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Sagot :
Let's work through each part of the problem step by step.
### Part (a): Finding the Inverse of [tex]\( f \)[/tex]
Given the function [tex]\( f(x) = 2x + 1 \)[/tex], we need to find its inverse.
1. Start by writing [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = 2x + 1 \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = 2y + 1 \][/tex]
3. Solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ x - 1 = 2y \][/tex]
[tex]\[ y = \frac{x - 1}{2} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x - 1}{2} \][/tex]
### Part (b): Domain and Range of [tex]\( f \)[/tex]
The function [tex]\( f(x) = 2x + 1 \)[/tex] is a linear function, and for linear functions:
- The domain is all real numbers.
- The range is also all real numbers.
Thus,
[tex]\[ \text{Domain of } f: (-\infty, \infty) \][/tex]
[tex]\[ \text{Range of } f: (-\infty, \infty) \][/tex]
### Part (c): Domain and Range of [tex]\( f^{-1} \)[/tex]
The domain and range of the inverse function [tex]\( f^{-1} \)[/tex] will be the same as the range and domain of the original function [tex]\( f \)[/tex].
- The domain of [tex]\( f^{-1}(x) \)[/tex] is the range of [tex]\( f(x) \)[/tex], which is all real numbers.
- The range of [tex]\( f^{-1}(x) \)[/tex] is the domain of [tex]\( f(x) \)[/tex], which is all real numbers.
Thus,
[tex]\[ \text{Domain of } f^{-1}: (-\infty, \infty) \][/tex]
[tex]\[ \text{Range of } f^{-1}: (-\infty, \infty) \][/tex]
### Part (d): Graph of [tex]\( f \)[/tex], [tex]\( f^{-1} \)[/tex], and [tex]\( y = x \)[/tex]
We need to graph [tex]\( f(x) = 2x + 1 \)[/tex], its inverse [tex]\( f^{-1}(x) = \frac{x - 1}{2} \)[/tex], and the line [tex]\( y = x \)[/tex].
- The graph of [tex]\( f(x) = 2x + 1 \)[/tex] is a straight line with a slope of 2 and a y-intercept of 1.
- The graph of [tex]\( f^{-1}(x) = \frac{x - 1}{2} \)[/tex] is a straight line with a slope of [tex]\(\frac{1}{2}\)[/tex] and a y-intercept of [tex]\(-\frac{1}{2}\)[/tex].
- The line [tex]\( y = x \)[/tex] is a diagonal line passing through the origin with a slope of 1.
Here is a detailed way to represent the data for graphing:
- We use a set of points for [tex]\( x \)[/tex] values ranging from -10 to 10.
- Calculate the corresponding [tex]\( y \)[/tex] values for [tex]\( f(x) \)[/tex], [tex]\( f^{-1}(x) \)[/tex], and [tex]\( y = x \)[/tex].
Based on the calculations, the arrays of [tex]\( x \)[/tex] values, and corresponding [tex]\( y \)[/tex] values for [tex]\( f(x) \)[/tex], [tex]\( f^{-1}(x) \)[/tex], and [tex]\( y = x \)[/tex] are plotted on the same set of axes.
Below is the visual representation to illustrate the graph:
1. Graph of [tex]\( f(x) = 2x + 1 \)[/tex]: A straight line with points ranging from [tex]\((-10, -19) \)[/tex] to [tex]\((10, 21) \)[/tex].
2. Graph of [tex]\( f^{-1}(x) = \frac{x - 1}{2} \)[/tex]: A straight line with points ranging from [tex]\((-10, -5.5) \)[/tex] to [tex]\((10, 4.5) \)[/tex].
3. Graph of [tex]\( y = x \)[/tex]: A straight line from [tex]\((-10, -10) \)[/tex] to [tex]\((10, 10) \)[/tex].
### Final Answer for Part (a)
[tex]\[ f^{-1}(x) = \frac{x - 1}{2} \][/tex]
### Summary of Domain and Range
- Domain of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
- Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
### Part (a): Finding the Inverse of [tex]\( f \)[/tex]
Given the function [tex]\( f(x) = 2x + 1 \)[/tex], we need to find its inverse.
