f(x)=2x^5(x-3)^4f(x)=2x5(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.