Showing there exists just one $x \in \mathbb{R}$ such that $x^3 + ax + b = 0$ In my course on real analysis, I have this problem I just can't figure out:
Problem: Let $a, b \in \mathbb{R}$ with $a > 0$. Show that there exists just one $x \in \mathbb{R}$ such that $x^3 + ax + b = 0$.
I don't know how to begin. I was looking for a theorem to use but there's not enough conditions given for them to able to apply. Any suggestions or help?
 A: Let $f(x) = x^3 + ax + b$ which is continuous since it is a polynomial. Now, it is clear that
$$\lim_{x \to -\infty} f(x) < 0$$
and
$$\lim_{x \to \infty} f(x) > 0.$$
By the Intermediate Value Theorem, these exists $-\infty < c < \infty$ such that $f(c) = 0$.
Furthermore, $f'(x) = 3x^2 + a > 0$ since $a > 0$. Thus, $f$ is a strictly increasing function, so $c$ can be the only root.
A: This is a partially intuitive proof. I'll leave the rigor up to you.
Let $p(x) = x^3 + ax+ b$
For $p(x)$ to have only 1 "zero", it can only cross the X-axis once. 
Let's look at $p'(x) = 3x^2 + a$
As $a > 0$,
$p'(x) > 0$ for all $x$
If the derivative of $p(x)$ is always more than 0, that means $p(x)$ is an increasing function for increasing values of x. Which means once it crosses the X-axis, it can't go back down.
A: First of all note that it is an odd degree polynomial (and obviously a continuous function of $x$). Thus:
$$
\lim_{x\rightarrow+\infty}(x^3 + ax + b = 0)=+\infty \\ 
\lim_{x\rightarrow-\infty}(x^3 + ax + b = 0)=-\infty
$$
By Bolzano's theorem, we get that there exists at least one real root. 
On the other hand, $f'(x)=3x^2+a>0$ for any $x\in\mathbb{R}$. This implies that $f(x)=x^3 + ax + b $ is a strictly increasing function, so it has at most one real root. 
Consequently, $f(x)=x^3 + ax + b =0$ has exactly one real root. 
A: You may find useful to see a proof not using derivatives:
Let $f(x)=x^3+ax+b,$ with $a\ge0$. As already said in the other answers, by Bolzano's theorem there exists at least one root $x_0$ of $f$. We have \begin{align}f(x_0+\varepsilon)&=(x_0+\varepsilon)^3+a(x_0+\varepsilon)+b \\ &=x_0^3+ax_0+b+3\varepsilon x_0^2+3\varepsilon^2x_0+a\varepsilon+\varepsilon^3\\&=3\varepsilon x_0^2+3\varepsilon^2x_0+a\varepsilon+\varepsilon^3,\end{align} thus $\Delta=9\varepsilon^4-12\varepsilon^4-12a\varepsilon^2=-3\varepsilon^2(4a+\varepsilon^2)$ being negative for all $\varepsilon\ne0$ yields $f(x_0+\varepsilon)\ne0$ for all $\varepsilon\ne0$, whence the uniqueness of $x_0$.
A: Another proof without derivatives, allowing the value $a=0$: The function $x^3$ is strictly increasing, and $ax$ is increasing. It follows that $x^3 + ax$ is strictly increasing. The addition of the constant $b$ does not change this. Thus if there is a zero, it must be unique. Set $f(x) = x^3+ax + b.$ You can check that $f(|b|+1) > 0, f(-(|b|+1)) < 0.$ By the intermediate value theorem, $f$ must have have a zero, and we're done.
