What are the values and types of the critical points, if any, of f(x)=2x^5(x-3)^4f(x)=2x5(x3)4?

1 Answer
Feb 11, 2016

At x=0x=0, there is a saddle point (neither min nor max). At x=3x=3 there is a local minimum (which is 00). At x=5/3x=53, there is a local maximum (which is f(5/3)f(53)).

Explanation:

f(x)=2x^5(x-3)^4f(x)=2x5(x3)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.