## Calculus (3rd Edition)

$f'(x)=3x^{2}-24x+45$ $=3(x^{2}-8x+15)=3(x-3)(x-5)$ $f'(x)=0$ at x= 3 and x=5. That is, x=3 and x=5 are the critical points. Now, $f''(x)= 6x-24$ $f''(3)= -6$ $f''(5)= 6$ Therefore, by second derivative test, x=3 is a point of local maxima and local maximum value of f at x=3 is $f(3)= 3^{3}-(12\times3^{2})+(45\times3)=54$ while x=5 is the point of local minima and local minimum value of f at x=5 is $f(5)= 5^{3}-(12\times5^{2})+(45\times5)= 50$