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January 30, 2015

January 30, 2015

Posted by **H** on Sunday, December 25, 2011 at 4:02pm.

(a) -9

(b) - 6

(c) -3

(d) 6

(e) 9

- calculus -
**drwls**, Sunday, December 25, 2011 at 7:52pmdf/dx = 1 - c/x^2 = 0

If the minimum is at x = 3, then

c/x^2 = c/9 = 1

c = 9

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