Question

What intervals is this function: `f(x)= x^5(lnx)` increasing and decreasing?

Answer 1

`f` strictly decreasing in `(0,e^(-1/5)]` & `f` strictly increasing in `[e^(-1/5),+oo)`

Explanation:

`f(x)=x^5lnx` , `D_f=(0,+oo)`

For `x` `in` `D_f`,
`f'(x)=(x^5lnx)' =`

`(x^5)'lnx+x^5(lnx)'` `=`

`5x^4lnx+x^5/x` `=`

`5x^4lnx+x^4` `=`

`x^4(5lnx+1)`

• `f'(x)=0` `<=>` `x^4=0` `<=>` `x=0` -> Impossible because `x>0`

`5lnx+1=0` `<=>` `5lnx=-1` `<=>` `lnx=-1/5` `<=>` `x=e^(-1/5)` `~=` `0.81`

For `x=1/2=0.5` ---> `f'(1/2)=5*1/2^4ln(1/2)+1/2^4` `=`
`5/16(ln1-ln2)+1/16` `=` `-ln2*5/16+1/16` `<0`

because `ln2~=0.7`
`1/16=0.0625`

For `x=1` ----> `f'(1)=1>0`

As a result we have,

• `f` continuous in `(0,e^(-1/5)]` with `f'(x)<0` when `x` `in` `(0,e^(-1/5))` so `f` strictly decreasing in `(0,e^(-1/5)]`

• `f` continuous in `[e^(-1/5),+oo)` with `f'(x)>0` when `x` `in` `(e^(-1/5),+oo)` so `f` strictly increasing in `[e^(-1/5),+oo)`