Answer :
Let's approach the problem step-by-step and provide a thorough explanation in complete sentences.
Part A: Comparing the Domain and Range of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]
The function [tex]\( f(x) \)[/tex] is a polynomial: [tex]\( f(x) = x^4 - 2x^3 - 3x^2 + 4x + 4 \)[/tex]. Polynomial functions are defined for all real numbers, so the domain of [tex]\( f(x) \)[/tex] is all real numbers, denoted as [tex]\( (-\infty, \infty) \)[/tex]. In terms of range, since [tex]\( f(x) \)[/tex] is a polynomial of degree 4 with a leading coefficient (positive), the range is also all real numbers. This is because as [tex]\( x \)[/tex] approaches [tex]\(\pm \infty\)[/tex], [tex]\( f(x) \)[/tex] tends to [tex]\(\infty\)[/tex], covering all real values in between.
In contrast, the function [tex]\( g(x) \)[/tex] is a radical function: [tex]\( g(x) = \sqrt{x^2 - x - 2} \)[/tex]. To determine the domain of [tex]\( g(x) \)[/tex], we need [tex]\( x^2 - x - 2 \geq 0 \)[/tex]. Solving this inequality, we factorize [tex]\( x^2 - x - 2 = (x + 1)(x - 2) \)[/tex]. The solutions to the inequality [tex]\( (x + 1)(x - 2) \geq 0 \)[/tex] are [tex]\( x \)[/tex] in the intervals [tex]\( (-\infty, -1] \)[/tex] and [tex]\( [2, \infty) \)[/tex]. Thus, the domain of [tex]\( g(x) \)[/tex] is [tex]\( (-\infty, -1] \cup [2, \infty) \)[/tex].
For the range of [tex]\( g(x) \)[/tex], since the minimum value within the domain [tex]\( (-\infty, -1] \cup [2, \infty) \)[/tex] occurs when evaluating [tex]\( g(x) = \sqrt{2} \)[/tex], the function is increasing without bound as [tex]\( x \)[/tex] approaches [tex]\(\pm \infty\)[/tex]. Therefore, the range of [tex]\( g(x) \)[/tex] is [tex]\( [\sqrt{2}, \infty) \)[/tex].
In summary:
- The domain of [tex]\( f(x) \)[/tex] is all real numbers [tex]\((-\infty, \infty)\)[/tex], and its range is also all real numbers [tex]\((-\infty, \infty)\)[/tex].
- The domain of [tex]\( g(x) \)[/tex] is [tex]\( (-\infty, -1] \cup [2, \infty) \)[/tex], and its range is [tex]\( [\sqrt{2}, \infty) \)[/tex].
Part B: Breaks in the Domain of [tex]\( h(x) \)[/tex]
The function [tex]\( h(x) \)[/tex] is defined as: [tex]\( h(x) = \frac{-x^2 + 1}{x^2 - x - 2} \)[/tex]. The breaks in the domain of [tex]\( h(x) \)[/tex] occur where the denominator is zero since division by zero is undefined. To find these breaks, we solve the equation [tex]\( x^2 - x - 2 = 0 \)[/tex].
Factoring the quadratic equation, we get [tex]\( (x + 1)(x - 2) = 0 \)[/tex]. Thus, the zeros of the denominator are [tex]\( x = -1 \)[/tex] and [tex]\( x = 2 \)[/tex]. These values cause breaks in the domain of [tex]\( h(x) \)[/tex].
In Part A, when examining the domain of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], we didn't need to consider the breaks. However, for [tex]\( h(x) \)[/tex], these specific points are crucial as they denote where [tex]\( h(x) \)[/tex] is not defined.
To summarize:
- The breaks in the domain of [tex]\( h(x) \)[/tex] occur at [tex]\( x = -1 \)[/tex] and [tex]\( x = 2 \)[/tex], which are the zeros of the denominator [tex]\( x^2 - x - 2 \)[/tex].
