Find the local maximum and local minimum values of the function given by $f(x) = x^{3} - 6x^{2} + 9x + 15$.

(N/A) Given $f(x) = x^{3} - 6x^{2} + 9x + 15$.
First,find the derivative $f^{\prime}(x) = 3x^{2} - 12x + 9$.
Set $f^{\prime}(x) = 0$ to find critical points: $3(x^{2} - 4x + 3) = 0 \Rightarrow 3(x - 1)(x - 3) = 0$.
Thus,the critical points are $x = 1$ and $x = 3$.
Now,find the second derivative $f^{\prime \prime}(x) = 6x - 12 = 6(x - 2)$.
For $x = 1$,$f^{\prime \prime}(1) = 6(1 - 2) = -6 < 0$. Since the second derivative is negative,$x = 1$ is a point of local maxima.
The local maximum value is $f(1) = (1)^{3} - 6(1)^{2} + 9(1) + 15 = 1 - 6 + 9 + 15 = 19$.
For $x = 3$,$f^{\prime \prime}(3) = 6(3 - 2) = 6 > 0$. Since the second derivative is positive,$x = 3$ is a point of local minima.
The local minimum value is $f(3) = (3)^{3} - 6(3)^{2} + 9(3) + 15 = 27 - 54 + 27 + 15 = 15$.