Solve number 3

Step 1: Understand the concept of domain. The domain of a function is the set of all possible input values (x-values) for which the function is defined. • For rational functions (fractions), the denominator cannot be zero. • For square root functions, the expression inside the square root must be greater than or equal to zero. • If a square root is in the denominator, the expression inside the square root must be strictly greater than zero.

Step 2: Solve part a). a) Find the domain of $f(x) = \frac{x+1}{x-2}$. This is a rational function. The denominator cannot be zero. $x-2 \neq 0$ $x \neq 2$ The domain is all real numbers except $2$. In interval notation: $\boxed{(-\infty, 2) \cup (2, \infty)}$.

Step 3: Solve part b). b) Find the domain of $g(x) = \sqrt{x-3}$. This is a square root function. The expression inside the square root must be non-negative. $x-3 \ge 0$ $x \ge 3$ The domain is all real numbers greater than or equal to $3$. In interval notation: $\boxed{[3, \infty)}$.

Step 4: Solve part c). c) Find the domain of $h(x) = \frac{x}{\sqrt{x+4}}$. This function has a square root in the denominator.

1. The expression inside the square root must be non-negative: $x+4 \ge 0 \implies x \ge -4$.
2. The denominator cannot be zero: $\sqrt{x+4} \neq 0 \implies x+4 \neq 0 \implies x \neq -4$. Combining these conditions, $x$ must be strictly greater than $-4$. $x > -4$ The domain is all real numbers strictly greater than $-4$. In interval notation: $\boxed{(-4, \infty)}$.

Step 5: Solve part d). d) Find the domain of $k(x) = \frac{x^2+1}{x^2-1}$. This is a rational function. The denominator cannot be zero. $x^2-1 \neq 0$ Factor the denominator: $(x-1)(x+1) \neq 0$ This implies: $x-1 \neq 0 \implies x \neq 1$ $x+1 \neq 0 \implies x \neq -1$ The domain is all real numbers except $1$ and $-1$. In interval notation: $\boxed{(-\infty, -1) \cup (-1, 1) \cup (1, \infty)}$.

Step 6: Solve part e). e) Find the domain of $m(x) = \sqrt{x^2-4}$. This is a square root function. The expression inside the square root must be non-negative. $x^2-4 \ge 0$ Factor the expression: $(x-2)(x+2) \ge 0$ To solve this inequality, we find the critical points where the expression equals zero: $x-2=0 \implies x=2$ and $x+2=0 \implies x=-2$. We test intervals: • For $x < -2$ (e.g., $x=-3$): $(-3-2)(-3+2) = (-5)(-1) = 5 \ge 0$. This interval is included. • For $-2 < x < 2$ (e.g., $x=0$): $(0-2)(0+2) = (-2)(2) = -4 < 0$. This interval is not included. • For $x > 2$ (e.g., $x=3$): $(3-2)(3+2) = (1)(5) = 5 \ge 0$. This interval is included. The points $x=-2$ and $x=2$ are also included because the inequality is $\ge$. So, the solution is $x \le -2$ or $x \ge 2$. The domain is all real numbers less than or equal to $-2$ or greater than or equal to $2$. In interval notation: $\boxed{(-\infty, -2] \cup [2, \infty)}$.