1. **Problem Statement:** Define even and odd functions, then find the domain and range of the function $f(x) = 1 + x^2$. 2. **Definitions:** - An **even function** satisfies $f(-x) = f(x)$ for all $x$ in its domain. - An **odd function** satisfies $f(-x) = -f(x)$ for all $x$ in its domain. 3. **Check if $f(x) = 1 + x^2$ is even or odd:** Calculate $f(-x) = 1 + (-x)^2 = 1 + x^2 = f(x)$. Since $f(-x) = f(x)$, $f$ is an **even function**. 4. **Find the domain:** The function $f(x) = 1 + x^2$ is defined for all real numbers because $x^2$ is defined for all real $x$. So, domain is $(-\infty, \infty)$. 5. **Find the range:** Since $x^2 \geq 0$ for all real $x$, the minimum value of $f(x)$ is at $x=0$: $$f(0) = 1 + 0 = 1$$ As $x^2$ increases without bound, $f(x)$ increases without bound. Thus, range is $[1, \infty)$. **Final answer:** - $f(x) = 1 + x^2$ is an **even function**. - Domain: $(-\infty, \infty)$. - Range: $[1, \infty)$.