Answer
Critical points: x=3 and x=5.
Local maximum value= f(3)= 54.
Local minimum value= f(5)= 50.
Work Step by Step
$f'(x)=3x^{2}-24x+45$
$=3(x^{2}-8x+15)=3(x-3)(x-5)$
$f'(x)=0$ 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
$f(5)= 5^{3}-(12\times5^{2})+(45\times5)= 50$
