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I was wondering if my proof sufficiently answers this question.

A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $f(x) = 2x^3+3x^2-4.$ Find the range of $f$. Is $f$ one-to-one (injective)? Is $f$ onto (surjective)? Is $f$ a bijection? Give reasons for all your answers.

My solution:

  1. Range = All real $x$. (This is obvious - but what reason could I give for this?)

  2. $f$ is not injective. For an f to be injective, for every $y \in \mathbb{R}$ (codomain of $f$), there must be at most one $x \in \mathbb{R}$ (domain of $f$) such that $y = f(x)$. [Definition of injection].

    As $f(-\frac{3}{2}) = f(0) = -4$ and $-\frac{3}{2} \neq 0$, for a $y \in \mathbb{R}$ (codomain) there exists two different $x \in \mathbb{R}$ (domain). Hence $f$ is not injective.

    I just guessed the values, surely there must be another way?

  3. Clearly $f$ is continuous (It is obvious but what reason can I give?). Hence for every $y \in \mathbb{R}$ (codomain), there is at least one $x \in \mathbb{R}$ (domain) such that $y=f(x)$. [How can I better word this?]

  4. Can I now automatically say that since $f$ is not injective then $f$ is therefore not a bijection?

Feel free to comment on anything that can improve my reasoning skills. Thanks!

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  • $\begingroup$ A couple of comments; firstly, you wrote the definition of injection incorrectly (the word "unique" is missing), and secondly, surjectivity has nothing to do with continuity (although $f$ is indeed continuous). Saying that the range is $\mathbb{R}$ is the same thing as saying that $f$ is surjective. $\endgroup$ – mdp Aug 19 '13 at 13:46
  • $\begingroup$ Thank you Matt, I fixed my definition of injection (should be correct now right)? So if you are saying for any function whose range is R, it will always be a surjective one? $\endgroup$ – Bobby Aug 19 '13 at 13:48
  • $\begingroup$ Your definition is now correct. Any function to $\mathbb{R}$ with range $\mathbb{R}$ is surjective - but the same formula defines a function $\mathbb{R}\to\mathbb{C}$ which still has range $\mathbb{R}$ so is not surjective. In general, surjectivity means the range is equal to the codomain. $\endgroup$ – mdp Aug 19 '13 at 13:49
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I don't think it is obvious that the range is all real $x$. I would argue continuity first as polynomials are continuous, then say it goes large and positive for large positive $x$ and large and negative for large negative $x$ and appeal to the intermediate value theorem. Maybe you have a theorem that polynomials of odd degree are surjective.

The unnumbered statement does show $f$ is not injective.

In 3, continuity does prove surjection onto the range, but I think the argument I gave in 1 is preferable.

For 4, yes, bijection requires both injection and surjection. Since you don't have injection you don't have bijection.

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  • $\begingroup$ Are you sure? I'm pretty sure my definition of injection is correct? Would it also suffice to just say that all polynomials are continuous? $\endgroup$ – Bobby Aug 19 '13 at 13:56
  • $\begingroup$ I misread. You are correct. Fixed. Thanks $\endgroup$ – Ross Millikan Aug 19 '13 at 14:01
  • $\begingroup$ I guess in light of your comments on 3, I should clarify mine. While it is true that using continuity together with knowledge of the limits at each infinity provides a proof of surjectivity, it looks from the original question like the OP is trying to claim that continuous functions are necessarily surjective, hence my objections. $\endgroup$ – mdp Aug 19 '13 at 14:07
  • $\begingroup$ @Bobby Continuity of polynomials follows from the fact that the identity is continuous, together with the fact that continuity is a linear property (you may know this as the sum rule) and the product rule. So hopefully this is something you can just use without needing to say too much. $\endgroup$ – mdp Aug 19 '13 at 14:09
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Try to show that for any arbitrary value there exists a number greater and less than it and then use the IVT. For example, $f(0)<0<f(2)$ and then use IVT to conclude there is an $x$ such that $f(x)=0$. Do this for an arbitrary number.

Is this a calculus class? If so, another way to show it is not injective is by using the derivative

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    $\begingroup$ Thank you! Well, it's a discrete mathematics class. I know you could differentiate it but I wanted to learn a different approach. Thanks! $\endgroup$ – Bobby Aug 19 '13 at 15:20

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