Find The Domain And Range Of The Following Functions: (a) $3x^2 - 5x + 7$ (b) $\frac{x-3}{x+2}$ (c) $\sqrt{(x-1)(x+1)}$ (d) $\sqrt[3]{\frac{x-4}{x-2}}$ (e) $\sqrt{\frac{x-2}{x+1}}$

In mathematics, understanding the domain and range of a function is crucial for a comprehensive analysis. The domain represents the set of all possible input values (often denoted as 'x') for which the function is defined, while the range represents the set of all possible output values (often denoted as 'y' or f(x)) that the function can produce. Determining the domain and range helps us understand the behavior and limitations of a function. This guide will explore various types of functions and provide detailed explanations on how to find their domains and ranges, ensuring a strong grasp of these fundamental concepts. This includes polynomial functions, rational functions, radical functions, and more complex combinations. By mastering these techniques, you will be well-equipped to analyze and understand a wide array of mathematical functions. Understanding the domain and range of functions is a cornerstone of mathematical analysis, enabling us to predict and interpret function behavior accurately.

Let's dive into the first function, analyzing polynomial functions such as $f(x) = 3x^2 - 5x + 7$. This is a quadratic function, a type of polynomial function, which is defined for all real numbers. Polynomial functions, in general, have domains that include all real numbers because there are no restrictions on the input values 'x'. You can substitute any real number into a polynomial, and you will always get a real number as an output. This is because polynomials involve only addition, subtraction, and multiplication of variables and constants, none of which introduce domain restrictions like division by zero or square roots of negative numbers. To determine the range, we need to consider the shape of the quadratic function. The graph of a quadratic function is a parabola. Since the coefficient of the $x^2$ term (3 in this case) is positive, the parabola opens upwards, meaning it has a minimum value. To find the minimum value, we can complete the square or use the vertex formula. The vertex formula for a quadratic function in the form $f(x) = ax^2 + bx + c$ is given by $x = -b/(2a)$. Applying this to our function, we get $x = -(-5) / (2 * 3) = 5/6$. Now, we substitute this value back into the function to find the minimum y-value: $f(5/6) = 3(5/6)^2 - 5(5/6) + 7 = 3(25/36) - 25/6 + 7 = 25/12 - 50/12 + 84/12 = 59/12$. Thus, the minimum value of the function is $59/12$. Since the parabola opens upwards, the range includes all values greater than or equal to this minimum. Therefore, the range is $[59/12, ext{∞})$. In summary, for the function $f(x) = 3x^2 - 5x + 7$, the domain is all real numbers $(- ext{∞}, ext{∞})$, and the range is $[59/12, ext{∞})$. Understanding this process is vital for analyzing various quadratic and polynomial functions, ensuring a strong foundation in function analysis. The nature of polynomial functions ensures a continuous domain, but the specific characteristics of the polynomial, such as the leading coefficient and degree, significantly influence the range.

Next, let’s consider the rational function f(x) = rac{x-3}{x+2}. Rational functions are functions that can be written as the ratio of two polynomials. The key consideration for the domain of a rational function is that the denominator cannot be zero, as division by zero is undefined. To find the values of x that make the denominator zero, we set $x+2 = 0$ and solve for x, which gives us $x = -2$. Therefore, the domain of this function is all real numbers except $x = -2$. In interval notation, this can be written as $(- ext{∞}, -2) ext{∪} (-2, ext{∞})$. To find the range, we need to determine the possible output values of the function. A helpful approach for rational functions is to consider the horizontal asymptote. The horizontal asymptote describes the behavior of the function as x approaches positive or negative infinity. For rational functions where the degree of the numerator and the denominator are the same (in this case, both are degree 1), the horizontal asymptote is the ratio of the leading coefficients. Here, the leading coefficient of the numerator is 1, and the leading coefficient of the denominator is also 1, so the horizontal asymptote is $y = 1$. This means that as x approaches infinity or negative infinity, the function approaches 1. However, we need to check if the function actually takes the value 1. To do this, we set the function equal to 1 and solve for x: rac{x-3}{x+2} = 1. Multiplying both sides by $(x+2)$, we get $x-3 = x+2$, which simplifies to $-3 = 2$, a contradiction. This means that the function never actually equals 1. Therefore, the range of the function is all real numbers except 1, which can be written in interval notation as $(- ext{∞}, 1) ext{∪} (1, ext{∞})$. In summary, for the function f(x) = rac{x-3}{x+2}, the domain is $(- ext{∞}, -2) ext{∪} (-2, ext{∞})$, and the range is $(- ext{∞}, 1) ext{∪} (1, ext{∞})$. Understanding how to find the domain and range of rational functions involves identifying potential discontinuities and analyzing asymptotic behavior, essential skills in calculus and function analysis.

