1. **Problem 1: For the function $f(x) = 2x + 5$** 1. a) The domain of $f(x)$ is all real numbers because there are no restrictions on $x$ in a linear function. \[ \text{Domain of } f = (-\infty, \infty) \] 1. b) The range of $f(x)$ is also all real numbers because as $x$ takes any real value, $f(x)$ can take any real value. \[ \text{Range of } f = (-\infty, \infty) \] 1. c) To find $f(3)$, substitute $x=3$ into the function: $$ f(3) = 2(3) + 5 = 6 + 5 = 11 $$ 1. d) The question repeats $f(3)$, so the answer is the same: $$ f(3) = 11 $$ 2. **Problem 2: For the function $g(x) = \sqrt{x - 1}$** 2. a) The domain of $g(x)$ requires the expression inside the square root to be non-negative: $$ x - 1 \geq 0 \implies x \geq 1 $$ \[ \text{Domain of } g = [1, \infty) \] 2. b) The range of $g(x)$ is all non-negative real numbers because the square root function outputs values $\geq 0$: \[ \text{Range of } g = [0, \infty) \] 2. c) To find $g(10)$, substitute $x=10$: $$ g(10) = \sqrt{10 - 1} = \sqrt{9} = 3 $$ 2. d) The function $g(x)$ does not have an inverse for $x < 1$ because the expression inside the square root becomes negative, which is not defined for real numbers. Therefore, $g(x)$ is not defined for $x < 1$, so no inverse exists there. **Final answers:** - $f(3) = 11$ - $g(10) = 3$ - Domains and ranges as stated above.