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anonymous
 5 years ago
Suppose that the function f has a continuous second derivative for all x, and that f(0)=2, f'(0)=3, and f"(0) =0. Let g be a function whose derivative is given by g'(x)=e^2x(3f(x)+2f'(x)) for all x.
Show that g"(x)=e^2x(6f(x)f'(x)+2f"(x)) Does g have a local maximum at x=0?
anonymous
 5 years ago
Suppose that the function f has a continuous second derivative for all x, and that f(0)=2, f'(0)=3, and f"(0) =0. Let g be a function whose derivative is given by g'(x)=e^2x(3f(x)+2f'(x)) for all x. Show that g"(x)=e^2x(6f(x)f'(x)+2f"(x)) Does g have a local maximum at x=0?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that e to the (2x )then back down, times the rest of that, or is it all raised from e?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0back down then the rest of it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me look at some more, I'll come back in a few minutes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just to make sure, we are trying to prove that the second derivative of g(x) has the local max at x= zero?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we're trying to see if g(x) has a local maximum at x=0 i think,..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, that make a whole lot more sense. brb

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay ill try and do it & let you know if i get stuck. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know why they gave you that other g'(x), but if g''(x)= wht you worte, then go back from there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so do the antiderivative of the first derivative to get my g(x)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. ima finish working it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay and how do i do that using only the g'?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0only 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.
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