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 February 11th, 2016, 11:57 AM #1 Member   Joined: Jan 2016 From: / Posts: 43 Thanks: 1 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.
 February 11th, 2016, 12:53 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra $$\mathrm e ^{(1-\ln x)\ln x}=\left(\mathrm e ^{\ln x}\right)^{(1-\ln x)}=x^{(1-\ln x)}$$ Thanks from SuperNova1250
 February 11th, 2016, 01:10 PM #3 Member   Joined: Jan 2016 From: / Posts: 43 Thanks: 1 Now tell me how am I supposed to see that! And it's so simple, but I always somehow miss these things...
February 11th, 2016, 01:17 PM   #4
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Quote:
 Originally Posted by SuperNova1250 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.

 February 11th, 2016, 01:27 PM #5 Member   Joined: Jan 2016 From: / Posts: 43 Thanks: 1 Sure, I know that! Just wrote it wrong, but all this in the exponent is the right form so : e^((1-lnx)*lnx)
 February 11th, 2016, 01:42 PM #6 Banned Camp   Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191 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?
 February 12th, 2016, 08:54 AM #7 Member   Joined: Jan 2016 From: / Posts: 43 Thanks: 1 Hopefully! thank you!

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