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I know how to do a quadratic version but how do I find the absolute minimum and absolute maximum values of f on the given interval. $f(x) = x − \ln 2x$, $[\frac{1}{2}, 2]$

I keep getting the wrong answers

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You do it the exact same way as your previous question! You find the critical points, you evaluate at the endpoints and at the critical points, and you compare the values. Remember that the critical points are the points where the derivative is either defined and equal to $0$, or undefined. As to "keep getting the wrong answers", if you bothered to show us what you did, perhaps we could tell you what you are doing wrong so you don't keep doing it. –  Arturo Magidin Jun 26 '12 at 4:57
    
Ah okay i thought there was something different to lnx questions. Thanks! –  soniccool Jun 26 '12 at 5:03
    
Why would you think that? Did you not cover finding extreme values in your course? The method does not depend on what goes into the function. –  Arturo Magidin Jun 26 '12 at 5:06
    
Im trying to learn calc on my own so im allover the place but im getting it slowly. –  soniccool Jun 26 '12 at 5:07
    
I am on the EVT today looking into it. –  soniccool Jun 26 '12 at 5:08
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2 Answers 2

up vote 2 down vote accepted

The derivative is $1-\frac{2}{2x}$. It is hard to go wrong after that. The contenders are $x=1/2$, $x=1$, and $x=2$.

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First you find the critical points of $f(x)$. That would be done by finding $f'(x)$ and setting that equal to 0. So $f'(x) = 0$, solve for $x$.

After checking the critical points, you must also check the boundaries, which are $\frac{1}{2}$ and $2$.

EDIT: Also, you have to find places where $f(x)$ is not differentiable to find all critical points.

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Not quite accurate; your method only finds the stationary points, not all the critical points –  Arturo Magidin Jun 26 '12 at 4:59
    
Out of curiosity, why do you call the usual ones stationary critical points? –  MGN Jun 26 '12 at 5:01
    
@MGNewman: A stationary point is a point where the derivative is zero (it comes from thinking about the derivative as representing speed... when the speed is $0$, you are stationary). Critical points are stationary points and points where the function is defined but not differentiable. –  Arturo Magidin Jun 26 '12 at 5:07
    
@ArturoMagidin: The other critical points would be found where $f'(x)$ is not differentiable. –  Salazar Jun 26 '12 at 5:08
    
@Salazar: No, it doesn't matter where $f'(x)$ is not differentiable. It's the points where $f$ itself is not differentiable; so your description of how to find the critical points of $f$ is incomplete, as I noted. –  Arturo Magidin Jun 26 '12 at 5:09
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