#f(x)=2x^5(x-3)^4#

#f'(x)=10x^4(x-3)^4 + 2x^5*4(x-3)^3(1)#

# = 10x^4(x-3)^4+8x^5(x-3)^3#

# = 2x^4(x-3)^3[5(x-3) + 4x]#

# = 2x^4(x-3)^3(9x-15)#

#f'(x)# is never undefined and it is #0# at #0#, #3#, and #15/9=5/3#.

All of these are in the domain of #f#, so they are critical points for #f#.

**Sign of #f'(x)#**

#{: (bb"Interval:",(-oo,0),(0,5/3),(5/3,3),(3,oo)),
(darrbb"Factors"darr,"=======","======","=====","====="),
(2x^4," +",bb" +",bb" +",bb" +"),
((x-3)^3, bb" -",bb" -",bb" -",bb" +"),
(9x-15,bb" -",bb" -",bb" +",bb" +"),
("==========","========","======","=====","======"),
(bb"Product"=f'(x),bb" +",bb" +",bb" -",bb" +")
:}#

The sign of #f'# does not change at #0#, so #f(0)# is neither minimum nor maximum.

#f'# changes from + to - at #5/3#, so #f(5/3)# is a local maximum.

#f'# changes from - to + at #3#, so #f(3)# is a local minimum.