# Intervals where the function is increasing

1. Jun 5, 2007

### t_n_p

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

For the function f(x)=xln(x²), x≠0

3. The attempt at a solution

f'(x)=2+ln(x²)

For the function to be increasing f'(x)>0

2+ln(x²)>0
ln(x²)>-2
x²>e^(-2)

Taking square roots of both sides

±x>±(1/e)

How can I find the intervals from this?
(Somebody has suggested to me that after taking the square root of 1/e the answer should be positive only rather than plus/minus, but I can't seem to find a reason why such would be the case)

2. Jun 5, 2007

### mjsd

note that if you accept that $$\log (x^2) = 2 \log (x)$$, then now RHS implies that x should be non-negative. That's the only possible reason I can think of that may dictate that you take the +ve root. But, it is clear that for your f(x) both +/- roots are good solutions and your interval is
$$(-\infty,-1/e) \cup (1/e, \infty)$$

3. Jun 5, 2007

### t_n_p

yeah I got it!
If I bring the two down to the front of the natural logarithm I will get
2ln|x|.

With the modulus then I get |x|>(1/e) and hence the intervals (-∞, -1/e) and (1/e, ∞)

Thanks! :tongue2: