Question 666136
Use the first derivative test to find the local maxima and minima.

y' = 3x^2 - 21x + 30

Set 3x^2 -21x + 30 = 0

Using quadratic formula we get

{{{x=((21 +-sqrt(21^2 - 30*3*4))/6)}}}

x = 2 and x =5.

These are our candidates for local maxima and minima.

Plug in f(2) and f(5).

f(2) = 46

f(5) = 32.5

Hence, (5,32.5) is the minimum and (2,46) is the maximum.