Answer
See explanations.
Work Step by Step
Step 1. To show that the function $f(x)=x^3+\frac{4}{x^2}+7$ has exactly one zero in the interval of $(-\infty,0)$, we will need to show that the function crosses the x-axis only once.
Step 2. Take the derivative of the function; we get $f'(x)=3x^2-\frac{8}{x^3}$. As $x\lt0$, we have $f'(x)\gt0$, which means that the curve will only go in one direction (increases as $x$ increases).
Step 3. Let $x=-1$; we have $f(-1)=(-1)^3+\frac{4}{(-1)^2}+7=10$ which is positive. Let $x=-3$; we have $f(-3)=(-3)^3+\frac{4}{(-3)^2}+7=-27+\frac{4}{9}+7=\frac{4}{9}-20\lt0$, thus the curve should cross the x-axis between $-3$ and $-1$. As a matter of fact, we can find the zero as $x=0$ as shown in the figure.
Step 4. Because $f'(x)\gt0$, the function has only one solution at $x=2$.
