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Math Help - Functions

  1. #1
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    Functions

    OK, the problem i have here i can't even begin to solve, so im in need of some help. Here's the exact problem:

    If the following conditions are true
    1. f '(x) = f(x) - g(x) [the apostrophe means the derivitive of f of x]
    2. g '(x) = g(x) - f(x)
    3. f(0) = 5
    4. g(0) = 1
    5. f and g are continuous are differentiable functions

    Find the constant value of f(x) + g(x) for all x. Explain your work and your steps.


    There it is. I don't get it at all, but my math teacher thinks i should for whatever reason. Please help! Thanks in advance!
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  2. #2
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    f'(x) = f(x) - g(x); g'(x) = g(x) - f(x)=-f'(x)

    f(x) = f'(x) + g(x); f(0)=5

    g(x) = g'(x) + f(x); g(0)=1

    f'(0)=4 \implies f(0)=4x+C_1 and g'(0)=-4 \implies g(0)=-4x+C_2

    f(0)=5 \implies C_1=5 and g(0)=1 \implies C_2=1

    then we have f(x)=4x+5 and g(x)=-4x+1. So the constant value of f(x)+g(x)=4x+5+(-4x+1)=6 You can use the two equations to check that all of the given equations hold for all x.
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  3. #3
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    Thanks a lot, i get most of what you said. But can you explain how you came up with the second part:

    f(x)= f '(x) + g(x)

    and how you used that to find out that f '(x) = 4

    thats the one part i still dont get
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  4. #4
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    Quote Originally Posted by Blue Griffin View Post
    Thanks a lot, i get most of what you said. But can you explain how you came up with the second part:

    f(x)= f '(x) + g(x)
    I got this from the first equation you gave me f'(x) = f(x) - g(x). All I did was add g(x) on both sides. I did the same thing for the second equation you gave me to get g'(x). It is actually not even necessary to re-write it this way.
    and how you used that to find out that f '(x) = 4

    thats the one part i still dont get
    how about i show you how to do it with JUST the 4 equations you gave me without any reordering. you are given:

    f'(x) = f(x) - g(x)

    g'(x) = g(x) - f(x)

    f(0) = 5

    g(0) = 1

    Using this, you can solve for f'(0) and g'(0):

    f'(0) = f(0) - g(0) = 5 - 1 = 4 and likewise g'(0) = g(0) - f(0) = 1 - 5 = -4

    Now repeating my earlier argument, f'(0)=4 \implies f(0)=4x+C_1 and likewise g'(0)=-4 \implies g(0)=-4x+C_2

    Now, f(0)=5=4(0)+C_1 \implies C_1=5 and likewise g(0)=1=-4(0)+C_2 \implies C_2=1

    So then f(x)=4x+5 and g(x)=-4x+1. Adding these two together yields f(x)+g(x)=4x+5+(-4x+1)=4x+5-4x+1=6. The constant you are looking for is 6.
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