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 Calculus Calculus Math Forum

 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: 7,689 Thanks: 2669 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!  Tags euler Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post uniquesailor Real Analysis 6 January 30th, 2012 04:38 AM FalkirkMathFan Calculus 1 November 5th, 2011 12:57 AM FalkirkMathFan Real Analysis 0 November 4th, 2011 04:08 AM FalkirkMathFan Calculus 0 November 3rd, 2011 04:52 PM brangelito Number Theory 18 August 9th, 2010 11:58 PM

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