derivative of xe^(-3x)

I got e^(-3x)-3e^(-3x)*x

how do I figure out the critical points from that^

Question

derivative of xe^(-3x)

I got e^(-3x)-3e^(-3x)*x

how do I figure out the critical points from that^

abb0t · Answer

take the derivative.
it's said that 
f'(c) = 0 OR f'(c) D.N.E 
then is a critical pt.

abb0t · Answer

That means, take the derivative of the function and set it equal to zero and solve.

anonymous · Answer

$$e^{-3x}-3e^{-3x}x=0$$
how in the world do you solve that^

abb0t · Answer

You can factor out your exponential term since that's a common factor firstly.
$$e^{-3x}(1-3x) = 0$$
notice you have two functions which you can set equal to zero

anonymous · Answer

I see the common term there.. but how does -3x turn into 1-3x?

abb0t · Answer

I bascially just factored out the exponential term. If you factor ONLY the exponential term, you're left with (1-3x) I left the 3 inside. If you take out the 3 along with the exponent, you'd then have
$$3e^{-3x}(\frac{ 1 }{ 3 }-x) = 0$$
and personally, I try to avoid fractions. But you can do that, too.