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 September 17th, 2012, 09:33 AM #1 Newbie   Joined: Sep 2012 Posts: 6 Thanks: 0 Critical point What is the value of the x-coordinate of the critical point of f to 2 decimal places? f(x) = ln(x) + x^-150 Any ideas guys? Thanks
 September 17th, 2012, 09:45 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,888 Thanks: 1836 The value is 1.03, but you would need to use calculus to find it, so you've posted the question in the wrong subforum.
 September 17th, 2012, 09:56 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: Critical point We are given: $f(x)=\ln(x)+x^{-150}$ To find the critical point, differentiate and equate to zero: $f'(x)=\frac{1}{x}-150x^{-151}=0$ $x^{150}-150=0$ $x=150^{\small{\frac{1}{150}}}$ To approximate this root, we could use Newton's method: $x_{n+1}=x_n-\frac{x_n^{150}-150}{150x_n^{149}}=\frac{149x_n^{150}+150}{150x_n^ {149}}$ Using an initial guess of $x_0=1$, we find after many iterations: $x_n\approx1.03$
 September 17th, 2012, 10:51 AM #4 Newbie   Joined: Sep 2012 Posts: 6 Thanks: 0 Re: Critical point Thanks skipjack and MarkFL, I've made a typing error... it's supposed to be -140 instead -150. Any ideas what the answer will be?
 September 17th, 2012, 10:58 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: Critical point Then you would have: $x=140^{\small{\frac{1}{140}}}\approx1.04$
 September 17th, 2012, 11:05 AM #6 Newbie   Joined: Sep 2012 Posts: 6 Thanks: 0 Re: Critical point Wow you guys are brilliant, thanks!

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