# Increasing or decreasing functions proof without derivation

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?

If $$0\leq x then $$x^2 then $$3x^2<3y^2$$ and finally $$3x^2+1<3y^2+1.$$ Conclusion: From $$0\leq x follows $$g(x) The function is strictly increasing in $$[0,\infty).$$

Similarly, if $$x we get $$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 Oct 14 '18 at 16:57

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 $$x. Then, $$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 '18 at 21:35
• @JoyRocha see update – gt6989b Oct 11 '18 at 12:26