Now, let’s examine the radical function $f(x) = ext{√(x-1)(x+1)}$. Radical functions, particularly those involving square roots, have restrictions on their domain because the expression inside the square root must be non-negative (i.e., greater than or equal to zero) to yield a real number. Thus, we need to solve the inequality $(x-1)(x+1) ≥ 0$. To do this, we first find the critical points where the expression equals zero, which are $x = 1$ and $x = -1$. These points divide the number line into three intervals: $(- ext{∞}, -1)$, $(-1, 1)$, and $(1, ext{∞})$. We test a value from each interval in the inequality $(x-1)(x+1) ≥ 0$. For $(- ext{∞}, -1)$, let’s test $x = -2$: $(-2-1)(-2+1) = (-3)(-1) = 3$, which is positive. For $(-1, 1)$, let’s test $x = 0$: $(0-1)(0+1) = (-1)(1) = -1$, which is negative. For $(1, ext{∞})$, let’s test $x = 2$: $(2-1)(2+1) = (1)(3) = 3$, which is positive. Thus, the inequality $(x-1)(x+1) ≥ 0$ is satisfied for the intervals $(- ext{∞}, -1]$ and $[1, ext{∞})$. These intervals form the domain of the function. Now, let’s determine the range. Since the function involves a square root, the output will always be non-negative. The minimum value of the function occurs at $x = -1$ and $x = 1$, where the expression inside the square root is zero, so $f(x) = 0$. As x moves away from these points, the value of $(x-1)(x+1)$ increases, and so does the square root. Therefore, the function can take any non-negative value. In summary, for the function $f(x) = ext{√(x-1)(x+1)}$, the domain is $(- ext{∞}, -1] ext{∪} [1, ext{∞})$, and the range is $[0, ext{∞})$. Understanding the restrictions imposed by square roots is crucial in determining the domain and range of radical functions, a skill fundamental to advanced mathematical analysis.

Consider the function $f(x) = ext{³√((x-4)/(x-2))}$, a cube root function. Cube root functions, unlike square root functions, do not have the same domain restrictions because the cube root of any real number (positive, negative, or zero) is defined. However, since we have a rational expression inside the cube root, we must consider the denominator. The denominator, $x-2$, cannot be zero, so $x ≠ 2$. Therefore, the domain of this function is all real numbers except $x = 2$, which can be written in interval notation as $(- ext{∞}, 2) ext{∪} (2, ext{∞})$. To find the range, we need to analyze the behavior of the function. Since the cube root function is defined for all real numbers, the range will depend on the values that the rational expression rac{x-4}{x-2} can take. Let y = rac{x-4}{x-2}. We can rewrite this equation to solve for x in terms of y: $y(x-2) = x-4$, which gives $yx - 2y = x - 4$. Rearranging the terms, we get $yx - x = 2y - 4$, so $x(y-1) = 2y - 4$. If $y ≠ 1$, then x = rac{2y-4}{y-1}. This shows that for any $y$ except $y = 1$, we can find a corresponding $x$. Thus, the rational expression rac{x-4}{x-2} can take any value except 1. Since we are taking the cube root of this expression, the range of $f(x)$ will be all real numbers except the cube root of 1, which is 1. Therefore, the range of $f(x)$ is all real numbers except 1, which can be written as $(- ext{∞}, 1) ext{∪} (1, ext{∞})$. In summary, for the function $f(x) = ext{³√((x-4)/(x-2))}$, the domain is $(- ext{∞}, 2) ext{∪} (2, ext{∞})$, and the range is $(- ext{∞}, 1) ext{∪} (1, ext{∞})$. Cube root functions, coupled with rational expressions, require careful consideration of both the denominator's restrictions and the cube root's expansive nature to accurately determine the domain and range.