This concludes the detailed step-by-step solution for the given question.
Part A: Comparing the Domain and Range of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]
The function [tex]\( f(x) \)[/tex] is a polynomial: [tex]\( f(x) = x^4 - 2x^3 - 3x^2 + 4x + 4 \)[/tex]. Polynomial functions are defined for all real numbers, so the domain of [tex]\( f(x) \)[/tex] is all real numbers, denoted as [tex]\( (-\infty, \infty) \)[/tex]. In terms of range, since [tex]\( f(x) \)[/tex] is a polynomial of degree 4 with a leading coefficient (positive), the range is also all real numbers. This is because as [tex]\( x \)[/tex] approaches [tex]\(\pm \infty\)[/tex], [tex]\( f(x) \)[/tex] tends to [tex]\(\infty\)[/tex], covering all real values in between.
In contrast, the function [tex]\( g(x) \)[/tex] is a radical function: [tex]\( g(x) = \sqrt{x^2 - x - 2} \)[/tex]. To determine the domain of [tex]\( g(x) \)[/tex], we need [tex]\( x^2 - x - 2 \geq 0 \)[/tex]. Solving this inequality, we factorize [tex]\( x^2 - x - 2 = (x + 1)(x - 2) \)[/tex]. The solutions to the inequality [tex]\( (x + 1)(x - 2) \geq 0 \)[/tex] are [tex]\( x \)[/tex] in the intervals [tex]\( (-\infty, -1] \)[/tex] and [tex]\( [2, \infty) \)[/tex]. Thus, the domain of [tex]\( g(x) \)[/tex] is [tex]\( (-\infty, -1] \cup [2, \infty) \)[/tex].
For the range of [tex]\( g(x) \)[/tex], since the minimum value within the domain [tex]\( (-\infty, -1] \cup [2, \infty) \)[/tex] occurs when evaluating [tex]\( g(x) = \sqrt{2} \)[/tex], the function is increasing without bound as [tex]\( x \)[/tex] approaches [tex]\(\pm \infty\)[/tex]. Therefore, the range of [tex]\( g(x) \)[/tex] is [tex]\( [\sqrt{2}, \infty) \)[/tex].
In summary:
- The domain of [tex]\( f(x) \)[/tex] is all real numbers [tex]\((-\infty, \infty)\)[/tex], and its range is also all real numbers [tex]\((-\infty, \infty)\)[/tex].
- The domain of [tex]\( g(x) \)[/tex] is [tex]\( (-\infty, -1] \cup [2, \infty) \)[/tex], and its range is [tex]\( [\sqrt{2}, \infty) \)[/tex].
Part B: Breaks in the Domain of [tex]\( h(x) \)[/tex]
The function [tex]\( h(x) \)[/tex] is defined as: [tex]\( h(x) = \frac{-x^2 + 1}{x^2 - x - 2} \)[/tex]. The breaks in the domain of [tex]\( h(x) \)[/tex] occur where the denominator is zero since division by zero is undefined. To find these breaks, we solve the equation [tex]\( x^2 - x - 2 = 0 \)[/tex].
Factoring the quadratic equation, we get [tex]\( (x + 1)(x - 2) = 0 \)[/tex]. Thus, the zeros of the denominator are [tex]\( x = -1 \)[/tex] and [tex]\( x = 2 \)[/tex]. These values cause breaks in the domain of [tex]\( h(x) \)[/tex].
In Part A, when examining the domain of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], we didn't need to consider the breaks. However, for [tex]\( h(x) \)[/tex], these specific points are crucial as they denote where [tex]\( h(x) \)[/tex] is not defined.
To summarize:
- The breaks in the domain of [tex]\( h(x) \)[/tex] occur at [tex]\( x = -1 \)[/tex] and [tex]\( x = 2 \)[/tex], which are the zeros of the denominator [tex]\( x^2 - x - 2 \)[/tex].
This concludes the detailed step-by-step solution for the given question.
