the real valued function $f(x)=x^3+15x^2-33x-36=0$
1) find $f’(x)=0$
$\implies3x^2+30x-33=0$
$\implies x^2+10x-11=0$
$\implies x^2+11x-x-11=0$
$\implies (x+11)(x-1)=0$
$\implies x=-11,1$
2) find $f^”(x)$ we get : $f^”(x)=6x+30$
3 ) $f^”(1)=6+30=36>0$ it is gives local minima
4 $f^”(-11)=-66+30=-36<0$ it is gives local maxima.
so given function $f(x)$ will give local maxima at $x=-11$ and local minima at $x=1$
$\therefore$ Option $A,C$ is correct.