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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?

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  1. anonymous
    • 5 years ago
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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?

  2. anonymous
    • 5 years ago
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    back down then the rest of it.

  3. anonymous
    • 5 years ago
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    let me look at some more, I'll come back in a few minutes

  4. anonymous
    • 5 years ago
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    okay thankyou!

  5. anonymous
    • 5 years ago
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    just to make sure, we are trying to prove that the second derivative of g(x) has the local max at x= zero?

  6. anonymous
    • 5 years ago
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    we're trying to see if g(x) has a local maximum at x=0 i think,..

  7. anonymous
    • 5 years ago
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    okay, that make a whole lot more sense. brb

  8. anonymous
    • 5 years ago
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    okay!

  9. anonymous
    • 5 years ago
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    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.

  10. anonymous
    • 5 years ago
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    okay ill try and do it & let you know if i get stuck. :)

  11. anonymous
    • 5 years ago
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    I don't know why they gave you that other g'(x), but if g''(x)= wht you worte, then go back from there.

  12. anonymous
    • 5 years ago
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    wrote

  13. anonymous
    • 5 years ago
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    so do the antiderivative of the first derivative to get my g(x)?

  14. anonymous
    • 5 years ago
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    right

  15. anonymous
    • 5 years ago
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    okay. ima finish working it.

  16. anonymous
    • 5 years ago
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    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.

  17. anonymous
    • 5 years ago
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    okay and how do i do that using only the g'?

  18. anonymous
    • 5 years ago
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    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.

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