Step 1 :The function is \(f(x)=-x^{3}-9 x^{2}-15 x-2\).
Step 2 :The derivative of the function is \(f'(x)=-3x^{2}-18x-15\).
Step 3 :Setting the derivative equal to zero gives us the equation \(-3x^{2}-18x-15=0\).
Step 4 :We can factor out a -3 to simplify the equation: \(x^{2}+6x+5=0\).
Step 5 :Factoring the quadratic gives us \((x+1)(x+5)=0\).
Step 6 :Setting each factor equal to zero gives us the critical points \(x=-1\) and \(x=-5\).
Step 7 :The second derivative of the function is \(f''(x)=-6x-18\).
Step 8 :Substituting \(x=-1\) into the second derivative gives us \(f''(-1)=-6(-1)-18=6-18=-12\), which is less than zero, so \(x=-1\) is a local maximum.
Step 9 :Substituting \(x=-5\) into the second derivative gives us \(f''(-5)=-6(-5)-18=30-18=12\), which is greater than zero, so \(x=-5\) is a local minimum.
Step 10 :For \(x=-1\), \(f(-1)=-(-1)^{3}-9(-1)^{2}-15(-1)-2=1-9+15-2=5\).
Step 11 :For \(x=-5\), \(f(-5)=-(-5)^{3}-9(-5)^{2}-15(-5)-2=-125-225+75-2=-277\).
Step 12 :\(\boxed{\text{So, the local minimum is -277 at } x=-5 \text{ and the local maximum is 5 at } x=-1}\).