Find the absolute maximum and minimum values of each of the following in the given intervals :
`f(x)=x^(3)-3x, -3 le x le3`

Answer

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

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

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

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

\[ 3x^2 - 3 = 0 \]

### Step 3: Solve for \( x \)
Now we solve for \( x \):

\[ 3x^2 = 3 \\ x^2 = 1 \\ x = \pm 1 \]

So, the critical points are \( x = 1 \) and \( x = -1 \).

### Step 4: Evaluate the function at the critical points and endpoints
Next, we will evaluate the function \( f(x) \) at the critical points and the endpoints of the interval \([-3, 3]\):

1. Evaluate at \( x = -3 \):
\[ f(-3) = (-3)^3 - 3(-3) = -27 + 9 = -18 \]

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

3. Evaluate at \( x = 1 \):
\[ f(1) = (1)^3 - 3(1) = 1 - 3 = -2 \]

4. Evaluate at \( x = 3 \):
\[ f(3) = (3)^3 - 3(3) = 27 - 9 = 18 \]

### Step 5: Determine the absolute maximum and minimum
Now we compare the values we have calculated:

- \( f(-3) = -18 \)
- \( f(-1) = 2 \)
- \( f(1) = -2 \)
- \( f(3) = 18 \)

From these values, we can see:
- The absolute maximum value is \( f(3) = 18 \).
- The absolute minimum value is \( f(-3) = -18 \).

### Conclusion
Thus, the absolute maximum value of \( f(x) \) on the interval \([-3, 3]\) is \( 18 \) and the absolute minimum value is \( -18 \).