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3 answers

f(x)=x^3(x+3)^2
f'(x)=x^3*2(x+3)+(x+3)^2*3x^2
=x^2(x+3)[2x+(3(x+3)]
=x^2(x+3)(5x+9)
now the critical numbers are x=0,-3,-9/5

2006-08-20 01:38:38 · answer #1 · answered by raj 7 · 0 0

x=0
x=-3
x=-9/5

2006-08-20 02:06:37 · answer #2 · answered by Anonymous · 0 0

zzzzzzzzzzzz........... i'm gonna go answer some beauty & style questions

2006-08-20 04:56:10 · answer #3 · answered by i'm bored 1 · 0 0

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