To determine which of the given functions is an even function, we need to check if each function satisfies the condition for being even:

A function $g(x)$ is even if: $g(-x) = g(x)$

Let's analyze each option.

Option (A): $g(x) = \sqrt{3 - x} + 5$

To check if $g(x)$ is even, we substitute $-x$ for $x$: $g(-x) = \sqrt{3 - (-x)} + 5 = \sqrt{3 + x} + 5$ This does not equal $g(x)$, so (A) is not an even function.

Option (B): $g(x) = \frac{x}{\sqrt{x}}$

For $g(x)$ to be defined, $x$ must be positive (otherwise, $\sqrt{x}$ would be undefined for real numbers). Testing $g(-x)$ would involve an undefined square root for $-x$, so (B) cannot be an even function.

Option (C): $g(x) = \frac{1}{x^2} + 8$

Let’s substitute $-x$ for $x$: $g(-x) = \frac{1}{(-x)^2} + 8 = \frac{1}{x^2} + 8 = g(x)$ This matches the original function, so (C) is an even function.

Option (D): $g(x) = -|x + 4| - 3$

Substitute $-x$ for $x$: $g(-x) = -|(-x) + 4| - 3 = -| -x + 4| - 3$ This does not simplify to $g(x)$, so (D) is not an even function.

Option (E): $g(x) = \frac{(x - 2)^2}{3}$

Substitute $-x$ for $x$: $g(-x) = \frac{((-x) - 2)^2}{3} = \frac{(x + 2)^2}{3}$ This does not match the original function $g(x)$, so (E) is not an even function.

Conclusion

The correct answer is: (C) $g(x) = \frac{1}{x^2} + 8$ is the even function.

Would you like more details on this solution or have any other questions?

Here are five related questions to expand your understanding:

1. What defines a function as odd?
2. Can a function be both even and odd? If so, what type of function would that be?
3. How can we determine if a polynomial is an even or odd function?
4. What role does symmetry play in identifying even and odd functions?
5. Are there specific types of functions that are more likely to be even?

Tip: An even function is symmetric about the y-axis, which means its graph looks the same on both sides of the y-axis.