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July 29, 2014

July 29, 2014

Posted by **victoria** on Monday, February 20, 2012 at 9:23am.

a)-9.004

b)-4.734

c)1.029

d)1.277

e)4.797

- calculus -
**Steve**, Monday, February 20, 2012 at 11:52amjust differentiate and solve. What's the trouble?

f = 3x^3*e^-x

f' = (9x^2 - 3x^3)e^-x

= 3x^2(3-x)e^-x

Now, x^2 >= 0 and e^-x >= 0

so the only way we can have f' < 0 is x>3

That means choice (e)

Check: 3*4.797^2*(3-4.797)e^-4.797 = -1.024

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