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Let $f$: $\mathbb{R}$ to $\mathbb{R}$ defined by $f(x) = x^3+x^2$. Prove that $f(x)$ is a surjective function.

I'm not quite sure how to approach this problem. If the function had an inverse, I could show that it would be bijective and therefore surjective, but this function does not have an inverse.

I was given the hint to use the Intermediate Value Theorem, but I don't see how this would help on the interval (-1, 0).

Any assistance is greatly appreciated.

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    Hint: You can come up with a proof that works for every polynomial of odd degree. – Torsten Schoeneberg Feb 12 '20 at 01:25
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    when you ask a question about function, state the domain, the codomain, and the rule. – Siong Thye Goh Feb 12 '20 at 01:28
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    You mention a particular interval, which suggests that you are not thinking of this as a function from $\mathbb R$ to $\mathbb R$. Can you clarify? – lulu Feb 12 '20 at 01:30
  • @L. B. Assuming you mean the roots of the polynomial, at least one of them is real (that is a special case of what is to prove here). It can happen that all except one are not real, but that is irrelevant to the question asked. – Torsten Schoeneberg Feb 12 '20 at 01:32
  • Finding the root to $x^3 + x^2 = y$ is not easy. But $(\sqrt[3]y)^3+(\sqrt[3]y)^2 \ge y$. Is if we can find a $w$ so that $y \ge w^3 + w^2$ the then by IVT there must be an $x$ between $w$ and $\sqrt[3]y$ where $x^2 + x^2 = y$. – fleablood Feb 12 '20 at 02:26

3 Answers3

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For a given value of $b$ we need to find an $x$ such that $f(x)=b$

Note that your function is continuous and $$\lim _{x\to \infty } f(x) = \infty $$ and $$\lim _{x\to -\infty } f(x) = -\infty $$ Thus you can find a value say $x_1$ such that $f(x_1)>b$

Similarly you can find a value say $x_0$ such that $f(x_0) <b$

Now use the intermediate value theorem to the interval $[x_0, x_1]$ to find an $x$ such that $f(x)=b$

Mohammad Riazi-Kermani
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A function $f:\textbf{R}\rightarrow\textbf{R}$ is called surjective if, for every $y_{0}\in\textbf{R}$, there exists a $x_{0}\in\textbf{R}$ such that $f(x_{0}) = y_{0}$. Now consider the equation \begin{align*} x^{3} + x^{2} = y_{0} \Longleftrightarrow x^{3} + x^{2} - y_{0} = 0 \end{align*} According to the fundamental theorem of algebra, there are three complex roots for each value of $y_{0}$. However, the coefficients of $p_{y_{0}}(x) = x^{3} + x^{2} - y_{0}$ are all real. Thus, if there is a complex root $z$, its conjugate $\overline{z}$ is also a root, which means that at least one of the roots is real, independently of $y_{0}$. Therefore it has been shown that, for every $y_{0}\in\textbf{R}$, there is a $x_{0}\in\textbf{R}$ such that $f(x_{0}) = y_{0}$, as desired.

user0102
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Note that as $x \to \infty$ than $f(x) =x^3 +x^2\to \infty$ and as $x\to -\infty$ then $x^3 + x^2\to -\infty$.

Now let's take those statement's literally.

Let $y_0 \in \mathbb R$.

If $y_0 > 0$ then there is an $N>0$ so that if $x > N$ then $f(x) > y_0$. SO let $b > N$ and and $f(b) > y_0$.

We have $f(0)= 0 < y_0 < f(b)$....

So by intermediat value theorem wyou have ther must be a $c: 0\le c \le b$ where $f(c) = y_0$.

If $y_0 = 0$ then $f(0) = 0$.

And if $y_0 < 0$ there is an $N< 0$ so that if $x < N$ then $f(x) < y_0$ so if $a < N$ we have $f(a) < y_0 < f(0)$ so by $IVT$ there is an $c: a < c < 0$ so that $f(c) = y_0$.

fleablood
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