The given function is:

$g(x) = x^3 - 4x^2$

Analyzing the Function:

1. Type of Function: This is a cubic function because the highest degree of $x$ is 3.

2. General Behavior:

   • As $x \to \infty$, $g(x) \to \infty$.
   • As $x \to -\infty$, $g(x) \to -\infty$.

3. Zeros of the Function: To find the zeros, set $g(x) = 0$: $x^3 - 4x^2 = 0$ Factor out $x^2$: $x^2(x - 4) = 0$ This gives two solutions:

   • $x = 0$
   • $x = 4$

4. Critical Points and Local Extrema: To find the critical points, take the derivative of $g(x)$: $g'(x) = 3x^2 - 8x$ Set $g'(x) = 0$ to find the critical points: $3x^2 - 8x = 0$ Factor out $x$: $x(3x - 8) = 0$ This gives:

   • $x = 0$
   • $x = \frac{8}{3}$

   These are the critical points. To determine if they are minima or maxima, use the second derivative test.

5. Second Derivative: The second derivative is: $g''(x) = 6x - 8$ Evaluate at the critical points:

   • $g''(0) = 6(0) - 8 = -8$ (Negative, so $x = 0$ is a local maximum)
   • $g''\left(\frac{8}{3}\right) = 6\left(\frac{8}{3}\right) - 8 = 16 - 8 = 8$ (Positive, so $x = \frac{8}{3}$ is a local minimum)

Summary:

• Zeros: $x = 0$ and $x = 4$
• Local Maximum: At $x = 0$
• Local Minimum: At $x = \frac{8}{3}$

Would you like further details or have any questions?

Here are 5 relative questions:

1. What is the behavior of the function as $x \to \infty$ and $x \to -\infty$?
2. How would you find the inflection points of $g(x)$?
3. Can you graph this function based on its critical points and zeros?
4. How does the cubic term $x^3$ influence the shape of the graph?
5. What is the significance of the second derivative test in determining local extrema?

Tip: Understanding how to derive and analyze critical points is key in sketching and interpreting the behavior of cubic functions.