#y = x^3 - 3x^2 - 9x +15#
#y' = 3x^2 - 6x - 9 = 0# when
#3(x^2 - 2x - 3) = 0#
#3(x - 3)(x+1) = 0# at #x=-1,3#
on #(-oo, -1)# test some number to see that #y'# is positive
on #(-1, 3)# test some number to see that #y'# is negative
So when #x= -1#, #y# has a local maximum of #20#
(Put #x=-1# in #y = x^3 - 3x^2 - 9x +15# to get #20#)
on #(-1, 3)# e already found that #y'# is negative
on #(3, oo)# test some number to see that #y'# is positive
So when #x= 3#, #y# has a local minimum of #-12#.
(Put #x=3# in #y = x^3 - 3x^2 - 9x +15# to get #-12#)
