I have this function: $$f(x) = \frac{x^2}{log(x)-1}$$ and I want to find increasing intervals, and decreasing intervals. Here's what I did:
- I've found domain in which the function is defined, and this is what I got: $f: ] 0, +\infty[ (without\space 1) -> R$.
$0$ and $1$ are critical points in which the function isn't defined (i.e if x = 0, the log is undefined, and if x = 1, the fraction is undefined).
- I've computed first derivative, and this is what I got (I've used quotient rule):
$$\frac{2x * (log(x) – 1) – x^2 * 1/x}{(log(x)−1)^2}$$
I've rewritten it, as follows:
$$\frac{2x * (log(x) – 1) – x}{(log(x)−1)^2}$$
the denominator is always positive so the sign depends only on the numerator. (I think the problem is in this step)
on a line, I've chosen 4 different random numbers, $1/4$, $1/2$, $3/2$, $2$, and I've put each number in the first derivative, and this is what I got: - in the first interval (from 0, to $1/4$), the original function is decreasing (minus sign)
- same for the other intervals
I don't have the full solution of the exercise, therefore I don't knwo if my solution is correct, but I think is strange that in this function there aren't increasing intervals.
- I think the problem is in the first derivative, but I'm sure I've used the quotient rule correctly.
Best Answer
Assuming $\log$ stands for the natural logarithm, you can write the derivative as
$$f'(x)=\frac{x(2\log(x)-3)}{(\log(x)-1)^2}.$$
As $x>0$, its sign depends only on the sign of the parenthesis in the numerator and we have that $x<e^{3/2}\implies f'(x)<0$ so the function is decreasing in $]0,1[\cup]1,e^{3/2}[$ and $x>e^{3/2}\implies f'(x)>0$ so it's increasing in $]e^{3/2},+\infty[.$