ok so I did all the revision problems and noted the ones I couldn't do today and Im posting them together, hope thats not a problem with the power that be?
I have exhibit A:
$e^{-x} -x + 2 $
So I differentiate to find where the derivative hits 0:
$-e^{-x} -1 = 0 $
Now HOW do I figure when this hits zero!?
$-1 = e^{-x} $
$\ln(-1) = \ln(e^{-x})$ ???
More to come ... as one day rests between me and my final exam/attempt at math!