Question

How do you determine if $y={{x}^{2}}+1$ is an even or odd function?

Answer 1

Hint: We know by our prior knowledge of basic calculus that there are three types of functions namely, even functions, odd functions and neither even nor odd functions. We can identify whether a function is even or odd by substituting the negative value of the variable used in the function. Thus, we shall substitute $-x$ in place of variable-$x$ in the given function.

Complete step by step answer:
In order to identify a function as an even function or an odd function, we substitute the variable with respect to which the function has been expressed with the negative value of the same variable. Let the function be $f\left( x \right)$.
If this modified function is equal to the original function, then the function is called an even function i.e., $f\left( -x \right)=f\left( x \right)$.
However, if this modified function is equal to the negative of the original function, then the function is said to be an odd function i.e., $f\left( -x \right)=-f\left( x \right)$.
In the given function, $y={{x}^{2}}+1$ or $f\left( x \right)={{x}^{2}}+1$, we shall substitute variable- $x$ by $-x$.
$\Rightarrow f\left( -x \right)={{\left( -x \right)}^{2}}+1$
We know that $-x\times \left( -x \right)={{x}^{2}}$. Putting this value, we get
$\Rightarrow f\left( -x \right)={{x}^{2}}+1$
$\Rightarrow f\left( -x \right)=f\left( x \right)$
Therefore, we determine that $y={{x}^{2}}+1$ is an even function.

Note: There are some cases in which the modified functions are neither positive nor negative of the original given functions. Such functions are classified as neither even nor odd functions. Also, the even functions expressed in terms of the x-variable are symmetrical with respect to the y-axis and vice-versa.