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February 11th, 2016, 11:57 AM   #1
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Euler again!

Ok here's sort of a problem, actually I am asking this:

I have this function:

e^(1-lnx)*lnx

And have to graph it, graphic calculator not allowed!

So first I multiply this, use logarithm properties and I get this:

x * e^(-lnx)^2

Now I know the procedure, with extrema and so on, but this function looks really hard to do so like 3-5 minutes later I finally derive it, get e, 0 and -1 as critical points, leave out -1 as ln is not defined there anyway.

Then I test the critical point, something should be at the point (1,1).

I take the second derivative again 10 minutes later this time and I get 0.

So if it's less than zero it's maxima if it's higher it's minima, and now this looks
like I have to take the third derivative, my question is where is this going?

Am I making this harder than it should be, or the problem really wants me to derive the function for the third time?

What if I get another zero?

Last edited by skipjack; February 11th, 2016 at 07:54 PM.
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February 11th, 2016, 12:53 PM   #2
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$$\mathrm e ^{(1-\ln x)\ln x}=\left(\mathrm e ^{\ln x}\right)^{(1-\ln x)}=x^{(1-\ln x)}$$
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February 11th, 2016, 01:10 PM   #3
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Now tell me how am I supposed to see that!

And it's so simple, but I always somehow miss these things...
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February 11th, 2016, 01:17 PM   #4
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Quote:
Originally Posted by SuperNova1250 View Post
Ok here's sort of a problem, actually I am asking this:

I have this function:

e^(1-lnx)*lnx

And have to graph it, graphic calculator not allowed!

So first I multiply this, use logarithm properties and I get this:

x * e^(-lnx)^2 No, that isn't.
e^(1 - lnx)*lnx $\displaystyle \ \ne \ e ^{(1-\ln x)\ln x} $


e^(1 - lnx)*lnx = $\displaystyle \ e^{1 - \ln x}*\ln x \ \ \ or \ \ \ \ln x*e^{1 - \ln x} \ \ $ by the Order of Operations.


If it were e^[(1 - lnx)*lnx], then that would equal $$ \ \mathrm e ^{(1-\ln x)\ln x}=\left(\mathrm e ^{\ln x}\right)^{(1-\ln x)}=x^{(1-\ln x)}$$


.

Last edited by skipjack; February 11th, 2016 at 07:58 PM.
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February 11th, 2016, 01:27 PM   #5
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Sure, I know that!

Just wrote it wrong, but all this in the exponent is the right form so :

e^((1-lnx)*lnx)
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February 11th, 2016, 01:42 PM   #6
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Let y = x^(1 - lnx)

lny = ln[x^(1 - lnx)]

lny = (1 - lnx)*lnx

Take derivatives:

(1/y)(y') = (-1/x)(lnx) + (1/x)(1 - lnx)

(1/y)(y') = (1/x)(-lnx + 1 - lnx)

(1/y)(y') = (1/x)(1 - 2lnx)

Multiply each side by what y equals:

y' = [x^(1 - lnx)]*[(1/x)(1 - 2lnx)]


Can you continue?
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February 12th, 2016, 08:54 AM   #7
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Hopefully! thank you!
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