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To solve problem 7, we need to find the absolute maximum and minimum values of the function \( f(x) \) on the given interval \([-2, 2]\).

Given:
\[ f(x) = 2x^3 - 3x + 5 \]

Let's follow these steps:

1. **Find the critical points by setting \( f'(x) = 0 \).**
2. **Evaluate \( f(x) \) at the critical points and the endpoints of the interval.**
3. **Determine the absolute maximum and minimum values.**

### Step 1: Find the Critical Points

First, we find the first derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 - 3x + 5) = 6x^2 - 3 \]

Set the first derivative equal to zero to find the critical points:
\[ 6x^2 - 3 = 0 \]
\[ 6x^2 = 3 \]
\[ x^2 = \frac{3}{6} = \frac{1}{2} \]
\[ x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \]

The critical points are \( x = \frac{\sqrt{2}}{2} \) and \( x = -\frac{\sqrt{2}}{2} \).

### Step 2: Evaluate \( f(x) \) at the Critical Points and Endpoints

Next, we evaluate \( f(x) \) at the critical points and the endpoints of the interval \([-2, 2]\).

1. Evaluate at \( x = -2 \):
\[ f(-2) = 2(-2)^3 - 3(-2) + 5 = 2(-8) + 6 + 5 = -16 + 6 + 5 = -5 \]

2. Evaluate at \( x = 2 \):
\[ f(2) = 2(2)^3 - 3(2) + 5 = 2(8) - 6 + 5 = 16 - 6 + 5 = 15 \]

3. Evaluate at \( x = \frac{\sqrt{2}}{2} \):
\[ f\left(\frac{\sqrt{2}}{2}\right) = 2\left(\frac{\sqrt{2}}{2}\right)^3 - 3\left(\frac{\sqrt{2}}{2}\right) + 5 = 2\left(\frac{2\sqrt{2}}{8}\right) - \frac{3\sqrt{2}}{2} + 5 = \frac{\sqrt{2}}{2} - \frac{3\sqrt{2}}{2} + 5 = -\sqrt{2} + 5 \]

4. Evaluate at \( x = -\frac{\sqrt{2}}{2} \):
\[ f\left(-\frac{\sqrt{2}}{2}\right) = 2\left(-\frac{\sqrt{2}}{2}\right)^3 - 3\left(-\frac{\sqrt{2}}{2}\right) + 5 = 2\left(-\frac{2\sqrt{2}}{8}\right) + \frac{3\sqrt{2}}{2} + 5 = -\frac{\sqrt{2}}{2} + \frac{3\sqrt{2}}{2} + 5 = \sqrt{2} + 5 \]

### Step 3: Determine the Absolute Maximum and Minimum Values

Now we compare the values we obtained:

1. \( f(-2) = -5 \)
2. \( f(2) = 15 \)
3. \( f\left(\frac{\sqrt{2}}{2}\right) = -\sqrt{2} + 5 \approx 3.586 \)
4. \( f\left(-\frac{\sqrt{2}}{2}\right) = \sqrt{2} + 5 \approx 6.414 \)

The absolute maximum value of \( f(x) \) on the interval \([-2, 2]\) is \( 15 \) at \( x = 2 \).

The absolute minimum value of \( f(x) \) on the interval \([-2, 2]\) is \( -5 \) at \( x = -2 \).

Would you like any further details or have any questions?

Here are five questions you might want to ask:

1. How do I find the derivative of a function?
2. Can you explain how to solve other problems from the provided image?
3. What are the critical points, and why are they important?
4. How do you evaluate functions at given points?
5. What is the difference between absolute and local maximum and minimum values?

**Tip:** When solving for absolute maximum and minimum values on a closed interval, always evaluate the function at the critical points and the endpoints of the interval.

Learn how to find the absolute maximum and minimum values of the function f(x) = 2x^3 - 3x + 5 on the interval [-2, 2]. This problem covers calculus concepts such as critical points, derivatives, and evaluating functions at given points. Suitable for advanced high school students.

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