How to show that f(x) = $e^x + x^3 - 19$ has only one real root?

Question

There's a root around $2.17$ and I wanna know how to show that this is the only real root when $f(x) = 0$

Thanks in advance!

Answer 1

$$f'(x) = e^x+3x^2$$

Since $f'(x)>0$ for all $x$ and $f(x)$ is continuous for all $x$, then let $r$ be the root such that $f(r) = 0$. Then for all $x<r, f(x)<0$ and for all $x>r, f(x)>0$.

Answer 2

Note that $f'(x) = e^x + 3x^2 > 0$, meaning $f(x)$ is monotonically increasing.

Moreover, we have $f(-\infty) = -\infty$ and $f(\infty) \to \infty$, which means that $f(x)$ much cross zero and can only cross once due to its monotonicity. Thus, $f(x)$ has only one root.