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is that e to the (-2x )then back down, times the rest of that, or is it all raised from e?
back down then the rest of it.
let me look at some more, I'll come back in a few minutes
just to make sure, we are trying to prove that the second derivative of g(x) has the local max at x= zero?
we're trying to see if g(x) has a local maximum at x=0 i think,..
okay, that make a whole lot more sense. brb
so, take that second derivative equation and sub in 0 for x, and all of the f(0) values. this should give you g''(0)= -15, because e^0 = 1. Whne you integrate g''(0)=15, g'(x) = 15x, @ 0 = 0, therefore you have a critical point. so sub in 1 and -1 on a number line for your first derivative and you can see that the slopes change from + to - at 0, therefore there is a local max.
okay ill try and do it & let you know if i get stuck. :)
I don't know why they gave you that other g'(x), but if g''(x)= wht you worte, then go back from there.
so do the antiderivative of the first derivative to get my g(x)?
okay. ima finish working it.
now it looks like the second deriv and the first derivs you're given are different, but if you find the first antideriv of the second, then the antideriv of that, it would give you g(x) from g''(x). It looks like you just have to prove that g(x) has a max, so you only need g'(x) to do it.
okay and how do i do that using only the g'?
only the g'(x) that they gave you doesn't work out. It only works using the g''(x) from what I could see. The g'(x) is = 12 at 0, and therefore could not be a critical value. It looks like you have to get g'(x) from g''(x), unless the g'(x) you wrote is different from what you have in the book or what not.