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Where is the first derivative of y = 3xe−x equal to 0?
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okay so iv got the derivative (-3e^(-x))(x-1)
So they gave you \[\Large y = 3xe^{-x}\] right?
yes
I'm getting \[\Large \frac{dy}{dx} = -3e^{-x}(x-1)\] as well
now set dy/dx equal to 0 and solve for x
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yea well thats what i cant do
\[\Large \frac{dy}{dx} = -3e^{-x}(x-1)\] \[\Large 0 = -3e^{-x}(x-1)\] \[\Large -3e^{-x} = 0 \text{ or } x-1 = 0\]
I'm using the idea A*B = 0 means either A = 0 or B = 0
Does that make sense? @Jdosio
okay sorry just fell asleep in my desk yes iv got it thank you verry much:D
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