# Evaluating a function

• Apr 4th 2013, 03:30 AM
Duffman
Evaluating a function
I am evaluating the function $f(x)=x^2-x-lnx$

The derivative should be $f'(x)=2x-1-1/x$

I am trying to find the critical numbers:

$0=2x-1-1/x$

Here comes my problem... I used two different methods but my results are different (Doh)

Method one:
$0=2x-1-1/x$

$0=(2x^2/x)-(x/x)-(1/x)$ (common denominator)

$0=(2x^2-x-1)/x$

And so on...

Method two:

$0=2x-1-1/x$

$1=2x-1/x$

$1=(2x^2)/x-(1/x)$

$1=(2x^2-1)/x$

$1*x=(2x^2-1)$

$x=(2x^2-1)$

$0=(2x^2-x-1)$

And so on...

When I now compare method one with method two I see that $0=(2x^2-x-1)/x$ is different from $0=(2x^2-x-1)$. (Why is my result in method two not devided by x?)
My question is now, why? And what am I doing wrong? :/

• Apr 4th 2013, 03:32 AM
Gusbob
Re: Evaluating a function
There is nothing wrong. Both expressions give identical solutions for x.
• Apr 4th 2013, 03:42 AM
Duffman
Re: Evaluating a function
Thank you Gusbob I see that you are correct :)
• Apr 4th 2013, 03:52 AM
Duffman
Re: Evaluating a function
If I now continue to find the critical numbers (using both method one and two):

Method one:

$0=(2x^2-x-1)/x$

$0=((2x+1)(x-1))/x$

The critical numbers should be:
$x=-1/2$
$x=1$
and $x=0$ (from the denominator)
?

Method two:

$0=(2x^2-x-1)$

$0=((2x+1)(x-1))$

The critical numbers should be:
$x=-1/2$
$x=1$
?

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?
• Apr 4th 2013, 04:27 AM
astartleddeer
Re: Evaluating a function
Only the numerator will make the quotient equate to zero. Division by zero is undefined so you can neglect the denominator.

So $\frac{2x^2 - x -1}{x} = 0$ reduces to $(2x + 1)(x - 1) = 0$
• Apr 4th 2013, 04:58 AM
HallsofIvy
Re: Evaluating a function
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: $y= \sqrt{x}$ is one.
• Apr 4th 2013, 05:12 AM
MINOANMAN
Re: Evaluating a function
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 :point 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
• Apr 4th 2013, 05:14 AM
MINOANMAN
Re: Evaluating a function
Sorry Hall
your post was not there when I started mine.