Example 1.2 (b): f(x) = sqrt(3x+2)−1

Here are the domain and range for the functions:

• Example 1.2 (b): $f(x) = \sqrt{3x+2}-1$

  • Step 1: Find the domain. For the function to be defined in real numbers, the expression under the square root must be non-negative.

    $$3x+2 \ge 0$$ $$3x \ge -2$$ $$x \ge -\frac{2}{3}$$

    The domain is $D(f) = \left[-\frac{2}{3}, \infty\right)$.

  • Step 2: Find the range. Since $\sqrt{3x+2} \ge 0$ for all $x$ in the domain, we can determine the range of $f(x)$.

    $$\sqrt{3x+2}-1 \ge 0-1$$ $$f(x) \ge -1$$

    The range is $R(f) = [-1, \infty)$.

  • Final Answer for (b): Domain: $\boxed{\left[-\frac{2}{3}, \infty\right)}$ Range: $\boxed{[-1, \infty)}$

• Example 1.2 (c): $f(x) = \frac{x}{x-2}$

  • Step 1: Find the domain. For a rational function, the denominator cannot be zero.

    $$x-2 \ne 0$$ $$x \ne 2$$

    The domain is $D(f) = (-\infty, 2) \cup (2, \infty)$ or $\mathbb{R} \setminus \{2\}$.

  • Step 2: Find the range. Let $y = f(x)$. We solve for $x$ in terms of $y$.

    $$y = \frac{x}{x-2}$$ $$y(x-2) = x$$ $$yx - 2y = x$$ $$yx - x = 2y$$ $$x(y-1) = 2y$$ $$x = \frac{2y}{y-1}$$

    For $x$ to be defined, the denominator $y-1$ cannot be zero.

    $$y-1 \ne 0$$ $$y \ne 1$$

    The range is $R(f) = (-\infty, 1) \cup (1, \infty)$ or $\mathbb{R} \setminus \{1\}$.

  • Final Answer for (c): Domain: $\boxed{(-\infty, 2) \cup (2, \infty)}$ Range: $\boxed{(-\infty, 1) \cup (1, \infty)}$