Show that for any real number $y$, there exists $x$ such that $f(x)=y$ Let $f(x)= \frac{x^3}{x^2+1}$. Show that for any real number $y$, there exists $x$ such that $f(x)=y$.
Step-by-step solution is needed as I don't know where to start solving this problem.
 A: Since $f$ is continuous, $\lim_{x \to \infty} f(x) = \infty$ and $\lim_{x \to -\infty} = -\infty$ the result follows from the intermediate value theorem.
A: The problem is to find a solution to the equation $y={x^3\over x^2+1}$ and you can rewrite the equation as
$$x^3-yx^2-y=0$$
any cubic with real coefficients has at least one real solution because the cubic polynomial is continuous and has $-\infty$ as limit at $-\infty$ and $+\infty$ at $+\infty$
A: $f'(x)=\frac{3x^2(x^2+1)-2x.x^3}{(x^2+1)^2}=\frac{x^4+3x^2}{(x^2+1)^2}$
$f'(x)>0$ for $x \ne 0$. Therefore $f(x)$ is strictly increasing. Moreover,
$\lim_{x \to \pm \infty} x\frac{x^2}{x^2+1} = \pm \infty$
Therefore for any $y$ there is only 1 $x$ such that $f(x)=y$.
A: What this question is essentially asking is to show that the function $f(x)$ is surjective. What you can do is demonstrate that this has to hold by contradiction, i.e. assume there exists some real number $a$ such that $a \neq \frac{x^3}{x^2+1}$. This implies that $x^3-ax^2-a=0$ has no solution. Graphically this tells us that $x^3-ax^2-a=0$ does not have any zeroes, but what do we know about the shape of cubic polynomials? (i.e. the number of turns).
A: Note that
$${x^2\over x^2+1}=1-{1\over x^2+1}\geq{1\over2}\qquad\bigl(|x|\geq1\bigr)\ .$$
Now let an $y\in{\mathbb R}$ be given. For $b:=2\bigl(1+|y|\bigr)\geq1$ one has
$$f(b)=b\>{b^2\over b^2+1}\geq{1\over2} b>|y|\ .$$
Similarly, for $a=-2\bigl(1+|y|\bigr)$ one gets $f(a)<-|y|.\ $ Since $-|y|\leq y\leq|y|$  the intermediate value theorem then guarantees  an $x\in\ ]a,b[\ $ with $f(x)=y$.
