Answer:
To apply the first derivative test to the function f(x) = 2x^3 + 3x^2 - 12x + 5, we need to find its derivative:
f'(x) = 6x^2 + 6x - 12
Then, we need to find the critical points by solving for f'(x) = 0:
6x^2 + 6x - 12 = 0
Dividing both sides by 6, we get:
x^2 + x - 2 = 0
Factoring the left side, we get:
(x + 2)(x - 1) = 0
So the critical points are x = -2 and x = 1.
To determine the intervals where f(x) is increasing and decreasing, we need to evaluate f'(x) on each side of the critical points. We can use a sign chart to do this:
x | -2 | 1 |
------|-----|----|
f'(x) | -12 | 0 |
Since f'(x) is negative to the left of x = -2 and positive to the right of x = -2, the function is decreasing to the left of x = -2 and increasing to the right of x = -2. Since f'(x) changes sign at x = 1, there is a local minimum at x = 1.
Therefore, using the first derivative test, we can conclude that the function has a local minimum at x = 1.
Explanation: