How do you tell if a function is odd even or neither?

Hint: In these types of problems first we have to learn about the even and odd function. We have to substitute the value of the variable with the negative of the variable and then simplify the function. After that we check whether it follows the condition of even, odd or neither.

Complete step by step answer:
A function \[f(x)\] is known to be even if we substitute \[x\] with \[-x\] and if we get as:
\[f(-x)=f(x)\] (I.e. after substituting the value, we will get our original function).
A function \[f(x)\] is known to be odd if we substitute \[x\] with \[-x\] and if we get as:
\[f(-x)=-f(x)\] (I.e. after substituting the value, we will get the negative of the original function).
And if we do not find either of these results, then the function is neither odd nor even.
Let’s understand this concept more with help of some examples-
For example, there is a function \[f(x)=\cos x\] . Now check whether this given function is odd even or neither.
For checking, we have to substitute \[x\] with \[-x\] and we will get as:
\[\Rightarrow f(-x)=\cos (-x)\]
\[\Rightarrow f(-x)=\cos x\]
\[\Rightarrow f(-x)=f(x)\]
As we can observe, this is the condition for an even function.
Hence, the given function \[f(x)=\cos x\] is an even function.
Let's say there is a function \[f(x)=-3{{x}^{3}}+2x\] . We have to check whether it is even, odd or neither.
We will substitute \[x\] with \[-x\] and we will get function as:
 \[f(x)=-3{{x}^{3}}+2x\]
\[\Rightarrow f(-x)=-3{{(-x)}^{3}}+2(-x)\]
\[\Rightarrow f(-x)=-3{{(-1)}^{3}}{{(x)}^{3}}+2(-x)\]
\[\Rightarrow f(-x)=-3(-1){{(x)}^{3}}+2(-x)\]
\[\Rightarrow f(-x)=3{{x}^{3}}-2x\]
So, now we take the negative common factor out in above:
\[\Rightarrow f(-x)=-(-3{{x}^{3}}+2x)\]
Now we get as:
\[\Rightarrow f(-x)=-f(x)\]
As we can observe, this is the condition for an odd function.
So, it implies that the given function \[f(x)=-3{{x}^{3}}+2x\] is an odd function.
Let’s take one more example, there is a function \[h(x)={{x}^{3}}-{{x}^{2}}-1\] .Now check whether the given function is even odd or neither.
After substituting value of \[x\] with \[-x\], we will get function as:
\[h(x)={{x}^{3}}-{{x}^{2}}-1\]
\[\Rightarrow h(-x)={{(-x)}^{3}}-{{(-x)}^{2}}-1\]
\[\Rightarrow h(-x)={{(-1)}^{3}}{{(x)}^{3}}-{{(-1)}^{2}}{{(x)}^{2}}-1\]
\[\Rightarrow h(-x)={{(-1)}^{3}}{{x}^{3}}-{{(-1)}^{2}}{{x}^{2}}-1\]
\[\Rightarrow h(-x)=-{{x}^{3}}-{{x}^{2}}-1\]
As, we can see \[h(-x)\ne h(x)\]
Now we will take the negative common factor out:
\[\Rightarrow h(-x)=-({{x}^{3}}+{{x}^{2}}+1)\]
As, we can also observe \[h(-x)\ne -h(x)\]
Hence it does not satisfy the condition of even or odd function.
Therefore, we can say that a given function is neither even nor odd.

Note: The zero function \[f(x)=0\] is the only function which is both even and odd. If we draw the graph of even and odd function then, the graph of even function is always symmetric with respect to \[y\] axis and the graph of odd function is always symmetric with respect to origin.