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 Post subject: CalculusPosted: Fri, 19 Nov 2010 15:31:19 UTC
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I heard this problem from a colleague recently, and I thought her solution was sufficiently clever to merit a calculus-level proposed problem. If you're forcing the answer, or using a calculator/graphing device, you're trying too hard.

The problem:

Find the minimum value of:

.

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 Post subject: Posted: Mon, 22 Nov 2010 11:32:58 UTC
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 Post subject: Re: CalculusPosted: Tue, 23 Nov 2010 04:17:10 UTC
 S.O.S. Oldtimer

Joined: Fri, 10 Oct 2008 15:35:23 UTC
Posts: 179
Location: Clarksville, ARkansas
I heard this problem from a colleague recently, and I thought her solution was sufficiently clever to merit a calculus-level proposed problem. If you're forcing the answer, or using a calculator/graphing device, you're trying too hard.

The problem:

Find the minimum value of:

.

man111,

Please don't post mathematics via images. Edit: Sorry, Shadow said it's OK.

I like your work down to the line before equation 2.

I would then find the extrema of , and apply the result to . Since has no real zeros, the absolute value will not introduce additional critical points.

. is not in the domain of .

f(t) has relative extrema at .

.

.

.

.

Last edited by Sam Snyder on Tue, 23 Nov 2010 04:43:12 UTC, edited 2 times in total.

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 Post subject: Re: CalculusPosted: Tue, 23 Nov 2010 04:21:32 UTC
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Location: Austin, TX
Sam Snyder wrote:
I heard this problem from a colleague recently, and I thought her solution was sufficiently clever to merit a calculus-level proposed problem. If you're forcing the answer, or using a calculator/graphing device, you're trying too hard.

The problem:

Find the minimum value of:

.

man111,

Please don't post mathematics via images.

You final answer looks correct to me.

I like your work down to the line before equation 2.

I would then find the extrema of , and apply the result to . Since has no real zeros, the absolute value will not introduce additional critical points

. is not in the domain of .

f(t) has relative extrema at .

.

.

.

.

He can post however he feels like, there's no reason he shouldn't.

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.

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 Post subject: Re: CalculusPosted: Tue, 23 Nov 2010 04:37:26 UTC
 S.O.S. Oldtimer

Joined: Fri, 10 Oct 2008 15:35:23 UTC
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Location: Clarksville, ARkansas

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.

But the absolute value can introduce new critical points at zeros of a real function.

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 Post subject: Re: CalculusPosted: Tue, 23 Nov 2010 04:47:43 UTC
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Sam Snyder wrote:

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.

But the absolute value can introduce new critical points at zeros of a real function.

Yes, but if you want to be situation specific, zeros of the function represent the minimum, so "new critical values" would all be moot anyways, just check for zeros of the function. I mean, it's okay to be thorough with what you're doing, but if you're going that far you need to go all the way, otherwise it just sounds like you're looking for a reason to add another post to your total post-count. The most qualitative thing you could have said there was not the lack of zeros making new critical points, but rather if it were zero we would already be done.

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 Post subject: Re: CalculusPosted: Tue, 23 Nov 2010 05:07:33 UTC
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Joined: Fri, 10 Oct 2008 15:35:23 UTC
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Sam Snyder wrote:

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.

But the absolute value can introduce new critical points at zeros of a real function.

Yes, but if you want to be situation specific, zeros of the function represent the minimum, so "new critical values" would all be moot anyways, just check for zeros of the function.

has two critical points in addition to the one has.

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 Post subject: Re: CalculusPosted: Tue, 23 Nov 2010 05:12:46 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
Sam Snyder wrote:
Sam Snyder wrote:

Also, the absolute value won't ever introduce new critical points if you're looking at zeros of the derivative, the only complex number of norm 0 is 0.

But the absolute value can introduce new critical points at zeros of a real function.

Yes, but if you want to be situation specific, zeros of the function represent the minimum, so "new critical values" would all be moot anyways, just check for zeros of the function.

has two critical points in addition to the one has.

See the edit to my prior post.

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