3. The maximum value of \( f(x)=2 x^{3}-15 x^{2}+36 x \) on the closed interval \( [0,4] \) is \( \begin{array}{llll}\text { (A) } 28 & \text { (B) } 30 & \text { (C) } 32 & \text { (D) } 48\end{array} \)

To find the maximum value of the function \( f(x) = 2x^3 - 15x^2 + 36x \) on the closed interval \( [0, 4] \), we first calculate the derivative \( f'(x) \) to find critical points. Setting the derivative \( f'(x) = 6x^2 - 30x + 36 \) equal to zero, we solve for \( x \) which gives us potential maximums or minimums. Calculating the roots gives us \( x = 2 \) as a critical point. Next, we evaluate \( f(x) \) at the endpoints and the critical point: - \( f(0) = 0 \) - \( f(4) = 32 \) - \( f(2) = 28 \) Comparing these values, we find that the maximum occurs at \( f(4) = 32 \). So the maximum value of \( f(x) \) on the interval \( [0, 4] \) is \( 32 \) which corresponds to option \( (C) \).