The minimum value of the function f(x) = x3 – 3×2 – 24x + 100 in the interval [-3, 3] is A. 20 B. 28 C. 16 D. 32

The minimum value of the function $f(x) = x^3 – 3x^2 – 24x + 100$ in the interval $[-3, 3]$ is $16$.

To find the minimum value, we can use the following steps:

1. Find the derivative of $f$, i.e. $f'(x)$.
2. Set $f'(x) = 0$ and solve for $x$.
3. Evaluate $f(x)$ at the critical points found in step 2 and at the endpoints of the interval, i.e. $x = -3$ and $x = 3$.
4. The minimum value is the smallest value among the values found in step 3.

In this case, we have:

$$f'(x) = 3x^2 – 6x – 24 = (x – 4)(3x + 6)$$

Solving $f'(x) = 0$, we get $x = 4$ or $x = -6$.

Evaluating $f(x)$ at the critical points and at the endpoints of the interval, we get:

$$f(-3) = 27 – 27 – 72 + 100 = 16$$
$$f(-6) = -216 – 108 – 144 + 100 = -162$$
$$f(4) = 64 – 48 – 96 + 100 = 16$$
$$f(3) = 27 – 27 – 72 + 100 = 16$$

Therefore, the minimum value of $f(x)$ in the interval $[-3, 3]$ is $16$.