if $f(x)$ is even , $g(x)$ is odd, $h(x)$ is not even and not odd.
Is it possible that $f(x) = g(x) + h(x)$ ?
I tried playing with the functions, changing variables and even more... but I don't know how to continue.
Asked
Active
Viewed 49 times
0
-
Does uneven function mean odd function, that is $-f(x)=f(-x)$? – Justin Benfield Jun 12 '16 at 19:04
-
or does it mean not even? – Olivier Oloa Jun 12 '16 at 19:05
-
Does "uneven" mean "odd"? If so, then any function $h(x)$ can be written as the sum of an even and an odd function so, yes. – lulu Jun 12 '16 at 19:05
-
Edited! I meant odd – Erica J. Jun 12 '16 at 19:06
-
Hint: Subtract $g(x)$ from both sides, and note that if $g(x)$ is odd, so is $-g(x)$. What can you say now? – Justin Benfield Jun 12 '16 at 19:08
-
See this question: every real-valued function on $\Bbb R$ is the sum of an even and an odd function. – Brian M. Scott Jun 12 '16 at 19:17
1 Answers
3
For every function $h$ it is the case that $$ h(x) = \underbrace{\tfrac12h(x)+\tfrac12h(-x)}_{f(x)} + \underbrace{\tfrac12h(x)-\tfrac12h(-x)}_{g(x)} $$ where the terms marked $f(x)$ constitute an even unction of $x$ and those marked $g(x)$ an odd one.
So you can get your desired situation by choosing an arbitrary neither-odd-nor-even function $h$ and then define $f$ and $g$ as above (and then negate $g$ if you want the precise equation in your question).
hmakholm left over Monica
- 286,031