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