Answer
See below.
Work Step by Step
$f(x)=x^{3}+4x^{-2}+7$
The only possible problem, x=0 is not in the domain, so f is continuous on $(-\infty,0).$
$f(-10)\lt 0, \qquad f(-1)=2\gt 0$,
so the Intermediate Value Theorem guarantees that there is a zero between $-10$ and $-1$.
Check whether f is increasing/decreasing over $(-\infty,0)$.
(If it is, the mentioned zero would be the only zero.)
$f'(x)=3x^{2}+4(-2x^{-3})=3x^{2}-\displaystyle \frac{8}{x^{3}}=\frac{3x^{5}-8}{x^{3}}$
The numerator is a sum of negative numbers; thus, negative.
The denominator is negative on the interval.
So, $f'$ is positive on $(-\infty,0) \Rightarrow$ f is increasing on $(-\infty,0)$.
The graph of $f(x)$ crosses the x-axis only once for $ x\in (-\infty,0)$; that is, $f$ has only one zero in $(-\infty,0)$.
