My Math Forum f(x)= log(sqrt(x)-qwt(x-1)), extrema

 Calculus Calculus Math Forum

 August 30th, 2013, 01:13 AM #1 Senior Member   Joined: Oct 2012 Posts: 460 Thanks: 0 f(x)= log(sqrt(x)-qwt(x-1)), extrema Dear All! Please, can we find maximum and minimum from: $f(x)=log(\sqrt{x}-\sqrt{x-1)}$ ? Many thanks! p.s. I got this pic on wa: http://www.wolframalpha.com/input/?i=gr ... 8x-1%29%29 But , I got completely other graph in maxima! Please what is the problem? Many thanks!
August 31st, 2013, 04:45 AM   #2
Math Team

Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,372
Thanks: 234

Re: f(x)= log(sqrt(x)-qwt(x-1)), extrema

Quote:
 Originally Posted by ungeheuer Dear All! Please, can we find maximum and minimum from: $f(x)=log(\sqrt{x}-\sqrt{x-1)}$ ? Many thanks! p.s. I got this pic on wa: http://www.wolframalpha.com/input/?i=gr ... 8x-1%29%29 But , I got completely other graph in maxima! Please what is the problem? Many thanks!
I'm going to assume natural logarithm base e because no index was given on log. The first thing to do is find the domain , the argument of the logarithm must be > 0 , this means

$\sqrt{x} \ - \ \sqrt{x-1} \ > \ 0$

Carefully looking at this inequality we can see the domain is $\ \ x \ \ge \ 1$

Let $\ \ u \= \ \sqrt{x} \ - \ \sqrt{x-1}$

$f(x) \= \ log \ u$

$f'(x) \= \ \frac{1}{u}u#39;$

So we need to find u' , then substitute , then set f'(x) to zero.

$u' \ = \ \frac{1}{2 \sqrt{x} } \ - \ \frac{1}{2 \sqrt{x \ - \ 1}} \ = \ \frac{1}{2} $$\frac{1}{\sqrt{x}} \ - \ \frac{1}{\sqrt{x-1}}$$$

Okay , substitute now into f'(x) and set equal to zero ,

$f'(x) \ = \ \frac{1}{\sqrt{x} \ - \ \sqrt{x-1} } \cdot \frac{1}{2} $$\frac{1}{\sqrt{x}} \ - \ \frac{1}{\sqrt{x-1} }$$ \ = \ 0$

For this to be 0 the expression inside parenthesis must be 0 , there is no other way... let's do this ,

$\frac{1}{\sqrt{x}} \ - \ \frac{1}{\sqrt{x-1} } \= \ 0$

$\frac{1}{\sqrt{x}} \= \ \frac{1}{\sqrt{x-1}$

Well , this is clearly impossible. The numeators are the same but the denominators are NOT the same so the fractions can never be equal.

Conclusion , the derivative is never zero or UNDEFINED for x > 1 so there can be no local max or min.

Note* The domain of f'(x) is x > 1 , we lost the equals sign for the derivative but x = 1 is still valid for f(x).

 Tags extrema, logsqrtxqwtx1

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post maluita659 Calculus 1 February 21st, 2014 01:02 PM mathhelp123 Calculus 5 December 3rd, 2009 06:57 PM nikkor180 Real Analysis 3 June 25th, 2009 04:18 AM ArmiAldi Calculus 1 March 19th, 2008 02:46 AM mia6 Calculus 1 October 21st, 2007 06:55 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top