For the record, I am sorry, I haven't yet learnt how to use LaTeX

I have a function $f(x) = 2x^3 - 1$

My proof of injection is as follows: $f$ is one to one for all $x_1,x_2$ element $X$, if $f(x_1) = f(x_2)$ then $x_1 = x_2$ Proof $f(x_1) = f(x_2)\\ 2x_1^3 - 1 = 2x_2^3 - 1\\ 2x_1^3 = 2x_2^3\\ x_1^3 = x_2^3$

Therefore $x_1 = x_2$ so $f(x)$ is one to one by direct proof - contraposition of 'if $x_1\neq x_2$, then $f(x_1)\neq f(x_2)$.

I am unsure how to approach the problem of surjection. I understand the concept, and I can show that it has a domain and a range which is an element of the real numbers, so it is definitely onto, but I don't know how to prove it.


You need to take $y\in\Bbb R,$ and show that there is some $x\in\Bbb R$ such that $f(x)=y$. In other words, you should show that $$y=2x^3-1$$ has a real solution $x$.

  • $\begingroup$ Okay, so I am just getting x by itself? Also was the proof of injection sufficient/correct? $\endgroup$ – Display Name Oct 10 '13 at 13:07
  • 1
    $\begingroup$ Yep, that's all you need to do! Your injection proof was just fine. $\endgroup$ – Cameron Buie Oct 10 '13 at 13:09

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