Final answer:
Option A. Using calculus methods, the function g(x) = x³ - 6x² + 9x has a local maxima at x = 1 and a local minima at x = 3, determined by setting the first derivative to zero and applying the second derivative test.
Step-by-step explanation:
To find the local maxima and minima of the function g(x) = x³ - 6x² + 9x, we can use calculus methods or a graphing calculator. We'll proceed with the calculus method for this example.
First, we need to find the first derivative of the function, g'(x) = 3x² - 12x + 9. To find the critical points where the local maxima and minima could occur, we set the derivative equal to zero:
3x² - 12x + 9 = 0.
Factoring out 3 gives us:
x² - 4x + 3 = 0.
This quadratic equation factors further into:
(x - 1)(x - 3) = 0,
giving us potential critical points at x = 1 and x = 3.
To determine whether these points are maxima or minima, we can use the second derivative test. The second derivative of the function is g''(x) = 6x - 12. Evaluating this at the critical points:
- g''(1) = 6(1) - 12 = -6 (negative, so x = 1 is a local maxima)
- g''(3) = 6(3) - 12 = 6 (positive, so x = 3 is a local minima)
Therefore, the answer is (a): Local maxima at x = 1, local minima at x = 3.