Answer :
Let's methodically work through each part of the given problem.
### (a) Finding the Inverse of [tex]\( f \)[/tex]
The function given is:
[tex]\[ f(x) = 8x + 1 \][/tex]
To find the inverse, [tex]\( f^{-1} \)[/tex], we need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = 8x + 1 \][/tex]
First, isolate [tex]\( x \)[/tex]:
[tex]\[ y - 1 = 8x \][/tex]
[tex]\[ x = \frac{y - 1}{8} \][/tex]
Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x - 1}{8} \][/tex]
### (b) Domain and Range of [tex]\( f \)[/tex]
For the function [tex]\( f(x) = 8x + 1 \)[/tex]:
- Since [tex]\( f(x) \)[/tex] is a linear function, it is defined for all real numbers. Thus, the domain of [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
- Because [tex]\( 8x + 1 \)[/tex] covers all real numbers as [tex]\( x \)[/tex] varies over all real numbers, the range of [tex]\( f \)[/tex] is also:
[tex]\[ (-\infty, \infty) \][/tex]
### (c) Domain and Range of [tex]\( f^{-1} \)[/tex]
For the inverse function [tex]\( f^{-1}(x) = \frac{x - 1}{8} \)[/tex]:
- The domain of [tex]\( f^{-1} \)[/tex] is the range of [tex]\( f \)[/tex], which is:
[tex]\[ (-\infty, \infty) \][/tex]
- The range of [tex]\( f^{-1} \)[/tex] is the domain of [tex]\( f \)[/tex], which is:
[tex]\[ (-\infty, \infty) \][/tex]
### (d) Graphs of [tex]\( f \)[/tex], [tex]\( f^{-1} \)[/tex], and [tex]\( y = x \)[/tex]
To graph [tex]\( f \)[/tex], [tex]\( f^{-1} \)[/tex], and [tex]\( y = x \)[/tex]:
1. Graph of [tex]\( f(x) = 8x + 1 \)[/tex]:
- It is a straight line with a slope of 8 and a y-intercept of 1.
- Points to plot could include [tex]\( (0, 1) \)[/tex] and [tex]\( (1, 9) \)[/tex].
2. Graph of [tex]\( f^{-1}(x) = \frac{x - 1}{8} \)[/tex]:
- It is a straight line with a slope of [tex]\(\frac{1}{8}\)[/tex] and a y-intercept of [tex]\( \frac{-1}{8} \)[/tex].
- Points to plot could include [tex]\( (0, -\frac{1}{8}) \)[/tex] and [tex]\( (8, 7) \)[/tex].
3. Graph of [tex]\( y = x \)[/tex]:
- This is a straight line with a slope of 1 and passes through the origin.
- Points to plot could include [tex]\( (0, 0) \)[/tex] and [tex]\( (1, 1) \)[/tex].
These three lines should be plotted on the same coordinate axes to visualize how [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] are reflections over the line [tex]\( y = x \)[/tex].
### Summary of Results
(a) The inverse of [tex]\( f \)[/tex]:
[tex]\[ f^{-1}(x) = \frac{x - 1}{8} \][/tex]
(b) Domain and Range of [tex]\( f \)[/tex]:
Domain of [tex]\( f \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
Range of [tex]\( f \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
(c) Domain and Range of [tex]\( f^{-1} \)[/tex]:
Domain of [tex]\( f^{-1} \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
Range of [tex]\( f^{-1} \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
Remember to carefully plot and check the proper scales when drawing the graphs to ensure accurate representation.
### (a) Finding the Inverse of [tex]\( f \)[/tex]
The function given is:
[tex]\[ f(x) = 8x + 1 \][/tex]
To find the inverse, [tex]\( f^{-1} \)[/tex], we need to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = 8x + 1 \][/tex]
First, isolate [tex]\( x \)[/tex]:
[tex]\[ y - 1 = 8x \][/tex]
[tex]\[ x = \frac{y - 1}{8} \][/tex]
Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x - 1}{8} \][/tex]
### (b) Domain and Range of [tex]\( f \)[/tex]
For the function [tex]\( f(x) = 8x + 1 \)[/tex]:
- Since [tex]\( f(x) \)[/tex] is a linear function, it is defined for all real numbers. Thus, the domain of [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
- Because [tex]\( 8x + 1 \)[/tex] covers all real numbers as [tex]\( x \)[/tex] varies over all real numbers, the range of [tex]\( f \)[/tex] is also:
[tex]\[ (-\infty, \infty) \][/tex]
### (c) Domain and Range of [tex]\( f^{-1} \)[/tex]
For the inverse function [tex]\( f^{-1}(x) = \frac{x - 1}{8} \)[/tex]:
- The domain of [tex]\( f^{-1} \)[/tex] is the range of [tex]\( f \)[/tex], which is:
[tex]\[ (-\infty, \infty) \][/tex]
- The range of [tex]\( f^{-1} \)[/tex] is the domain of [tex]\( f \)[/tex], which is:
[tex]\[ (-\infty, \infty) \][/tex]
### (d) Graphs of [tex]\( f \)[/tex], [tex]\( f^{-1} \)[/tex], and [tex]\( y = x \)[/tex]
To graph [tex]\( f \)[/tex], [tex]\( f^{-1} \)[/tex], and [tex]\( y = x \)[/tex]:
1. Graph of [tex]\( f(x) = 8x + 1 \)[/tex]:
- It is a straight line with a slope of 8 and a y-intercept of 1.
- Points to plot could include [tex]\( (0, 1) \)[/tex] and [tex]\( (1, 9) \)[/tex].
2. Graph of [tex]\( f^{-1}(x) = \frac{x - 1}{8} \)[/tex]:
- It is a straight line with a slope of [tex]\(\frac{1}{8}\)[/tex] and a y-intercept of [tex]\( \frac{-1}{8} \)[/tex].
- Points to plot could include [tex]\( (0, -\frac{1}{8}) \)[/tex] and [tex]\( (8, 7) \)[/tex].
3. Graph of [tex]\( y = x \)[/tex]:
- This is a straight line with a slope of 1 and passes through the origin.
- Points to plot could include [tex]\( (0, 0) \)[/tex] and [tex]\( (1, 1) \)[/tex].
These three lines should be plotted on the same coordinate axes to visualize how [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] are reflections over the line [tex]\( y = x \)[/tex].
### Summary of Results
(a) The inverse of [tex]\( f \)[/tex]:
[tex]\[ f^{-1}(x) = \frac{x - 1}{8} \][/tex]
(b) Domain and Range of [tex]\( f \)[/tex]:
Domain of [tex]\( f \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
Range of [tex]\( f \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
(c) Domain and Range of [tex]\( f^{-1} \)[/tex]:
Domain of [tex]\( f^{-1} \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
Range of [tex]\( f^{-1} \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
Remember to carefully plot and check the proper scales when drawing the graphs to ensure accurate representation.
