How do you determine if $f\left( x \right)={{e}^{-{{x}^{2}}}}$ is an even or odd function?
Answer
617.7k+ views
Hint: If a function satisfies the property $f\left( -x \right)=f\left( x \right),$ then this function is an even function. If a function satisfies the property $f\left( -x \right)=-f\left( x \right),$ then this function is an odd function. If there is a function that does not satisfy neither of the above conditions for odd functions and even functions, then such a function is said to be neither an even function nor an odd function.
Complete step by step answer:
Let us take the given function into consideration.
That is, $f\left( x \right)={{e}^{-{{x}^{2}}}}.$
We have to check if the given function is an even function or an odd function.
For that, we apply the properties of even functions and odd functions on the given function.
Let us first apply the property of odd functions so that we can check if the function $f\left( x \right)={{e}^{-{{x}^{2}}}}$ is an odd function.
The property to be verified is $f\left( -x \right)=-f\left( x \right).$
When we apply this property on the given function we get,
$\Rightarrow f\left( -x \right)={{e}^{-{{\left( -x \right)}^{2}}}}$
We have already learnt the rule that says ${{\left( -x \right)}^{2}}={{x}^{2}}.$
And from this we will get the following,
\[\Rightarrow f\left( -x \right)={{e}^{-{{x}^{2}}}}\]
And let us take a look at the definition of the given function which is given as $f\left( x \right)={{e}^{-{{x}^{2}}}}.$
Also, from what we have obtained earlier, we will be directed to the following equation,
$\Rightarrow f\left( -x \right)={{e}^{-{{x}^{2}}}}=f\left( x \right)$
The above equation leads us to what we are asked to find.
That is,
$\Rightarrow f\left( -x \right)=f\left( x \right)$
This is the property of even functions which is used to determine whether a function is even or not.
Since the given function satisfies the property of even functions, this is an even function.
Hence, $f\left( x \right)={{e}^{-{{x}^{2}}}}$ is an even function.
Note:
This does not mean all the exponential functions are even. ${{e}^{x}}$ is a function which is neither even nor odd. Before making any conclusion, one must check for both the properties.
Complete step by step answer:
Let us take the given function into consideration.
That is, $f\left( x \right)={{e}^{-{{x}^{2}}}}.$
We have to check if the given function is an even function or an odd function.
For that, we apply the properties of even functions and odd functions on the given function.
Let us first apply the property of odd functions so that we can check if the function $f\left( x \right)={{e}^{-{{x}^{2}}}}$ is an odd function.
The property to be verified is $f\left( -x \right)=-f\left( x \right).$
When we apply this property on the given function we get,
$\Rightarrow f\left( -x \right)={{e}^{-{{\left( -x \right)}^{2}}}}$
We have already learnt the rule that says ${{\left( -x \right)}^{2}}={{x}^{2}}.$
And from this we will get the following,
\[\Rightarrow f\left( -x \right)={{e}^{-{{x}^{2}}}}\]
And let us take a look at the definition of the given function which is given as $f\left( x \right)={{e}^{-{{x}^{2}}}}.$
Also, from what we have obtained earlier, we will be directed to the following equation,
$\Rightarrow f\left( -x \right)={{e}^{-{{x}^{2}}}}=f\left( x \right)$
The above equation leads us to what we are asked to find.
That is,
$\Rightarrow f\left( -x \right)=f\left( x \right)$
This is the property of even functions which is used to determine whether a function is even or not.
Since the given function satisfies the property of even functions, this is an even function.
Hence, $f\left( x \right)={{e}^{-{{x}^{2}}}}$ is an even function.
Note:
This does not mean all the exponential functions are even. ${{e}^{x}}$ is a function which is neither even nor odd. Before making any conclusion, one must check for both the properties.
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