1. Start by writing [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = 2x + 1 \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = 2y + 1 \][/tex]
3. Solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ x - 1 = 2y \][/tex]
[tex]\[ y = \frac{x - 1}{2} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x - 1}{2} \][/tex]
### Part (b): Domain and Range of [tex]\( f \)[/tex]
The function [tex]\( f(x) = 2x + 1 \)[/tex] is a linear function, and for linear functions:
- The domain is all real numbers.
- The range is also all real numbers.
Thus,
[tex]\[ \text{Domain of } f: (-\infty, \infty) \][/tex]
[tex]\[ \text{Range of } f: (-\infty, \infty) \][/tex]
### Part (c): Domain and Range of [tex]\( f^{-1} \)[/tex]
The domain and range of the inverse function [tex]\( f^{-1} \)[/tex] will be the same as the range and domain of the original function [tex]\( f \)[/tex].
- The domain of [tex]\( f^{-1}(x) \)[/tex] is the range of [tex]\( f(x) \)[/tex], which is all real numbers.
- The range of [tex]\( f^{-1}(x) \)[/tex] is the domain of [tex]\( f(x) \)[/tex], which is all real numbers.
Thus,
[tex]\[ \text{Domain of } f^{-1}: (-\infty, \infty) \][/tex]
[tex]\[ \text{Range of } f^{-1}: (-\infty, \infty) \][/tex]
### Part (d): Graph of [tex]\( f \)[/tex], [tex]\( f^{-1} \)[/tex], and [tex]\( y = x \)[/tex]
We need to graph [tex]\( f(x) = 2x + 1 \)[/tex], its inverse [tex]\( f^{-1}(x) = \frac{x - 1}{2} \)[/tex], and the line [tex]\( y = x \)[/tex].
- The graph of [tex]\( f(x) = 2x + 1 \)[/tex] is a straight line with a slope of 2 and a y-intercept of 1.
- The graph of [tex]\( f^{-1}(x) = \frac{x - 1}{2} \)[/tex] is a straight line with a slope of [tex]\(\frac{1}{2}\)[/tex] and a y-intercept of [tex]\(-\frac{1}{2}\)[/tex].
- The line [tex]\( y = x \)[/tex] is a diagonal line passing through the origin with a slope of 1.
Here is a detailed way to represent the data for graphing:
- We use a set of points for [tex]\( x \)[/tex] values ranging from -10 to 10.
- Calculate the corresponding [tex]\( y \)[/tex] values for [tex]\( f(x) \)[/tex], [tex]\( f^{-1}(x) \)[/tex], and [tex]\( y = x \)[/tex].
Based on the calculations, the arrays of [tex]\( x \)[/tex] values, and corresponding [tex]\( y \)[/tex] values for [tex]\( f(x) \)[/tex], [tex]\( f^{-1}(x) \)[/tex], and [tex]\( y = x \)[/tex] are plotted on the same set of axes.
Below is the visual representation to illustrate the graph:
1. Graph of [tex]\( f(x) = 2x + 1 \)[/tex]: A straight line with points ranging from [tex]\((-10, -19) \)[/tex] to [tex]\((10, 21) \)[/tex].
2. Graph of [tex]\( f^{-1}(x) = \frac{x - 1}{2} \)[/tex]: A straight line with points ranging from [tex]\((-10, -5.5) \)[/tex] to [tex]\((10, 4.5) \)[/tex].
3. Graph of [tex]\( y = x \)[/tex]: A straight line from [tex]\((-10, -10) \)[/tex] to [tex]\((10, 10) \)[/tex].
### Final Answer for Part (a)
[tex]\[ f^{-1}(x) = \frac{x - 1}{2} \][/tex]
### Summary of Domain and Range
- Domain of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
- Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
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