Posted by casidhe on Friday, April 20, 2012 at 8:40pm.
let's take the derivative
f'(x) = 3x^2 (e^(9x) ) + x^3 (9) e^(9x)
= 3x^2 e^(9x) [ 1 + 3x]
= 0 for a max/min
x = 0 , x = -1/3
f(0) = 0
f(-1/3) = (-1/27) e^-3 = -1/(27e^3) = appr -.00184
f(-2) = -8 e^-18 = -8/e^18 = appr -.0000000121
so the absolute minimum is -8/e^18 , when x = -2
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