I have this function: f(x)=x^2/(log(x)-1) and I want to find increasing intervals, and decreasing intervals.

Juan Leonard

Juan Leonard

Answered question

2022-10-16

First derivative test to find where the function is increasing, and decreasing
I have this function: f ( x ) = x 2 l o g ( 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 , + [ ( w i t h o u t   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):
2 x ( l o g ( x ) 1 ) x 2 1 / x ( l o g ( x ) 1 ) 2
I've rewritten it, as follows: 2 x ( l o g ( x ) 1 ) x ( l o g ( 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.

Answer & Explanation

Tirioliwo

Tirioliwo

Beginner2022-10-17Added 12 answers

Step 1
Assuming log stands for the natural logarithm, you can write the derivative as f ( x ) = x ( 2 log ( x ) 3 ) ( log ( x ) 1 ) 2 ..
Step 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 f ( x ) < 0 so the function is decreasing in ] 0 , 1 [ ] 1 , e 3 / 2 [ and x > e 3 / 2 f ( x ) > 0 so it's increasing in ] e 3 / 2 , + [ .

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