The function $$ f(x)=-x^{4}+72 x^{2} $$ has A. one relative minimum and two relative maxima. B. one relative minimum and one relative maximum. C. no relative minimum and two relative maxima. D. two relative minima and no relative maximum. E. two relative minima and one relative maximum.

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To determine the relative minima and maxima of the function $$ f(x) = -x^4 + 72x^2 $$, follow these steps:

1. **Find the First Derivative:**
   $$
   f'(x) = \frac{d}{dx}(-x^4 + 72x^2) = -4x^3 + 144x
   $$
   
2. **Determine Critical Points:**
   Set the first derivative equal to zero:
   $$
   -4x^3 + 144x = 0 \implies -4x(x^2 - 36) = 0
   $$
   Thus, the critical points are:
   $$
   x = 0, \quad x = 6, \quad x = -6
   $$

3. **Find the Second Derivative:**
   $$
   f''(x) = \frac{d}{dx}(-4x^3 + 144x) = -12x^2 + 144
   $$
   
4. **Apply the Second Derivative Test:**
   - **At $$ x = 0 $$:**
     $$
     f''(0) = -12(0)^2 + 144 = 144 > 0 \quad \text{(Relative Minimum)}
     $$
     
   - **At $$ x = 6 $$:**
     $$
     f''(6) = -12(6)^2 + 144 = -432 + 144 = -288 < 0 \quad \text{(Relative Maximum)}
     $$
     
   - **At $$ x = -6 $$:**
     $$
     f''(-6) = -12(-6)^2 + 144 = -432 + 144 = -288 < 0 \quad \text{(Relative Maximum)}
     $$
     
5. **Conclusion:**
   - $$ x = 0 $$ is a **relative minimum**.
   - $$ x = 6 $$ and $$ x = -6 $$ are **relative maxima**.

Therefore, the function $$ f(x) = -x^4 + 72x^2 $$ has **one relative minimum and two relative maxima**.

**Answer:** A. one relative minimum and two relative maxima.
The function $$ f(x) = -x^4 + 72x^2 $$ has one relative minimum and two relative maxima.

The function \( f(x)=-x^{4}+72 x^{2} \) has A. one relative minimum and two relative maxima. B. one relative minimum and one relative maximum. C. no relative minimum and two relative maxima. D. two relative minima and no relative maximum. E. two relative minima and one relative maximum.

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