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$f:x\mapsto x^3:\Bbb{R}\to\Bbb{R}$ is an injective, but not a surjective, function.

I have a question that asks whether the above state is true or false. The answer key (question 3(b)) says that this is a false statement.

As we all know, this cannot be a surjective function, since the range consists of all real values, but $f(x)$ can only produce cubic values.

Also from observing a graph, this function produces unique values; hence it is injective.

So I conclude that the given statement is true. Now my question is: Am I right? This is a sample question paper from a reputed institute, so I will not be surprised if there is something else to this question.

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    $\begingroup$ Your argument for not surjective is wrong. $\endgroup$ – DonAntonio Mar 21 '18 at 14:18
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    $\begingroup$ There is no difference between "cube real numbers" and "ordinary real numbers": any real number $\alpha$ is the cube of some real number, namely $\sqrt[3]\alpha$. This difference exist on rationals, integers or some other subfield, but not in $\Bbb R$ itself. $\endgroup$ – user228113 Mar 21 '18 at 14:20
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    $\begingroup$ Every real number is the cube of some real number. As a map of rationals, $x^3$ is not surjective. But as a map of reals, it is. $\endgroup$ – GEdgar Mar 21 '18 at 14:21
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    $\begingroup$ Note that the inverse exists $ f^{-1}(x)=\sqrt[3] x \quad \mathbb{R}\to\mathbb{R}$ thus $f$ is bijective. $\endgroup$ – user Mar 21 '18 at 14:21
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    $\begingroup$ @swarm Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Mar 22 '18 at 21:18
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Since the equation $x^3=a$ is solvable (in $\mathbb{R}$) for each $a\in \mathbb{R}$ given function is surjective.

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$$\lim_{x\to +\infty }x^3=+\infty \quad \text{and}\quad \lim_{x\to -\infty }x^3=-\infty .$$ By intermediate value theorem, you get $f(\mathbb R)=\mathbb R$ and thus it's surjective.

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HINT

Note that

  • $f(x)$ is continuos
  • consider limit for $x\to \pm \infty$ and IVT
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Let $x$ be a real number. Then $$ f\left(\sqrt[3]{x}\right)=\sqrt[3]{x}^3=x $$

Hence it IS surjective.

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A cubic value can be any real number. E.g. $f(2^\frac13)=2.$

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As we all know that this cannot be a surjective function; since the range consist of all real values, but f(x) can only produce cubic values.

This doesn't mean $f(x)$ is not surjective. Technically, every real number is a "cubic value" since every real number is the cube of some other real number.

Big hint:

To show surjectivity of $f(x) = x^3$, you basically want to show that for any real number $y$, there is some number $x$ such that $f(x) = y$. In other words, the goal is to fix $y$, then choose a specific $x$ that's defined in terms of $y$, and prove that your chosen value of $x$ works. Can you see how to do that? The level of rigor really depends on the course in general, and since this is for an M.Sc. entrance exam then I suspect an undergraduate-level proof (it's very short) is expected.

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Since the domain of $f(x)$ is $\mathbb{R}$, there exists only one cube root (or pre-image) of any number (image) and hence $f(x)$ satisfies the conditions for it to be injective.

However, for $f(x)$ to be surjective, you have to check whether the given codomain equals the range of $f(x)$ or not. Clearly, for $f(x) = x^3$, the function can return any value belonging to $\mathbb{R}$ for any input. Hence the range of $f(x) = x^3$ is $\mathbb{R}$. You can verify this by looking at the graph of the function.
And since the codomain is also $\mathbb{R}$, the function is surjective.

Therefore the statement is False , as very rightly mentioned in your answer key.

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