Remedy to $1-f(x) = f(-x)$

Can we locate $f(x)$ considered that $1-f(x) = f(-x)$ for all actual $x$?

I start by repositioning to: $f(-x) + f(x) = 1$. I can locate an instance such as $f(x) = |x|$ that benefits some values of $x$, yet not all. Exists a method below? Is this feasible?

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2019-05-06 02:31:09
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Answers: 6

Clearly, we just have relationships in between f (x) and also f (- x). The relationship suggests that 0 needs to have value 1/2. We can separate all non - absolutely no actual numbers right into disjoint sets of x and also - x and also specify the function f on each set independently. For each and every set, f (x) can be offered any kind of value and afterwards f (- x) has a solitary legitimate value. As stated by Grigory, the legitimate features can be qualified as any kind of weird function plus 1/2.

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2019-05-08 13:41:10
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WolframAlpha gives a remedy to this (and also several various other) reappearance formulas :
http://www.wolframalpha.com/input/?i=1-f(x)+%3D+f(-x)

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2019-05-08 12:57:26
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$$f(x)=\frac{1}{2}+\text{(any odd function)}.$$. As an example, $f(x)=\frac{1}{2}+x$ or, claim, $f(x)=\frac{1}{2}+99x^3+7x^5$.

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2019-05-08 10:31:09
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Let $$f(x) = \begin{cases} 1 \quad x>0, \\ 1/2 \quad x=0, \\ 0 \quad x<0\end{cases}.$$

If $x > 0$, after that $-x < 0$ to make sure that $f(x)+f(-x)=1+0=1$

Likewise with $x<0$.

If $x=0$, after that $f(x)+f(-x)=(1/2)+(1/2)=1$.

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2019-05-08 10:11:14
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Consider rather the features $g$ that please the identification $-g(x)=g(-x)$ for all $x$. If $(x,g(x))$ is a factor of the function $g$, after that $(-x,-g(x))$ is additionally a factor (given that $g(-x)=-g(x)$). Consequently, every function $g$ is symmetrical when revolved by $180$ levels concerning the factor $(0,0)$.

Just how do points transform for the identification $1-f(x)=f(-x)$? We just change the factor of proportion to $(0,1/2)$. Below the factor $(x,f(x))$ indicates the factor $(-x,1-f(x))$.

The function $f$ pleases the identification $1-f(x)=f(-x)$ for all actual numbers $x$ if and also just if it is symmetrical when revolved concerning the factor $(0,1/2)$ by $180$ levels.

There's mosting likely to be most of these features ; several of which will certainly be polynomials, several of which will certainly not.

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2019-05-08 10:06:51
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Usually straightforward troubles similar to this ask you to locate a function that values the problem, not all of them. And Also (once more) generally you start by examining if a straightforward polynomial function of the first level can be a remedy.

So, if

$$f(x) = ax + b$$

Then

$$f(x) + f(-x) = 1 \implies ax + b + a \cdot (-x) + b = 1 \implies 2b = 1 \implies b = 1/2$$

So the problem is pleased by any kind of function of the type :

$$f(x) = ax + 1/2$$

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2019-05-08 10:00:12
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