Step-by-step Solution:

The given piecewise function is: \[ f(x) = \begin{cases} x^2, & \text{if } x \leq 0 \\ 2\sin(x), & \text{if } x > 0 \end{cases} \] Continuity at \( x = 0 \): For \( f(x) \) to be continuous at \( x = 0 \), \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0). \] When \( x \leq 0 \), \( f(x) = x^2 \). \[ \lim_{x \to 0^-} f(x) = 0^2 = 0. \] When \( x > 0 \), \( f(x) = 2\sin(x) \). \[ \lim_{x \to 0^+} f(x) = 2\sin(0) = 0. \] Since \( f(0) = 0^2 = 0 \), all values agree, and \( f(x) \) is continuous at \( x = 0 \). \[\] Differentiability at \( x = 0 \): \[\] For differentiability, the left-hand and right-hand derivatives at \( x = 0 \) must be equal. \[\]Left-hand derivative: For \( x \leq 0 \), \( f(x) = x^2 \). \[ f'(x) = 2x, \quad f'(0^-) = 2(0) = 0. \] Right-hand derivative: For \( x > 0 \), \( f(x) = 2\sin(x) \). \[ f'(x) = 2\cos(x), \quad f'(0^+) = 2\cos(0) = 2. \] Since \( f'(0^-) \neq f'(0^+) \), \( f(x) \) is not differentiable at \( x = 0 \). \[\]Behavior near \( x = 0 \): \[\] - When \( x \leq 0 \), \( f(x) = x^2 \), which increases as \( x \to 0^- \). \[\] - When \( x > 0 \), \( f(x) = 2\sin(x) \), which also increases near \( x = 0 \). \[\] Thus, \( f(x) \) has a local minimum at \( x = 0 \). \[\] The function \( f(x) \) is continuous but not differentiable at \( x = 0 \). It has a **local minimum** at \( x = 0 \). The answer is option A: minima.