Get expert advice and community support on IDNLearn.com. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
Sagot :
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.
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.
