Let's go through each function to determine its domain and range.

1. $f(x) = \sqrt{2x + 1}$

• Domain: The expression under the square root must be non-negative, so: $2x + 1 \geq 0$ $x \geq -\frac{1}{2}$ Therefore, the domain is $x \in \left[ -\frac{1}{2}, \infty \right)$.
• Range: The square root function produces non-negative values. Since $\sqrt{2x + 1} \geq 0$ for all $x$ in the domain, and $2x + 1$ can grow indefinitely, the range is: $f(x) \in [0, \infty)$

2. $f(x) = 3x^2 + 1$

• Domain: This is a polynomial function, so it is defined for all real values of $x$: $x \in (-\infty, \infty)$
• Range: Since $3x^2 + 1$ is a quadratic function that opens upwards (coefficient of $x^2$ is positive), the minimum value occurs at the vertex. The vertex of $y = 3x^2 + 1$ is at $x = 0$: $f(0) = 3(0)^2 + 1 = 1$ As $x$ goes to $\pm \infty$, $3x^2 + 1$ increases without bound. Therefore, the range is: $f(x) \in [1, \infty)$

3. $f(x) = 2|x - 3|$

• Domain: The absolute value function is defined for all real $x$, so: $x \in (-\infty, \infty)$
• Range: The function $2|x - 3|$ is always non-negative since it is a scaled absolute value function. The minimum value occurs when $x = 3$: $f(3) = 2|3 - 3| = 0$ As $x$ moves away from 3, $2|x - 3|$ increases without bound. Thus, the range is: $f(x) \in [0, \infty)$

Summary Table

Here's the completed table for the domain and range of each function: