Others have mentioned the $f(x)=0$ is one example. In fact, we can prove that it is the only instance of a duty from $jajalger2018.orgbbR o jajalger2018.orgbbR$ (i.e a function which takes in actual values and outputs real values) the is both odd and even. Suppose $f(x)$ is any function which is both odd and even. Then $f(-x) = -f(x)$ by odd-ness, and $f(-x)=f(x)$ by even-ness. Hence $-f(x) = f(x)$, so $f(x)=0.$



Yes. The continuous function $f(x) = 0$ satisfies both conditions.

You are watching: What number is both even and odd

Even:$$f(-x) = 0 = f(x)$$

Odd:$$f(-x) = 0 = -f(x)$$

Furthermore, it"s the only real function that satisfies both conditions:

$$f(-x) = f(x) = -f(x) Rightarrow 2f(x) = 0 Rightarrow f(x) = 0$$


Hint $ m f:$ is even and odd $ miff f(x) = f(-x) = -f(x):Rightarrow: 2,f(x) = 0.:$ This is true if $ m:f = 0,:$ but may additionally have other solutions, e.g. $ m:f = n:$ in $ m:jajalger2018.orgbb Z/2n =:$ integers mode $ m 2n,$ whereby $ m: -n equiv n.$


Suppose $f$ odd one even. Let $x in D$ ( D is set meaning of $f$) climate you have actually : $ f(x)=f(-x)=-f(x)$. What can you conclude around $f$ ?


As other world have pointed out already, the real role $f(x)$ which maps every actual number come zero (i.e.$f(x) = 0 space forall x in jajalger2018.orgbbR$) is both even and also odd because $$f(x) - f(-x) = 0 space space , f(x)+f(-x) = 0space forall x in jajalger2018.orgbbR .$$ additionally it is the only role defined end $jajalger2018.orgbbR$to own this property.

Let $R$ it is in a Boolean ring and also $X$ be an arbitrarily set. Then every duty $f:R ightarrow X$ is both even and odd.

I write-up this as an addendum come the awesome answers currently present for this Q.

Most trivial example:

For $f(x)=0$ us have:

$f(x)= 0 =-f(-x)$ Hence, Odd

$f(x)= 0 =f(-x)$ Hence, Even

I was looking for an answer to my question.

$f(x) = sin(x) + sin(pi + x)$

Here, the role is additionally even and also odd at the very same time (another instance or depiction of the same). This is because of the value causing zero over the whole domain. So, it have the right to be concluded that all the features that have actually their $Range=$ have to be both even and also odd in ~ the exact same time regardless of their notation is the a constant function($f(x)=0$) or not.

Another example,

$f:1,-1\to jajalger2018.orgbbR$

$f(x) = x^2-1$

$f(x) = 0$ because that all values in the domain. So, it is both even and also odd in ~ the exact same time because while deciding also odd functions the domain come which the function is restricted by definition must be considered.

edited Jun 12 "20 at 10:38
reply Oct 25 "15 at 18:56
Aditya GuruAditya Guru
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We could additionally think even and odd functions as the following to acquire $f(x) = 0$ is a function that is both odd and even and go for some various functions:

Odd functions have actually graphs that room symmetric through respect come origin.

Even functions have actually graphs that are symmetric v respect to $y$-axis.

So, graph the $f(x) = 0$ satisfies both conditions, thus it is both odd and even. Furthermore, us can specify some piecewise functions that satisfies this condition using your graphs. Because that example, allow $f(x)$ not defined on intervals $(-1,-2)$ and also $(1,2)$; and whenever it is defined, $f(x) = 0$. Climate $f(x)$ is, again, both even and odd. And also as long as we define a piecewise duty by remove intervals that room symmetric v respect come $y$-axis, and also rest that the duty $= 0$ also though they room not continuous.

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answer Sep 2 "18 at 19:21
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