To find the maximum and minimum values of the function f(x) = 3x³ - 9x² - 27x + 15, we need to find the critical points by taking the derivative of the function and setting it equal to zero.

1. Find the first derivative:

f'(x) = d/dx (3x³ - 9x² - 27x + 15)

f'(x) = 9x² - 18x - 27

2. Set the first derivative equal to zero and solve for x:

9x² - 18x - 27 = 0

Divide by 9: x² - 2x - 3 = 0

Factor: (x - 3)(x + 1) = 0

So, x = 3 or x = -1

3. Find the second derivative:

f''(x) = d/dx (9x² - 18x - 27)

f''(x) = 18x - 18

4. Use the second derivative test to determine if the critical points are maximum or minimum:

Evaluate f''(x) at x = 3:

f''(3) = 18(3) - 18 = 54 - 18 = 36

Since f''(3) > 0, x = 3 is a local minimum.

Evaluate f''(x) at x = -1:

f''(-1) = 18(-1) - 18 = -18 - 18 = -36

Since f''(-1) < 0, x = -1 is a local maximum.

5. Find the values of the function at the critical points:

f(3) = 3(3)³ - 9(3)² - 27(3) + 15 = 3(27) - 9(9) - 81 + 15 = 81 - 81 - 81 + 15 = -66

f(-1) = 3(-1)³ - 9(-1)² - 27(-1) + 15 = 3(-1) - 9(1) + 27 + 15 = -3 - 9 + 27 + 15 = 30

6. Determine the maximum and minimum values:

The local maximum value is 30 at x = -1.

The local minimum value is -66 at x = 3.

In summary:

• Local Maximum: 30 at x = -1
• Local Minimum: -66 at x = 3