Total Number of Local Maxima and Local Minima of Piecewise Function

To find the total number of local maxima and local minima of the function

$$f(x) = \begin{cases} (2+x)^3, & -3 < x \le -1 \\ x^{2/3}, & -1 < x < 2 \end{cases}$$

we analyze the function across its domain $(-3, 2)$:

1. Interval $(-3, -1)$:
   Here, $f(x) = (2+x)^3$.
   Its derivative is:
   $f'(x) = 3(2+x)^2$

   • For all $x \in (-3, -1) \setminus \{-2\}$, $f'(x) > 0$.
   • At $x = -2$, $f'(x) = 0$, but the derivative does not change sign (it remains positive on both sides of $x = -2$). Thus, $x = -2$ is an inflection point and not a local extremum.
   • Therefore, $f(x)$ is strictly increasing on $(-3, -1]$.

2. At $x = -1$:
   First, we check the continuity at $x = -1$:

   • Left-hand value: $f(-1) = (2 - 1)^3 = 1$
   • Right-hand limit: $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} x^{2/3} = (-1)^{2/3} = 1$

   Since $\lim_{x \to -1} f(x) = f(-1) = 1$, the function is continuous at $x = -1$.

   Now, testing the behavior in an immediate neighborhood of $x = -1$:

   • For $x \in (-2, -1)$, we have $2+x < 1 \implies f(x) = (2+x)^3 < 1$.
   • For $x \in (-1, 0)$, we have $0 < |x| < 1 \implies f(x) = |x|^{2/3} < 1$.

   Since $f(x) < f(-1)$ for all $x$ in a deleted neighborhood of $-1$, the function achieves a local maximum at $x = -1$.

3. Interval $(-1, 2)$:
   Here, $f(x) = x^{2/3}$.

   • For $x \in (-1, 0)$, $f'(x) = \frac{2}{3}x^{-1/3} < 0$, which means $f(x)$ is strictly decreasing.
   • At $x = 0$, $f(0) = 0$.
   • For $x \in (0, 2)$, $f'(x) = \frac{2}{3}x^{-1/3} > 0$, which means $f(x)$ is strictly increasing.

   Since $f(x) \ge 0$ for all $x \in (-1, 2)$ and $f(0) = 0$, the function achieves a local minimum at $x = 0$.

4. Conclusion:

   • Local maxima: $1$ (at $x = -1$)
   • Local minima: $1$ (at $x = 0$)

The total number of local maxima and local minima is $1 + 1 = 2$.

Therefore, the correct option is C.