Prove that $f(x) + g(x) = 6$ Calculus theory problem I'm belaboring to solve... Tried it for an hour only spinning my wheels. Any hints/nudges in the right direction would be greatly appreciated. Here's the problem:
$f$ and $g$ are functions that are differentiable for all real x. They have the following properties:
i) $f'(x) = f(x) - g(x)$
ii) $g'(x) = g(x) - f(x)$
iii) $f(0) = 5$
iv) $g(0) = 1$
Prove that $f(x) + g(x) = 6$ for all $x$.
Good luck.
 A: Now that you've gotten the answer, here is a little bit of a post mortem clarifying the essential points, as may come up again in future problems.  (In general, once you've solved a problem, it's a really good idea to ask for the "moral(s)".  If you can't identify one, how are you better off now than before you solved it?)
In this case, I think the most important single step is the recognition that the conclusion
"$f(x) + g(x) = 6$ for all $x$" 
can be more usefully rephrased as
"$f(x) + g(x)$ is a constant function, with constant value $6$."
This is much more suggestive, because a function $f$ defined on an interval is constant if and only if its dervative is constantly equal to zero.  (Stationarity is equivalent to identically zero velocity.)  Note that half of this statement is obvious, but the other half is not: it is consequence of the Mean Value Theorem that should be emphasized both in the text and by your instructor.   
Thus you are clued in to the fact that the main thing you want to show is that $(f+g)' = f' + g' \equiv 0$ (i.e., constantly zero).  Since a constant function is determined by plugging in any point, once you know that $f+g$ is constant, seeing that it's constantly equal to SOMETHING -- in this case $6$ -- shouldn't be a problem.  So now you look and see how to get from what you're given to the conclusion that $f' + g' \equiv 0$, and you see that you're being given an expression for $f'$ and $g'$ separately.  So certainly you want to add them together and hope to get $0$; in this case, that hope is immediately fulfilled.
A: As a follow up to Dario... If we let $s(x) := f(x) + g(x)$, $s'(x) = f'(x)+g'(x)$, which by (i) and (ii) is equal to $f(x)-g(x)+g(x)-f(x) = 0$. Since $s(0) = 6$ and $s'(x) = 0$, $s(x)$ is constant; $s(x) = 6$ for all $x$, therefore $f(x)+g(x) = 6$.
A: Let's introduce their sum function 
$$s(x) := f(x) + g(x)$$
The statement to be proven thus becomes $s(x) = 6$. 
We know 
$$s(0) = 5 + 1 = 6$$
So now what derivative $s'$ do we need such that $s$ does never change it's constant value? 
A: HINT 
$$ \frac{d}{dx} \bigl( f(x) + g(x) \bigr) = ? $$
A: This problem should help you to understand and practice with two main facts about derivatives:


*

*Derivative is a linear function in fact if $f$ and $g$ are differentiable on $\mathbb{R}$ and $\alpha , \beta \in \mathbb{R}$ then $(\alpha f + \beta g)' = \alpha f' + \beta g'$

*Derivative is a rate of change this is more subtle, the essential fact is that if $f$ is differentiable on $\mathbb{R}$ and $f'(x) = 0$ for all $x \in \mathbb{R}$ then the rate of change of $f$ is null on all $\mathbb{R}$ and then the value of $f$ does not change on all $\mathbb{R}$ (pay attention, this is not always true if you break the domain of $f$ into pieces). This is a consequence of a more general and important result, the mean value theorem.
Observe that i) and ii) in the question give you global information on $f$ and $g$ (i. e. they refer to all $x \in \mathbb{R}$), while iii) and iv) give you local information (in the point $x = 0$). Your question is about a global propriety of the function $f+g$, so you should try to combine local and global information to reach the solution.
