Find the absolute maximum and minimum values of each of the following in the given intervals :
`f(x)=x^(3)" in "[-2, 2]`.

Answer

To find the absolute maximum and minimum values of the function \( f(x) = x^3 \) on the interval \([-2, 2]\), we will follow these steps:

### Step 1: Find the derivative of the function
To find the critical points, we first differentiate the function.

\[ f'(x) = \frac{d}{dx}(x^3) = 3x^2 \]

### Step 2: Set the derivative equal to zero
Next, we set the derivative equal to zero to find the critical points.

\[ 3x^2 = 0 \]

Solving this gives:

\[ x^2 = 0 \implies x = 0 \]

### Step 3: Evaluate the function at critical points and endpoints
Now we will evaluate the function \( f(x) \) at the critical point \( x = 0 \) and at the endpoints of the interval, \( x = -2 \) and \( x = 2 \).

1. **At the critical point \( x = 0 \)**:
\[ f(0) = 0^3 = 0 \]

2. **At the left endpoint \( x = -2 \)**:
\[ f(-2) = (-2)^3 = -8 \]

3. **At the right endpoint \( x = 2 \)**:
\[ f(2) = 2^3 = 8 \]

### Step 4: Compare the values
Now we compare the values obtained:

- \( f(-2) = -8 \)
- \( f(0) = 0 \)
- \( f(2) = 8 \)

### Step 5: Determine the absolute maximum and minimum
From the values calculated:

- The **absolute minimum** value is \( -8 \) at \( x = -2 \).
- The **absolute maximum** value is \( 8 \) at \( x = 2 \).

### Final Answer
- Absolute Maximum: \( 8 \) at \( x = 2 \)
- Absolute Minimum: \( -8 \) at \( x = -2 \)

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