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September 17th, 2012, 09:33 AM   #1
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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
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September 17th, 2012, 09:45 AM   #2
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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.
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September 17th, 2012, 09:56 AM   #3
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Re: Critical point

We are given:



To find the critical point, differentiate and equate to zero:







To approximate this root, we could use Newton's method:



Using an initial guess of , we find after many iterations:

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September 17th, 2012, 10:51 AM   #4
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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?
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September 17th, 2012, 10:58 AM   #5
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Re: Critical point

Then you would have:

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September 17th, 2012, 11:05 AM   #6
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Re: Critical point

Wow you guys are brilliant, thanks!
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