The derivative of the function $$ f $$ is defined by $$ f^{\prime}(x)=x^{2} \sin (3 x) $$. If $$ f(0)=5 $$, what is the absolute maximum value of the function $$ f $$ on the closed interval $$ [-2,2] $$ ? You may use a calculator and round your answer to the nearest thousandth.

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To find the absolute maximum value of the function $$ f $$ on the interval $$[-2, 2]$$, we'll follow these steps:

1. **Find $$ f(x) $$ by Integrating $$ f'(x) $$:**

Given:
   $$
   f'(x) = x^2 \sin(3x)
   $$
   
   Integrate $$ f'(x) $$ to find $$ f(x) $$:
   $$
   f(x) = \int x^2 \sin(3x) \, dx + C
   $$
   
   Using integration by parts twice, the antiderivative is:
   $$
   f(x) = -\frac{x^2 \cos(3x)}{3} + \frac{2x \sin(3x)}{9} + \frac{2 \cos(3x)}{27} + C
   $$
   
   Given $$ f(0) = 5 $$, we can solve for $$ C $$:
   $$
   5 = -\frac{0^2 \cos(0)}{3} + \frac{2 \cdot 0 \sin(0)}{9} + \frac{2 \cos(0)}{27} + C \implies C = 5 - \frac{2}{27} = \frac{133}{27}
   $$
   
   Thus, the function is:
   $$
   f(x) = 5 + \left( -\frac{x^2 \cos(3x)}{3} + \frac{2x \sin(3x)}{9} + \frac{2(\cos(3x) - 1)}{27} \right)
   $$

2. **Identify Critical Points:**

The critical points occur where $$ f'(x) = 0 $$:
   $$
   x^2 \sin(3x) = 0 \implies x = 0 \text{ or } \sin(3x) = 0
   $$
   
   Within $$[-2, 2]$$, the critical points are:
   $$
   x = -\frac{\pi}{3}, 0, \frac{\pi}{3}
   $$

3. **Evaluate $$ f(x) $$ at Critical Points and Endpoints:**

- **At $$ x = -2 $$:**
     $$
     f(-2) \approx 3.593
     $$
   
   - **At $$ x = -\frac{\pi}{3} $$:**
     $$
     f\left(-\frac{\pi}{3}\right) \approx 5.217
     $$
   
   - **At $$ x = 0 $$:**
     $$
     f(0) = 5
     $$
   
   - **At $$ x = \frac{\pi}{3} $$:**
     $$
     f\left(\frac{\pi}{3}\right) \approx 5.217
     $$
   
   - **At $$ x = 2 $$:**
     $$
     f(2) \approx 3.593
     $$

4. **Determine the Absolute Maximum:**

The highest value among these is approximately $$ 5.217 $$.

**Final Answer:**
The absolute maximum value of $$ f $$ on $$[-2, 2]$$ is approximately **5.217**.
The absolute maximum value of $$ f $$ on $$[-2, 2]$$ is approximately **5.217**.

The derivative of the function is defined by . If ,
what is the absolute maximum value of the function on the closed interval
? You may use a calculator and round your answer to the nearest
thousandth.

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The derivative of the function is defined by . If , what is the absolute maximum value of the function on the closed interval ? You may use a calculator and round your answer to the nearest thousandth.