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 follow these steps:

1. **Given Information:**
   - The derivative of the function is $$ f^{\prime}(x) = x^2 \sin(3x) $$.
   - The initial condition is $$ f(0) = 5 $$.

2. **Find the Function $$ f(x) $$:**
   - To find $$ f(x) $$, we integrate the derivative:
     $$
     f(x) = \int_0^x t^2 \sin(3t) \, dt + f(0) = 5 + \int_0^x t^2 \sin(3t) \, dt
     $$
   - Using integration by parts, the integral evaluates to:
     $$
     f(x) = 5 + \left[ -\frac{x^2 \cos(3x)}{3} + \frac{2x \sin(3x)}{9} + \frac{2 \cos(3x)}{27} - \frac{2}{27} \right]
     $$

3. **Identify Critical Points:**
   - Critical points occur where $$ f'(x) = 0 $$, i.e., $$ x = 0 $$ or $$ \sin(3x) = 0 $$.
   - Within the interval $$[-2, 2]$$, the critical points are $$ x = -\frac{\pi}{3} $$, $$ x = 0 $$, and $$ x = \frac{\pi}{3} $$.

4. **Evaluate $$ f(x) $$ at Critical Points and Endpoints:**
   - **At $$ x = -2 $$:**
     $$
     f(-2) \approx 3.841
     $$
   - **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 $$ x = 2 $$:**
     $$
     f(2) \approx 3.593
     $$

5. **Determine the Absolute Maximum:**
   - The highest value among these evaluations is approximately $$ 5.217 $$ at both $$ x = -\frac{\pi}{3} $$ and $$ x = \frac{\pi}{3} $$.

**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.