Lastly, let’s examine the radical rational function $f(x) = ext{√((x-2)/(x+1))}$. This function combines a square root with a rational expression, so we need to consider two restrictions. First, the expression inside the square root must be non-negative, i.e., rac{x-2}{x+1} ≥ 0. Second, the denominator cannot be zero, so $x + 1 ≠ 0$, which means $x ≠ -1$. To solve the inequality rac{x-2}{x+1} ≥ 0, we find the critical points where the expression equals zero or is undefined, which are $x = 2$ and $x = -1$. These points divide the number line into three intervals: $(- ext{∞}, -1)$, $(-1, 2)$, and $(2, ext{∞})$. We test a value from each interval in the inequality rac{x-2}{x+1} ≥ 0. For $(- ext{∞}, -1)$, let’s test $x = -2$: rac{-2-2}{-2+1} = rac{-4}{-1} = 4, which is positive. For $(-1, 2)$, let’s test $x = 0$: rac{0-2}{0+1} = rac{-2}{1} = -2, which is negative. For $(2, ext{∞})$, let’s test $x = 3$: rac{3-2}{3+1} = rac{1}{4}, which is positive. Thus, the inequality rac{x-2}{x+1} ≥ 0 is satisfied for the intervals $(- ext{∞}, -1)$ and $[2, ext{∞})$. Note that we include 2 because the expression can equal zero, but we exclude -1 because the denominator cannot be zero. Therefore, the domain of the function is $(- ext{∞}, -1) ext{∪} [2, ext{∞})$. To find the range, we observe that the function involves a square root, so the output will always be non-negative. As $x$ approaches $-1$ from the left, the expression rac{x-2}{x+1} approaches infinity, so the square root also approaches infinity. At $x = 2$, the function value is $ ext{√((2-2)/(2+1))} = ext{√0} = 0$. As $x$ becomes very large, the function approaches $ ext{√1} = 1$. We need to determine if the function can take the value 1. Set $ ext{√((x-2)/(x+1))} = 1$, then rac{x-2}{x+1} = 1, which gives $x-2 = x+1$, and $-2 = 1$, which is a contradiction. Thus, the function never equals 1. Therefore, the range is $[0, 1)$. In summary, for the function $f(x) = ext{√((x-2)/(x+1))}$, the domain is $(- ext{∞}, -1) ext{∪} [2, ext{∞})$, and the range is $[0, 1)$. The analysis of radical rational functions necessitates a comprehensive approach, considering both the square root’s non-negativity requirement and the rational expression’s restrictions to accurately define the domain and range.

In conclusion, finding the domain and range of various functions is a fundamental skill in mathematics. By understanding the restrictions imposed by different types of functions—such as polynomial, rational, and radical functions—we can accurately determine the set of possible input and output values. Whether dealing with polynomial functions with their broad domains, rational functions with their denominators to consider, or radical functions with their non-negativity requirements, a systematic approach is crucial. Each type of function demands specific techniques, and mastering these methods provides a solid foundation for more advanced mathematical concepts. The ability to find the domain and range of functions is not only essential for mathematical problem-solving but also for understanding the behavior and limitations of mathematical models in real-world applications. This guide has provided detailed explanations and examples to equip you with the necessary tools for confidently analyzing functions and determining their domains and ranges. Continued practice and application of these principles will solidify your understanding and enhance your mathematical proficiency.