The sum of these functions equals a constant

[kent davidge]

If I have a sum $f(x) + g(x) = c$, with $c$ a constant, does this imply that both $f(x)$ and $g(x)$ are also constants?
If I just solve this equation for $x$, I will find some values of $x$ which satisfy the equation. However, if I require that the equation be true for all $x$, there is no way other than the functions being equal to constants. Am I right?

 

[Mark44]

If I have a sum $f(x) + g(x) = c$, with $c$ a constant, does this imply that both $f(x)$ and $g(x)$ are also constants?

No, not at all. Consider $f(x) = \cos x$ and $g(x) = -\cos x + 1$. Neither function is constant, but their sum is always equal to 1.

If I just solve this equation for $x$, I will find some values of $x$ which satisfy the equation. However, if I require that the equation be true for all $x$, there is no way other than the functions being equal to constants. Am I right?

No. With my example functions, you can't solve the equation $f(x) + g(x) = 1$ for x. My equation is true for all real x.

Also, it depends on whether the equation you're working with is true only for a certain number of values of x, or is an identity, one that is true for all values in the domains of the two functions.

Can you be more specific about what you're trying to do?

 

[kent davidge]

Can you be more specific about what you're trying to do?

I just have two seemgly arbritary functions of $x$ whose sum is zero, and wanted to know if that implies that the functions themselves are equal to zero, but now with your answer I see that it doesn't.

 

[mfb]

If I just solve this equation for x, I will find some values of x which satisfy the equation.

All x should satisfy the equation if that equation is a general relation ("c is a constant" implies that).

 

[Svein]

I just have two seemgly arbritary functions of $x$ whose sum is zero, and wanted to know if that implies that the functions themselves are equal to zero, but now with your answer I see that it doesn't.

If $f(x) + g(x) = 0$ then $g(x) = -f(x)$.