Step-by-step Solution:

The function is defined as: \[ f(x) = \begin{cases} -x^3 + 3x^2 + 1 & \text{if } x \leq 2, \\ \cos x & \text{if } 2 < x < 4, \\ e^{-x} & \text{if } x > 4. \end{cases} \] The derivative of \( f(x) \) is given by: \[ f'(x) = \begin{cases} -3x^2 + 6x & \text{if } x < 2, \\ -\sin x & \text{if } 2 < x < 4, \\ -e^{-x} & \text{if } x > 4. \end{cases} \] Local Maximum Analysis To find critical points where \( f'(x) = 0 \): \[\] 1. For \( x < 2 \), \( f'(x) = -3x^2 + 6x = 0 \): \[ -3x(x - 2) = 0 \implies x = 0 \text{ or } x = 2. \] The second derivative for \( x \leq 2 \) is: \[ f''(x) = -6x + 6. \] At \( x = 2 \): \[ f''(2) = -6(2) + 6 = -6 \implies f(x) \text{ has a local maximum at } x = 2. \] 2. For \( 2 < x < 4 \), \( f'(x) = -\sin x \): \[ -\sin x = 0 \implies x = \pi. \] The second derivative is: \[ f''(x) = -\cos x. \] At \( x = \pi \): \[ f''(\pi) = -\cos \pi = 1 \implies f(x) \text{ has a local minimum at } x = \pi. \] Global and Local Maxima From the above analysis: \( f(x) \) has a local maximum at \( x = 2 \), but it is not the global maximum. \[\] \( f(x) \) has a local minimum at \( x = \pi \).