A function g(x) is given by g(x) = x^4 - 6x^2 + 9. Where does the local maximum occur?

To find where the local maximum of the function \( g(x) = x^4 - 6x^2 + 9 \) occurs, we need to follow these steps:

### 1. **Find the first derivative of the function:**
   \[
   g'(x) = \frac{d}{dx}(x^4 - 6x^2 + 9)
   \]
   \[
   g'(x) = 4x^3 - 12x
   \]

### 2. **Set the first derivative equal to zero to find critical points:**
   \[
   4x^3 - 12x = 0
   \]
   Factor the equation:
   \[
   4x(x^2 - 3) = 0
   \]
   Set each factor equal to zero:
   \[
   4x = 0 \quad \Rightarrow \quad x = 0
   \]
   \[
   x^2 - 3 = 0 \quad \Rightarrow \quad x = \pm\sqrt{3}
   \]

So, the critical points are \( x = 0 \), \( x = \sqrt{3} \), and \( x = -\sqrt{3} \).

### 3. **Determine the nature of each critical point using the second derivative:**
   Find the second derivative \( g''(x) \):
   \[
   g''(x) = \frac{d}{dx}(4x^3 - 12x)
   \]
   \[
   g''(x) = 12x^2 - 12
   \]

   Evaluate \( g''(x) \) at each critical point:
   - At \( x = 0 \):
     \[
     g''(0) = 12(0)^2 - 12 = -12 \quad (\text{Negative, indicating a local maximum})
     \]
   - At \( x = \sqrt{3} \) and \( x = -\sqrt{3} \):
     \[
     g''(\sqrt{3}) = 12(\sqrt{3})^2 - 12 = 12(3) - 12 = 36 - 12 = 24 \quad (\text{Positive, indicating local minima})
     \]

### 4. **Conclusion:**
   The local maximum occurs at \( x = 0 \).

Would you like to explore more details, or do you have any questions?

Here are some related questions:

1. How do you determine if a critical point is a local maximum, minimum, or a saddle point?
2. What is the significance of the second derivative test in finding local extrema?
3. Can a function have more than one local maximum or minimum?
4. How do the coefficients of a polynomial function influence the location and nature of critical points?
5. What is the relationship between the degree of a polynomial and the number of critical points?

**Tip:** The second derivative test is a quick way to determine the nature of a critical point: if \( g''(x) \) is negative at a critical point, it's a local maximum; if positive, it's a local minimum.

Discover where the local maximum occurs for the function g(x) = x^4 - 6x^2 + 9 using calculus. Learn how to find critical points, evaluate second derivatives, and apply the second derivative test to determine local extrema. Suitable for advanced high school and college students.

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