Question Detail

To find the x-coordinate of the maximum value of the function, we follow these steps:

1. Find the first derivative f'(x):f'(x) = d/dx(x³ + 2x² - 4x + 6) = 3x² + 4x - 4
2. Set the first derivative to zero to find critical points:3x² + 4x - 4 = 0Using the quadratic formula x = (-b ± √(b² - 4ac)) / (2a):x = (-4 ± √(4² - 4(3)(-4))) / (2(3))x = (-4 ± √(16 + 48)) / 6x = (-4 ± √64) / 6x = (-4 ± 8) / 6The critical points are: x1 = (-4 + 8) / 6 = 4/6 = 2/3 and x2 = (-4 - 8) / 6 = -12/6 = -2.
3. Find the second derivative f''(x):f''(x) = d/dx(3x² + 4x - 4) = 6x + 4
4. Use the second derivative test to classify the critical points:
   • At x = 2/3: f''(2/3) = 6(2/3) + 4 = 4 + 4 = 8. Since 8 > 0, x = 2/3 corresponds to a local minimum.
   • At x = -2: f''(-2) = 6(-2) + 4 = -12 + 4 = -8. Since -8 < 0, x = -2 corresponds to a local maximum.

The maximum value of the function occurs at x = -2. Although the question asks for the "maximum value", the options provided are x-coordinates. Therefore, the x-coordinate for the maximum is -2.