Question

How do you find the critical points for `x^4(lnx)`?

Answer 1

`P_min(e^(-1/4),-1/(4e))`

Explanation:

we get
`f'(x)=x^3(4ln(x)+1)`
and since we have `x>0`
we get
`f'(x)=0` if `x=e^(-1/4)`
so `f''(x)=3x^2(4ln(x)+1)+x^3+4/x`

`f''(x)=3x^2(4ln(x)+1)+4x^2`

and

`f''(e^(-1/4))=4(e^(-1/4))^2>0`
and
`f(e^(-1/4))=-1/4*e^(-1)`