For $f(x)=4x^3+12x^2-24x+10$ (so $f'(x)=12x(x-1)(x+2)$ after dividing constants), the second-derivative test at $x=0$ gives that $x=0$ is

1. With $f'(x)=12x(x-1)(x+2)=12x^3+12x^2-24x$, critical points are $x=0,1,-2$. 2. $f''(x)=36x^2+24x-24$. 3. $f''(0)=-24<0$. 4. By the second derivative test, $f''(0)<0$ means $x=0$ is a point of local maxima. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.19_