Question

Over what intervals is `f(x)=(9x^2-x)/(x-1)` increasing and decreasing?

Answer 1

`f` is increasing on `(-oo,(3-2sqrt2)/3)` and on `((3+2sqrt2)/3,oo)` and it is decreasing on `((3-2sqrt2)/3,1)` and on `(1,(3+2sqrt2)/3)`

Explanation:

`f'(x) = ((18x-1)(x-1)-(9x^2-x)(1))/(x-1)^2`

`= (9x^2-18x+1)/(x-1)^2`

`f'(x) = 0`, at solutions to `9x^2-18x+1=0`.
Solve by completing the square of by using the quadratic formula.
`x=(3+-2sqrt2)/3`

`f'(x)` is not defined at `x=0`

Test the sign of `f'` on each interval:

`{: (bb"Interval:",(-oo,(3-2sqrt2)/3),((3-2sqrt2)/3,1),(1,(3+2sqrt2)/3),((3+2sqrt2)/3,oo)), (darrbb"Factors"darr,"========","======","=====","======"), (9x^2-18x+1, bb" +",bb" -",bb" -",bb" +"), ((x-1)^2,bb" +",bb" +",bb" +",bb" +"), ("==========","========","======","=====","======"), (bb"Product"=f'(x),bb" +",bb" -",bb" -",bb" +") :}`