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Calc by Fry & Leela - Sun, 16 Jul 2017 12:43:38 EST ID:Rhgh4/nK No.15536 Ignore Report Quick Reply
File: 1500223418101.jpg -(12838B / 12.54KB, 398x66) Thumbnail displayed, click image for full size. 12838
Can anyone tell me where to begin with this problem? I'm clueless. Our professor didn't cover it. I imagine start by taking d/dx and plugging the values in at some point?
Thanks
>>
Cyril Clugglefoot - Mon, 17 Jul 2017 18:43:04 EST ID:tgwdoW8d No.15537 Ignore Report Quick Reply
So... when the derivative crosses the y-axis, that means that your function, x2x-ex, there has changed directions. So that means that ANY time your d/dx crosses the y-axis, it creates a "local" maximum value, which may not be the overall actual max peak.

I believe you:
  • find the derivative
  • find the zeros (what's that process called again?), i.e. solve for zero
  • plug in each x value you found into the original function, x2e-x -> compare all y values to find the greatest


Correct me if I'm wrong, it's been 3 years.
>>
Fuck Murdridge - Tue, 18 Jul 2017 16:12:50 EST ID:XBxBdy5C No.15538 Ignore Report Quick Reply
>>15537

Pretty much this, but check the values at the endpoints of the interval too not just at local Mac/mins.
>>
Martha Hellyford - Wed, 19 Jul 2017 00:10:34 EST ID:cezjQDuj No.15540 Ignore Report Quick Reply
>>15537
>So that means that ANY time your d/dx crosses the y-axis, it creates a "local" maximum value
Oops, I meant
>creates a "local" max OR min value, i.e. just a peak/trough somewhere
It's only a max if the derivative's slope is negative around that zero
>>
Jack Bimmleshit - Sat, 11 Nov 2017 17:51:42 EST ID:y7H0hQp+ No.15583 Ignore Report Quick Reply
>>15540
or, an inflection point if f''(x) = 0
>>
Jack Ciblinghood - Sat, 18 Nov 2017 17:25:36 EST ID:BDm+BNlx No.15586 Ignore Report Quick Reply
>>15583

The concavity has to change for it to be an inflection point. If it's concave up, then a place where f''(x)=0, then goes back to being concave up, it's not an inflection point.


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