he derivative of the function $$ f $$ is defined by $$ f^{\prime}(x)=x^{2} \sin (3 x) $$. If $$ f(0)=5 $$, hat 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 nousandth.

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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. **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, we find:
   $$
   f(x) = -\frac{1}{3}x^2 \cos(3x) + \frac{2}{9}x \sin(3x) + \frac{2}{27} \cos(3x) + C
   $$
   
   Apply the initial condition $$ f(0) = 5 $$:
   $$
   f(0) = -\frac{1}{3}(0)^2 \cos(0) + \frac{2}{9}(0) \sin(0) + \frac{2}{27} \cos(0) + C = 5
   $$
   $$
   \Rightarrow \frac{2}{27} + C = 5 \implies C = \frac{133}{27}
   $$
   
   Thus, the function is:
   $$
   f(x) = -\frac{1}{3}x^2 \cos(3x) + \frac{2}{9}x \sin(3x) + \frac{2}{27} \cos(3x) + \frac{133}{27}
   $$

2. **Identify Critical Points:**

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

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

- **At $$ x = -2 $$ and $$ x = 2 $$:**
     $$
     f(-2) \approx 3.597 \quad \text{and} \quad f(2) \approx 3.597
     $$
     
   - **At $$ x = 0 $$:**
     $$
     f(0) = 5
     $$
     
   - **At $$ x = -\frac{\pi}{3} $$ and $$ x = \frac{\pi}{3} $$:**
     $$
     f\left(\frac{\pi}{3}\right) \approx f\left(-\frac{\pi}{3}\right) \approx 5.219
     $$

4. **Determine the Absolute Maximum:**

Comparing the values:
   $$
   f\left(\frac{\pi}{3}\right) \approx 5.219 \quad \text{is the highest value}
   $$

**Therefore, the absolute maximum value of $$ f $$ on the interval $$[-2, 2]$$ is approximately $$ \mathbf{5.219} $$.**
The absolute maximum value of $$ f $$ on $$[-2, 2]$$ is approximately **5.219**.

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

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