There is nothing wrong. Both expressions give identical solutions for x.
I am evaluating the function
The derivative should be
I am trying to find the critical numbers:
Here comes my problem... I used two different methods but my results are different
Method one:
(common denominator)
And so on...
Method two:
And so on...
When I now compare method one with method two I see that is different from . (Why is my result in method two not devided by x?)
My question is now, why? And what am I doing wrong? :/
Please help
If I now continue to find the critical numbers (using both method one and two):
Method one:
The critical numbers should be:
and (from the denominator)
?
Method two:
The critical numbers should be:
?
My final question is.. why is there 3 critical numbers using method one but only 2 critical numbers using method two? Am I thinking worng?
No, you cannot "neglect the denominator" for this problem. astarledeer is correct that a fraction, and so the dervative, is 0 only when the numerator is 0, independent of the denominator. But this problem asked for "critical points" which are points at which the derivative is 0 or where the derivative does not exist. Going back to the original equation,
f'= 2x- 1- 1/x, f' does not exist at x= 0. However, now we note that the original function, f(x)= x^2- x- ln(x) is also not defined at x= 0. Therefore, the only critical points in the domain of the original function are x= 1/2 and x= 1. But there are examples where the function exists and some point but the derivative does not: is one.
Duffman
Mathematics is not just doing additions or subtractions and ..other operations but rather thinking and reasoning.
first of all do you know what a critical point is?
if not go here to revise.... Critical point (mathematics) - Wikipedia, the free encyclopedia
in your case the critical points are 3 as you said oint zero where the derivative does not exist.....this is obvious from the domain of your function...and points 1 and (-1/2)??? but you miss the point again..... the value x= -1/2 is out of the domain of your function...in other words your function is defined in the domain (0,+infinity)..
and x = -1/2 is not included....therefore the only available critical point to consider is the point x=1.
MINOAS