The problem asks to show that the function $f(x) = -x^3 + 3x^2 + x - 3$ has a real zero between $x = -2$ and $x = 0$.

To prove this, we can apply the Intermediate Value Theorem (IVT), which states that if a function $f$ is continuous on a closed interval $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there is at least one $c \in (a, b)$ such that $f(c) = 0$.

Step-by-step solution:

1. Check continuity:
   The function $f(x) = -x^3 + 3x^2 + x - 3$ is a polynomial, and all polynomials are continuous over the entire real line. Hence, $f(x)$ is continuous on the interval $[-2, 0]$.

2. Calculate $f(-2)$:

   = -(-8) + 3(4) - 2 - 3 = 8 + 12 - 2 - 3 = 15$$ So, $$f(-2) = 15$$.

3. Calculate $f(0)$:

   = -3$$ So, $$f(0) = -3$$.

4. Apply the Intermediate Value Theorem: Since $f(-2) = 15$ and $f(0) = -3$, and these values have opposite signs, the IVT guarantees that there is at least one real zero of $f(x)$ between $x = -2$ and $x = 0$.

Thus, the function $f(x)$ has a real zero between $x = -2$ and $x = 0$.

Would you like further clarification or additional details?

Here are five related questions to explore:

1. Can the Intermediate Value Theorem be applied to functions with discontinuities?
2. What other methods could be used to find roots of polynomials?
3. How can we refine the interval further to locate the exact zero?
4. How would you apply Newton's method to approximate the zero of this function?
5. How does the behavior of the derivative of $f(x)$ influence the number of real zeros?

Tip: The Intermediate Value Theorem is a useful tool for showing the existence of a zero but doesn't provide the exact location of the zero. Numerical methods like bisection can help find a more precise value.