Question

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

Answer 1

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

Explanation:

`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.