1. **Problem statement:** Find the range of the functions: i. $f(x) = 2x^2 - x$ for $x \in \mathbb{R}$ ii. $g(x) = \frac{1}{x}$ for $x \in \mathbb{R}, x \neq 0$ 2. **Range of $f(x)$:** - The function is quadratic: $f(x) = 2x^2 - x$. - Since the coefficient of $x^2$ is positive ($2 > 0$), the parabola opens upwards. - The vertex gives the minimum value of $f(x)$. 3. **Find vertex of $f(x)$:** - Vertex $x$-coordinate: $x_v = -\frac{b}{2a} = -\frac{-1}{2 \times 2} = \frac{1}{4}$. - Evaluate $f(x_v)$: $$f\left(\frac{1}{4}\right) = 2\left(\frac{1}{4}\right)^2 - \frac{1}{4} = 2 \times \frac{1}{16} - \frac{1}{4} = \frac{1}{8} - \frac{1}{4} = \frac{1}{8} - \frac{2}{8} = -\frac{1}{8}$$ 4. **Range of $f(x)$:** - Since parabola opens upward and minimum value is $-\frac{1}{8}$, - Range is $\left[-\frac{1}{8}, \infty\right)$. 5. **Range of $g(x) = \frac{1}{x}$:** - Domain excludes $x=0$. - $g(x)$ can take any real value except $0$ because $\frac{1}{x} = 0$ has no solution. - Range is $\mathbb{R} \setminus \{0\}$. **Final answers:** - Range of $f(x)$ is $\boxed{\left[-\frac{1}{8}, \infty\right)}$. - Range of $g(x)$ is $\boxed{\mathbb{R} \setminus \{0\}}$.