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Old 07-12-2014, 07:04 PM   #1 (permalink)
Those Pump Guyz
 
Join Date: Dec 2008
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who has some calculus chops?

There are some smart cookies on this board, so here goes!
I need to find the relative extrema using the second derivative test if applicable. I have solved to the second derivative. I'm trying to find the possible inflection points. Setting it equal to zero gives X = sqrt-1, so that doesn't exist since we aren't dealing with imaginary numbers. Leaving me to think, no inflection points. Next I tried setting the 1st derivative equal to 0 to find the critical numbers. Here is where I am truly confused. I think X = +/- 1 are the critical numbers. The book's solution in the back says only x=1 is a critical number. Why? Also, I have already tried Wolfram Alpha, it was no help. According the book, all of math up to the critical number is correct. They don't show how to solve for the critical number.
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Old 07-12-2014, 07:35 PM   #2 (permalink)
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I dont remember, but it might have to do with positive squares.

This may be relevant.
Square Root = Always Positive? : GMAT Quantitative Section
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Old 07-12-2014, 09:06 PM   #3 (permalink)
Those Pump Guyz
 
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thanks for looking, got the answer

in case you're curious, x=-1 is not an answer because you can't take a log of a negative number
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Old 07-17-2014, 10:40 AM   #4 (permalink)
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Complex logarithms will come about, and they are easy to deal with.
e.g. ln(-1)=i*pi which is just another way of writing Euler's identity. You can use this to compute any complex log; i.e. ln(-15) = ln(15) + i*pi

Plugging that back in leaves you with +1/2 as a real result of the equation, with the imaginary component of -pi; +1/2 is the same real result as x=+1, but there you only have the real solution.
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Last edited by Siress; 07-17-2014 at 10:43 AM.
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Old 07-19-2014, 10:54 PM   #5 (permalink)
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Quote:
Originally Posted by Siress View Post
Complex logarithms will come about, and they are easy to deal with.
e.g. ln(-1)=i*pi which is just another way of writing Euler's identity. You can use this to compute any complex log; i.e. ln(-15) = ln(15) + i*pi

Plugging that back in leaves you with +1/2 as a real result of the equation, with the imaginary component of -pi; +1/2 is the same real result as x=+1, but there you only have the real solution.
So yeah... you know a good bit more math than I do, give me another year and I may have learned that. I can see some of the logic to it (not getting why pi ended up in there) but, we haven't learned anything like that yet.
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