0) = 1$, so it seems decreasing first, and increasing afterwards, but not monotone across all real $x$.

If you want to limit it, you can prove that $g(x)$ is decreasing on $(-\infty,0]$ and increasing on $[0,\infty)$.

  • Thanks! I understood. Do you know what happen with the other one? $k(x)=3x^2+1$? Because it isn't injective so we have cases, right? – Joy Rocha Oct 10 at 21:35
  • @JoyRocha see update – gt6989b Oct 11 at 12:26

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