please help me :can we inverse this function: $f(x)=x^3$? I know that if a function is a bijective function only then it can be inversed. Is this a bijective function?
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What is the definition of Bijective? How do you go about proving whether a function has this property? – abiessu Oct 30 '13 at 16:08
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Bijective function is a function which is both onto function and one-one function – Fazla Rabbi Mashrur Oct 30 '13 at 16:11
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How do you prove that a function is one to one? How do you prove that it is onto? Do those hold for this function? – Ben Grossmann Oct 30 '13 at 16:12
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Yes, indeed, it is both one-to-one (injective) and onto (surjective), and hence it is bijective.
To show it is one-to-one, show that $f(x_1) = f(x_2) \implies x_1 = x_2$.
To show that it is onto, note that the image of $f$ is the entire codomain: the set of all real numbers. So for every $y \in \mathbb R$, there exists an $x\in \mathbb R$ such that $f(x) = y$.
So its inverse exists.
To find the inverse $f^{-1}$:
- Solve for $x$ as a function of $y$: $$x = \sqrt[\large 3] y$$
- swap $x$ for $y$: $$f^{-1}(x) = \sqrt[\large 3] x$$
amWhy
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Actually, isn't it every element $y$ in the codomain which must satisy $f(x)=y$ for some $x$ in the domain for a function to be onto? In this case of course the domain is the same as the codomain. – George Tomlinson Oct 30 '13 at 16:18
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Seems to me you should say 'the image of $f$ is the entire codomain'. – George Tomlinson Oct 30 '13 at 16:20
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...would I be too fussy if I were to ask for $f^{-1}(x)=\text{sgn}(x)\sqrt[3]{|x|}$?? – JP McCarthy Oct 30 '13 at 16:42
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It is interesting that $\sqrt[3]{x}$ is defined as the function that is a (right) inverse of $f$, so saying that $f^{-1}(x) = \sqrt[3]{x}$ doesn't say much. It certainly doesn't prove that the inverse exists, and it doesn't prove surjectivity. I think it would be more correct to prove surjectivity using the mean value theorem, without appealing to cubic roots. – Dan Shved Oct 31 '13 at 13:13
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Although maybe you don't mean to appeal to cubic roots. Sorry, I'm not criticizing, I'm only expressing a worry just in case :) – Dan Shved Oct 31 '13 at 13:19
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Hint Check whether your function $f:X\to Y$ satisfies
$1.f(x_1)=f(x_2)\Rightarrow x_1=x_2$
$2. \forall x\in Y\exists y\in X \ni f(y)=x$
$1\Rightarrow$ $f$ is injective and $2\Rightarrow f$ is surjective, A function is bijective if it is injective and surjective.
Myshkin
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