Determine Domain and Range - College Algebra

1. The function $m(x) = \frac{\sqrt{x^2-9}}{x-5}$ has which domain?
   Answer: $(-\infty, -3] \cup [3, \infty) \setminus \{5\}$ For the domain, we need two conditions: (1) $x^2 - 9 \geq 0$ for the square root, and (2) $x - 5 \neq 0$ for the denominator. From condition (1), $x^2 \geq 9$, so $x \leq -3$ or $x \geq 3$. This gives us $(-\infty, -3] \cup [3, \infty)$. From condition (2), $x \neq 5$. Combining both conditions, we get $(-\infty, -3] \cup [3, \infty) \setminus \{5\}$. Note that the endpoints $x = -3$ and $x = 3$ are included because $\sqrt{x^2-9} = 0$ is valid. Choice B incorrectly excludes the endpoints, choice C has the wrong interval, and choice D ignores the square root restriction.
2. Consider the function $r(x) = \sqrt{\frac{x-1}{x+2}}$. What is its domain?
   Answer: $(-\infty, -2) \cup [1, \infty)$ When finding the domain of a function involving square roots and fractions, you need to ensure two conditions are met: the expression under the square root must be non-negative, and the denominator cannot equal zero. For $r(x) = \sqrt{\frac{x-1}{x+2}}$, you need $\frac{x-1}{x+2} \geq 0$ and $x + 2 \neq 0$. First, note that $x = -2$ makes the denominator zero, so it's excluded from the domain. To solve $\frac{x-1}{x+2} \geq 0$, find where the numerator and denominator change signs. The numerator $x-1 = 0$ when $x = 1$, and the denominator $x+2 = 0$ when $x = -2$. Test the sign of the fraction in each interval: - For $x < -2$: both $x-1 < 0$ and $x+2 < 0$, so $\frac{x-1}{x+2} > 0$ ✓ - For $-2 < x < 1$: $x-1 < 0$ and $x+2 > 0$, so $\frac{x-1}{x+2} < 0$ ✗ - For $x > 1$: both $x-1 > 0$ and $x+2 > 0$, so $\frac{x-1}{x+2} > 0$ ✓ - At $x = 1$: $\frac{1-1}{1+2} = 0 \geq 0$ ✓ Therefore, the domain is $(-\infty, -2) \cup [1, \infty)$, which is answer C. Choice A incorrectly includes values between -2 and 1. Choice B only includes the interval between the critical points. Choice D excludes $x = 1$, but this value makes the expression equal zero, which is acceptable under a square root. Study tip: For rational expressions under square roots, always check both where the expression equals zero (include these points) and where it's undefined (exclude these points).
3. For the piecewise function $h(x) = \begin{cases} x^2-4 & \text{if } x < 2 \\ \sqrt{x-1} & \text{if } x \geq 2 \end{cases}$, what is the range?
   Answer: $[-4, 0) \cup [1, \infty)$ For $x < 2$, we have $h(x) = x^2 - 4$. As $x$ approaches 2 from the left, $h(x)$ approaches $4-4 = 0$. As $x \to -\infty$, $h(x) \to \infty$. The minimum value on this piece occurs at $x = 0$, giving $h(0) = -4$. So the first piece contributes $[-4, 0)$ to the range. For $x \geq 2$, we have $h(x) = \sqrt{x-1}$. At $x = 2$, $h(2) = \sqrt{1} = 1$. As $x \to \infty$, $h(x) \to \infty$. So the second piece contributes $[1, \infty)$. The complete range is $[-4, 0) \cup [1, \infty)$. Choice A misses the gap, choice C incorrectly fills the gap, and choice D has the wrong left endpoint.
4. The function $f(x) = \log_2(4-x^2)$ has domain and range, respectively:
   Answer: Domain: $(-2, 2)$; Range: $(-\infty, 2]$ For the domain, we need $4-x^2 > 0$, which means $x^2 < 4$, so $-2 < x < 2$. The domain is $(-2, 2)$. For the range, note that as $x$ approaches $\pm 2$, $4-x^2$ approaches 0, so $\log_2(4-x^2) \to -\infty$. The maximum value occurs at $x = 0$, where $f(0) = \log_2(4) = 2$. Since the argument $4-x^2$ can take any value in $(0, 4]$ and $\log_2$ is increasing, the range is $(-\infty, 2]$. Choice B incorrectly includes endpoints in domain, choice C has wrong range, and choice D has incorrect domain.
5. For the function $p(x) = 3 + \frac{2}{x^2+1}$, what is the range?
   Answer: $(3, 5]$ To find the range of a rational function like this, you need to analyze how the function behaves by examining its components and any restrictions on the output values. Start by examining the fraction $\frac{2}{x^2+1}$. Since $x^2 \geq 0$ for all real numbers, we know $x^2+1 \geq 1$. This means $\frac{2}{x^2+1} \leq \frac{2}{1} = 2$. The fraction approaches 2 as $x$ approaches 0, and approaches 0 as $|x|$ approaches infinity. However, since $x^2+1$ is never zero, the fraction is always positive and never actually reaches 0. Therefore, $\frac{2}{x^2+1}$ takes on all values in the interval $(0, 2]$. Adding 3 to this entire interval gives us $p(x) = 3 + \frac{2}{x^2+1}$, which takes on all values in $(3, 5]$. The function approaches but never reaches 3, and achieves its maximum value of 5 when $x = 0$. Choice A, $[5, \infty)$, incorrectly suggests the function can exceed 5 and includes values above the maximum. Choice B, $[3, 5]$, wrongly includes 3 as an achievable value, when the function only approaches it. Choice C, $(3, \infty)$, correctly excludes 3 but incorrectly suggests the function can take arbitrarily large values. Strategy tip: For rational functions, always check the behavior of denominators to find restrictions, then trace how those restrictions affect the overall function's output range.

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