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I would like to verify if I am doing this right, and if I am on the right track. I feel like my proof is a bit short, so I would like some verification.

Problem: Let $f(x) = 4x^3 + 5x$. Use $f'$ to prove $f$ is one-to-one $\forall x\in\mathbb{R}$.

Attempt: If we are going to be using $f'$ to show that $f$ is a one-to-one function $\forall x\in\mathbb{R}$, then one of the two conditions must be met.

  1. If $f'(x) > 0 \ \forall x\in\mathbb{R}$, then $f(x)$ is one-to-one $\forall x\in\mathbb{R}$.
  2. If $f'(x) < 0 \ \forall x\in\mathbb{R}$, then $f(x)$ is one-to-one $\forall x\in\mathbb{R}$.

The derivative of $f$ is simply: $f'(x) = 12x^2 + 5$. If $f'(x) = 0$, then there will be no solutions to $0 = 12x^2 + 5$. Therefore, there can only be one interval, which is $x\in(-\infty, \infty)$.

If we substitute any value of $x$ into $f'(x) = 12x^2 + 5$, then the value is positive, such that $f'(x) > 0$ which means that $f'$ is increasing on $x\in(-\infty, \infty)$.

And since the function is never decreasing, and is always going to be increasing, which proves that $f$ is a one-to-one function. $\square$

Let me know if this is correct, and if there are any suggestions that I can do to make this solution better. Thanks!

maraik2002
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  • Looks fine, just the conditions 1 & 2 above are sufficient, not necessary. For example $x \mapsto x^3$ is strictly increasing but satisfies neither condition. – copper.hat Feb 01 '21 at 21:50
  • @copper.hat Not quite, because at $x = 0$ we have that $f'(x) = 0$ which is not $> 0$ or $< 0$. So, as OP says, if $f$ is strictly monotone, as it is, then $f$ is injective. – Prime Mover Feb 01 '21 at 21:56
  • @PrimeMover Right, which means that condition is sufficient. But the example of $f(x) = x^3$ shows that's it doesn't have to monotonically increasing or decreasing, it just needs to be monotonically non-decreasing or non-increasing. – Jared Feb 01 '21 at 22:00
  • What's the paragraph where you determine that "there can only be one interval" doing? I don't know if that's part of a method that you've been taught, but it feels unnecessary to me. Are you allowed to quote that the conditions imply $f$ is one-to-one? In that case I would structure my proof as "hence it suffices to show $f' > 0$, and finish the proof once I'd done that. (PS: I think your statement about one of the conditions needing to be met isn't that flawed. It's not true interpreted literally as a mathematical statement, but it's true that it's necessary for you to solve the problem) – Izaak van Dongen Feb 01 '21 at 22:02
  • @PrimeMover I'm missing your point. I gave a smooth injective (in fact bijective) example that satisfies neither 1 nor 2. – copper.hat Feb 01 '21 at 22:29
  • @IzaakvanDongen That's First Derivative Test. You find where the first derivative DNE or equals $0$, then break up the real line into intervals between critical points, then the function is monotone increasing or decreasing on every such interval. OP is saying since the derivative always exists and is positive, there's only one such interval. – Rivers McForge Feb 01 '21 at 22:32
  • @copper.hat There was never an attempt being made to prove the converse. All OP was doing was proving that if $f$ satisfies 1 and 2 then it is injective. Indeed, what you say is true, but there is no indication that OP has the need to even make an observation about this. There exists the danger of confusing OP by making him question: okay, so what have I done wrong? – Prime Mover Feb 01 '21 at 22:51
  • @PrimeMover My comment was about the wording after Attempt above where it says that one of the two condition must be met. That is what I was correcting, not any attempt as the converse. – copper.hat Feb 01 '21 at 22:58
  • @copper.hat Yes of course, sorry, my bad. – Prime Mover Feb 01 '21 at 23:00

3 Answers3

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You can significantly reduce your working by noting that $12 x^2 + 5$ cannot be less than zero (and cannot even go as low as that) because $x^2 \ge 0$ for all reals.

The rest is indeed correct.

Prime Mover
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The proof is correct. However, it is not true that one of those two conditions must be met. For instance$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&x^3\end{array}$$is one-to-one, but you have $f'(0)=0$.

José Carlos Santos
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The fact that there is no sign change in the derivative is enough to prove. But I think you're missing the reasoning. What needs to exist to not be one-to-one? You need the curve to "turn around" at some point. And what causes a function to "turn around"? A local extrema. So in your case, you simply have no critical points, therefore no local extrema. In the example mentioned:

$$ f(x) = x^3 $$

Look at the derivative:

$$ f'(x) = 2x^2 \longrightarrow x = 0 \text{ is a critical point} $$

But is there a sign change? There isn't--it's positive on both sides. So again, there are no local extrema therefore this function is also one-to-one.

Jared
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