Question

In Exercises \(21-30\) , determine whether the function is even, odd, or neither. Try to answer without writing anything (except the answer). $$y=\sqrt{x^{2}+2}$$

Short answer

The function \( y=\sqrt{x^{2}+2} \) is an even function.

Step-by-step solution

Check whether the function is even

We need to determine if the function \( f(x) = \sqrt{x^{2}+2} \) is equal to \( f(-x) \). Substitute \( -x \) into the function in place of \( x \): \( f(-x)=\sqrt{(-x)^{2}+2}= \sqrt{x^{2}+2} \). We can see that \( f(x) = f(-x) \). This means we cannot make any determination about the function being odd or neither, but it is certainly even.

Check whether the function is odd

We need to determine if the function \( f(x) = \sqrt{x^{2}+2} \) is equal to \( -f(-x) \). Substitute \( -x \) into the function in place of \( x \): \( -f(-x)=-\sqrt{(-x)^{2}+2}= -\sqrt{x^{2}+2} \). We can see that \( f(x) \neq -f(-x) \). This means that the function is not odd.

Conclusion

Based on steps 1 and 2, we can conclude that the function \( f(x) = \sqrt{x^{2}+2} \) is even, not odd, and cannot be classified as neither since it fulfills the criterion of an even function perfectly.