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April 18, 2014

April 18, 2014

Posted by **Anonymous** on Wednesday, February 18, 2009 at 11:24am.

2. Find the coordinates of the turning point on the curve y=2e^3x+8e^-3x and determine the nature of this turning point.

For 1, I have sketched the curves, but i'm not sure how to find the length. I thought I could do y=y to find the x co-ordinates (and use those as my limits) for each but I don't know how to find them out with e^x.

For 2, I differentiated it to get y=6e^3x-24e^-3x and don't know what to do after that.

Please help. Thanks in advance for any help!

- Maths -
**bobpursley**, Wednesday, February 18, 2009 at 12:25pm2. find each x by

4=e^x

ln 4= x so the point is ln4,4

and the other point is

ln2=-2x

x=-ln4/2 and the boint B is -1/2 ln4,4

Use the distance formula to find the length.

- MATH -
**drwls**, Wednesday, February 18, 2009 at 12:27pm1. Since the y coordinates of A and B are the same, the distance between the points will be the difference in the X values of those two points.

2. The derivative is

y' = 6e^3x -24e^-3x

When that equals zero,

e^3x = 4e^-3x

e^6x = 4

6x = ln 4 = 1.3863

x = 0.23105 is an extreme point of the function. The value of y there is

2 e^0.693 + 8 e^-0.693 = 2*2 + 8/(1/2) = 20

Take the second derivative of y(x) to see if (0.23105,20) is a minumum or maximum. It is going to be positive. You also see that, since y becomes infinite as x gets much larger, it must be a minimum.

y''(x) =

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