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

1. **Find $$ f(x) $$ by Integrating $$ f'(x) $$:**
   
   Given:
   $$
   f'(x) = x^2 \sin(3x)
   $$
   
   Integrate $$ f'(x) $$ with respect to $$ x $$:
   $$
   f(x) = \int x^2 \sin(3x) \, dx + C
   $$
   
   Using integration by parts twice, the integral evaluates to:
   $$
   f(x) = -\frac{x^2 \cos(3x)}{3} + \frac{2x \sin(3x)}{9} + \frac{2 \cos(3x)}{27} + C
   $$
   
   Apply the initial condition $$ f(0) = 5 $$ to find $$ C $$:
   $$
   f(0) = -0 + 0 + \frac{2 \cdot 1}{27} + C = \frac{2}{27} + C = 5 \implies C = \frac{133}{27}
   $$
   
   Therefore, the function becomes:
   $$
   f(x) = -\frac{x^2 \cos(3x)}{3} + \frac{2x \sin(3x)}{9} + \frac{2 \cos(3x)}{27} + \frac{133}{27}
   $$
   
2. **Identify Critical Points:**
   
   Critical points occur where $$ f'(x) = 0 $$:
   $$
   f'(x) = x^2 \sin(3x) = 0 \implies x = 0 \text{ or } \sin(3x) = 0
   $$
   
   Solving $$ \sin(3x) = 0 $$ within $$[-2, 2]$$:
   $$
   x = \frac{k\pi}{3} \quad \text{for} \quad k = -1, 0, 1
   $$
   
   Thus, the critical points are $$ x = -\frac{\pi}{3}, 0, \frac{\pi}{3} $$.

3. **Evaluate $$ f(x) $$ at Critical Points and Endpoints:**
   
   - **At $$ x = -2 $$ and $$ x = 2 $$:**
     $$
     f(\pm 2) \approx 3.630
     $$
   
   - **At $$ x = 0 $$:**
     $$
     f(0) = 5
     $$
   
   - **At $$ x = \pm \frac{\pi}{3} $$:**
     $$
     f\left(\pm \frac{\pi}{3}\right) \approx \frac{\pi^2 + 131}{27} \approx 5.220
     $$
   
4. **Determine the Absolute Maximum:**
   
   Comparing the values:
   $$
   f(-2) \approx 3.630, \quad f\left(-\frac{\pi}{3}\right) \approx 5.220, \quad f(0) = 5, \quad f\left(\frac{\pi}{3}\right) \approx 5.220, \quad f(2) \approx 3.630
   $$
   
   The highest value is approximately $$ 5.220 $$, achieved at $$ x = \pm \frac{\pi}{3} $$.

**Final Answer:**
$$
\boxed{5.220}
$$
The absolute maximum value of the function $$ f $$ on the interval $$[-2, 2]$$ is approximately **5.220**.

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.