Final answer:
The absolute minimum value of the function f(x) = x^3 + 3x^2 + 7 on the closed interval [-4, 2] is -9, which occurs at x = -4 after evaluating the function at critical points and boundaries.
Step-by-step explanation:
To determine the absolute minimum value of the function f(x) = x^3 + 3x^2 + 7 on the closed interval [-4, 2], we need to evaluate the function at its critical points and endpoints. Critical points occur where the derivative of the function is zero or undefined. Therefore, we first calculate the derivative, f'(x) = 3x^2 + 6x. Setting the derivative equal to zero, we find the critical points: 3x^2 + 6x = 0, x(x + 2) = 0, resulting in x = 0 and x = -2.
Now, we evaluate the function at the critical points and the endpoints of the interval. These points are x = -4, x = -2, x = 0, and x = 2. We get:
- f(-4) = (-4)^3 + 3(-4)^2 + 7 = -64 + 48 + 7 = -9
- f(-2) = (-2)^3 + 3(-2)^2 + 7 = -8 + 12 + 7 = 11
- f(0) = (0)^3 + 3(0)^2 + 7 = 7
- f(2) = (2)^3 + 3(2)^2 + 7 = 8 + 12 + 7 = 27
Among these values, the absolute minimum value on the interval [-4, 2] is -9, which occurs at x = -4.