1. The problem asks us to determine which of the given functions is even. 2. Recall that a function $f(x)$ is even if it satisfies the condition: $$f(-x) = f(x)$$ for all $x$ in the domain. 3. Let's check each function: - a. $f(x) = \sin x$ $$f(-x) = \sin(-x) = -\sin x \neq \sin x$$ So, $f(x) = \sin x$ is not even. - b. $f(x) = x + 1$ $$f(-x) = -x + 1 \neq x + 1$$ So, $f(x) = x + 1$ is not even. - c. $f(x) = x^3$ $$f(-x) = (-x)^3 = -x^3 \neq x^3$$ So, $f(x) = x^3$ is not even. - d. $f(x) = x^2 + 1$ $$f(-x) = (-x)^2 + 1 = x^2 + 1 = f(x)$$ So, $f(x) = x^2 + 1$ is even. 4. Therefore, the only even function among the options is $f(x) = x^2 + 1$. **Final answer:** $f(x) = x^2 + 1$ is even.