Question

f(x) = x(lnx)^2
find the local max and mins

Answer 1

I think you find the first derivative and then find the zeros.

Answer 2

First derivative set equal to zero fr the crit. points. Plug them into the second derivative to determine if they are a max or min.

Answer 3

Yess i got up to there.
But i cannot solve 0=(lnx)(2x+lnx)

Answer 4

how do you know that?

Answer 5

You took the derivative incorrectly. You need to use the product and chain rule here.

Answer 6

suppoed to be lnx(2+lnx)

Answer 7

but how can you solve lnx=0?

Answer 8

ln(1)=0

Answer 9

in order to cancel the ln you have to e^

Answer 10

because of that you also have to do it to the right side.
e^(lnx) = e^(0)
x = 1
anything to the zero is 1.