# Functions

• Feb 21st 2008, 03:20 PM
Blue Griffin
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!
• Feb 21st 2008, 03:45 PM
xifentoozlerix
$f'(x) = f(x) - g(x)$; $g'(x) = g(x) - f(x)=-f'(x)$

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

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

$f'(0)=4 \implies f(0)=4x+C_1$ 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 $x$.
• Feb 21st 2008, 04:32 PM
Blue Griffin
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
• Feb 21st 2008, 05:37 PM
xifentoozlerix
Quote:

Originally Posted by Blue Griffin
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.
Quote:

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 $6$.