Question Detail
Question
Find the minimum value of the function
y = 2x³ - 21x² + 36x - 20.
Correct Answer
Option A is the correct answer.
Detailed Explanation
To find the minimum value of the function, we follow these steps:
- Find the first derivative
dy/dx:dy/dx = d/dx(2x³ - 21x² + 36x - 20) = 6x² - 42x + 36 - Set the first derivative to zero to find critical points:
6x² - 42x + 36 = 0x² - 7x + 6 = 0(divide by 6)(x - 1)(x - 6) = 0The critical points arex = 1andx = 6. - Find the second derivative
d²y/dx²:d²y/dx² = d/dx(6x² - 42x + 36) = 12x - 42 - Use the second derivative test to classify the critical points:
- At
x = 1:d²y/dx² = 12(1) - 42 = -30. Since-30 < 0,x = 1corresponds to a local maximum. - At
x = 6:d²y/dx² = 12(6) - 42 = 72 - 42 = 30. Since30 > 0,x = 6corresponds to a local minimum.
- At
- Calculate the function's value at the local minimum
x = 6:y(6) = 2(6)³ - 21(6)² + 36(6) - 20y(6) = 2(216) - 21(36) + 216 - 20y(6) = 432 - 756 + 216 - 20y(6) = 648 - 776 = -128
-128.
Hint
First, find the critical points by setting the first derivative to zero. Then use the second derivative test to determine whether each critical point is a local maximum or minimum. Finally, substitute the x-value of the local minimum back into the original function to find the minimum value.