The user wants me to find the total number of local maxima and local minima for the given piecewise function.

Function: $f(x) = \left\{ \begin{matrix} {{{(2 + x)}^3},} & { - 3 < x \le - 1} \\ {{x^{2/3}},} & { - 1 < x < 2} \\ \end{matrix} \right.$

To find the local maxima and minima, we need to find the critical points of the function. Critical points occur where the derivative f'(x) is zero or undefined.

Step 1: Analyze the first piece of the function, $f(x) = (2 + x)^3$ for -3 < x <= -1

1. Find the derivative: $f'(x) = d/dx [(2 + x)^3] = 3(2 + x)^2$

2. Find critical points by setting f'(x) = 0: $3(2 + x)^2 = 0$ 2 + x = 0 x = -2

3. Check if x = -2 is in the interval (-3, -1]: Yes, -3 < -2 <= -1.

4. Use the First Derivative Test to classify x = -2: We examine the sign of $f'(x) = 3(2 + x)^2$ around x = -2.

   • For x < -2 (e.g., x = -2.5), $f'(x) = 3(2 - 2.5)^2 = 3(-0.5)^2 > 0$.
   • For x > -2 (e.g., x = -1.5), $f'(x) = 3(2 - 1.5)^2 = 3(0.5)^2 > 0$. Since f'(x) does not change sign at x = -2 (it is positive on both sides), x = -2 is a point of inflection, not a local maximum or minimum.

Step 2: Analyze the second piece of the function, $f(x) = x^(2/3)$ for -1 < x < 2

1. Find the derivative: $f'(x) = d/dx [x^(2/3)] = (2/3)x^(-1/3) = 2 / (3 * x^(1/3))$

2. Find critical points:

   • f'(x) = 0: $2 / (3 * x^(1/3)) = 0$ has no solution.
   • f'(x) is undefined when the denominator is zero, which occurs at x = 0.

3. Check if x = 0 is in the interval (-1, 2): Yes, -1 < 0 < 2.

4. Use the First Derivative Test to classify x = 0: We examine the sign of $f'(x) = 2 / (3 * x^(1/3))$ around x = 0.

   • For x < 0 (e.g., x = -0.1), $x^(1/3)$ is negative, so f'(x) < 0. The function is decreasing.
   • For x > 0 (e.g., x = 0.1), $x^(1/3)$ is positive, so f'(x) > 0. The function is increasing. Since f'(x) changes from negative to positive at x = 0, the function has a local minimum at x = 0.

Step 3: Analyze the point where the function definition changes, x = -1

Points where the function definition changes can also be local extrema if the function is not differentiable there. We need to check the behavior of the function around x = -1.

1. Check for continuity at x = -1:

   • Value at x = -1: $f(-1) = (2 + (-1))^3 = 1^3 = 1$.
   • Limit from the right: $lim_{x->-1^+} f(x) = lim_{x->-1^+} x^(2/3) = (-1)^(2/3) = ((-1)^2)^(1/3) = 1^(1/3) = 1$. Since the value of the function and the limit from the right are equal, the function is continuous at x = -1.

2. Check differentiability at x = -1 (or use the first derivative test across the point):

   • For x in (-3, -1), we found $f'(x) = 3(2 + x)^2$. As x approaches -1 from the left, f'(x) approaches $3(2 - 1)^2 = 3$, which is positive. So, f(x) is increasing to the left of x = -1.
   • For x in (-1, 0), we found $f'(x) = 2 / (3 * x^(1/3))$. For x in this interval, $x^(1/3)$ is negative, so f'(x) is negative. So, f(x) is decreasing to the right of x = -1.

   Since the function is increasing to the left of x = -1 and decreasing to the right, x = -1 is a point of local maximum.

Step 4: Conclusion

We have found the following:

• x = -2 is a point of inflection.
• x = -1 is a local maximum.
• x = 0 is a local minimum.

The total number of local maxima and local minima is one maximum plus one minimum, which is 2.

This corresponds to option C.