*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**

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

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.

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edited Jun 12 "20 at 10:38

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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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