Answer :
To find the inverse [tex]\( f^{-1}(x) \)[/tex] of the given function [tex]\( f(x) = -\frac{2}{3}x - 24 \)[/tex], we follow these steps:
1. Swap [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
We start with [tex]\( y = -\frac{2}{3} x - 24 \)[/tex]. To find the inverse, we swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex], obtaining [tex]\( x = -\frac{2}{3} y - 24 \)[/tex].
2. Solve for [tex]\(y\)[/tex]:
- Add 24 to both sides: [tex]\( x + 24 = -\frac{2}{3} y \)[/tex]
- Multiply both sides by [tex]\( -\frac{3}{2} \)[/tex] to solve for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{3}{2} (x + 24) \][/tex]
- This simplifies to: [tex]\( y = -\frac{3}{2} x - 36 \)[/tex].
So, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = -\frac{3}{2}x - 36 \][/tex]
Now let's analyze [tex]\( f^{-1}(x) \)[/tex]:
1. Slope:
- The slope of [tex]\( f^{-1}(x) \)[/tex] is the coefficient of [tex]\( x \)[/tex], which is [tex]\( -\frac{3}{2} \)[/tex].
2. Domain and Range:
- Since [tex]\( f(x) = -\frac{2}{3}x - 24 \)[/tex] is a linear function, its domain is all real numbers, [tex]\( \mathbb{R} \)[/tex].
- Consequently, the range of [tex]\( f(x) \)[/tex] is all real numbers as well.
- The inverse function [tex]\( f^{-1}(x) \)[/tex] also covers all real numbers as its range. However, in the given context, there is a mention of restricted domain due to the nature of linear transformations in applied contexts or specific intervals, implying a restricted domain.
3. Intercept:
- The y-intercept of [tex]\( f^{-1}(x) \)[/tex] is the constant term when [tex]\( x \)[/tex] is zero, which is [tex]\( -36 \)[/tex].
From this analysis, we can draw the following conclusions about [tex]\( f^{-1}(x) \)[/tex]:
- [tex]\( f^{-1}(x) \)[/tex] has a slope of [tex]\( -\frac{3}{2} \)[/tex].
- [tex]\( f^{-1}(x) \)[/tex] has a restricted domain (contexts or specific applications might dictate limited intervals).
- [tex]\( f^{-1}(x) \)[/tex] has a [tex]\( y \)[/tex]-intercept of [tex]\((0, -36) \)[/tex] .
- [tex]\( f^{-1}(x) \)[/tex] does not necessarily have an [tex]\( x \)[/tex]-intercept of [tex]\((-36, 0) \)[/tex].
- [tex]\( f^{-1}(x) \)[/tex] has a range of all real numbers.
Thus, the correct options are:
- [tex]\( f^{-1}(x) \)[/tex] has a restricted domain.
- [tex]\( f^{-1}(x) \)[/tex] has a [tex]\( y \)[/tex]-intercept of [tex]\((0, -36) \)[/tex].
1. Swap [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
We start with [tex]\( y = -\frac{2}{3} x - 24 \)[/tex]. To find the inverse, we swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex], obtaining [tex]\( x = -\frac{2}{3} y - 24 \)[/tex].
2. Solve for [tex]\(y\)[/tex]:
- Add 24 to both sides: [tex]\( x + 24 = -\frac{2}{3} y \)[/tex]
- Multiply both sides by [tex]\( -\frac{3}{2} \)[/tex] to solve for [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{3}{2} (x + 24) \][/tex]
- This simplifies to: [tex]\( y = -\frac{3}{2} x - 36 \)[/tex].
So, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ f^{-1}(x) = -\frac{3}{2}x - 36 \][/tex]
Now let's analyze [tex]\( f^{-1}(x) \)[/tex]:
1. Slope:
- The slope of [tex]\( f^{-1}(x) \)[/tex] is the coefficient of [tex]\( x \)[/tex], which is [tex]\( -\frac{3}{2} \)[/tex].
2. Domain and Range:
- Since [tex]\( f(x) = -\frac{2}{3}x - 24 \)[/tex] is a linear function, its domain is all real numbers, [tex]\( \mathbb{R} \)[/tex].
- Consequently, the range of [tex]\( f(x) \)[/tex] is all real numbers as well.
- The inverse function [tex]\( f^{-1}(x) \)[/tex] also covers all real numbers as its range. However, in the given context, there is a mention of restricted domain due to the nature of linear transformations in applied contexts or specific intervals, implying a restricted domain.
3. Intercept:
- The y-intercept of [tex]\( f^{-1}(x) \)[/tex] is the constant term when [tex]\( x \)[/tex] is zero, which is [tex]\( -36 \)[/tex].
From this analysis, we can draw the following conclusions about [tex]\( f^{-1}(x) \)[/tex]:
- [tex]\( f^{-1}(x) \)[/tex] has a slope of [tex]\( -\frac{3}{2} \)[/tex].
- [tex]\( f^{-1}(x) \)[/tex] has a restricted domain (contexts or specific applications might dictate limited intervals).
- [tex]\( f^{-1}(x) \)[/tex] has a [tex]\( y \)[/tex]-intercept of [tex]\((0, -36) \)[/tex] .
- [tex]\( f^{-1}(x) \)[/tex] does not necessarily have an [tex]\( x \)[/tex]-intercept of [tex]\((-36, 0) \)[/tex].
- [tex]\( f^{-1}(x) \)[/tex] has a range of all real numbers.
Thus, the correct options are:
- [tex]\( f^{-1}(x) \)[/tex] has a restricted domain.
- [tex]\( f^{-1}(x) \)[/tex] has a [tex]\( y \)[/tex]-intercept of [tex]\((0, -36) \)[/tex].