Answer
See proof below.
Work Step by Step
$f(-2)=11,\qquad f(-1)=-1$
Since f is continuous, f(x) assumes all values between -1 and 11, including 0.
So, there is at least one zero in $(-2,-1)$.
$f'(x)=4x^{3}+3$
The graph of $f'$ is obtained by vertically stretching and shifting upwards the graph of $y=x^{3}$.
So, the graph of $f'(x)$ is always rising, while $f'(-1)=-1$ is negative.
Then for all values of x $\lt -1,\ f'(x)$ is negative as well.
Since $f'$ is negative on $(-2,-1)$,
$f$ is a decreasing continuous function on $(-2,-1)$,
which means that the graph of $f(x)$ crosses the x-axis only once for $x\in(-2,-1)$; that is, $f$ has only one zero in $(-2,-1)$.
