# Homework Help: Max or min of xlnx^2

1. Oct 30, 2012

### 939

1. The problem statement, all variables and given/known data

2. Relevant equations

Y = xlnx^2

3. The attempt at a solution

Y' = (lnx^2) + (x)(2x/x^2)

For the max/min, I would just set Y' = 0 by trial and error and then do y", but I don't want to do that until I know y' is correct.

2. Oct 30, 2012

### Ray Vickson

Your Y' is OK. Solving the equation Y' = 0 should be easy, and does not need "trial and error". Hint: look at the form of Y'; simplify it as much as possible.

RGV

3. Oct 30, 2012

### Dick

Looks ok so far if you mean x*ln(x^2). You could simplify the second term quite a bit.

4. Oct 30, 2012

### 939

Thanks for such quick anwers.

So I simplify y' a little bit and have:

y ' = (lnx^2) + (x)(2x/x^2)
y ' = (lnx^2) + (2x^2/x^2)
y ' = (2lnx) + (2x/x)
y = (2lnx) + (2)

But how can I set y ' = 0? 2lnx would have to be - 2, no? It would be invalid, wouldn't it? :-/

5. Oct 30, 2012

### Dick

Not at all. Some numbers have negative logs. Think about that a bit more.

6. Oct 30, 2012

### 939

Got it. Thanks a lot for the help! =)