# 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
\$\displaystyle f'(x) = f(x) - g(x)\$; \$\displaystyle g'(x) = g(x) - f(x)=-f'(x)\$

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

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

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

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

then we have \$\displaystyle f(x)=4x+5\$ and \$\displaystyle g(x)=-4x+1\$. So the constant value of \$\displaystyle 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 \$\displaystyle 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 \$\displaystyle f'(x) = f(x) - g(x)\$. All I did was add \$\displaystyle g(x)\$ on both sides. I did the same thing for the second equation you gave me to get \$\displaystyle 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:

\$\displaystyle f'(x) = f(x) - g(x)\$

\$\displaystyle g'(x) = g(x) - f(x)\$

\$\displaystyle f(0) = 5\$

\$\displaystyle g(0) = 1\$

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

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

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

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

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