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Sagot :
To solve the given problem step-by-step, let's break it down into its parts a), b), c), and d).
(a) Find the inverse of [tex]\( f \)[/tex].
To find the inverse of the function [tex]\( f(x) = 4x + 3 \)[/tex]:
1. Start with the equation [tex]\( y = 4x + 3 \)[/tex].
2. Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]: [tex]\( y = 4x + 3 \)[/tex].
3. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y - 3 = 4x \][/tex]
[tex]\[ x = \frac{y - 3}{4} \][/tex]
4. Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] and [tex]\( x \)[/tex] with [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}(x) = \frac{x - 3}{4} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x - 3}{4} \][/tex]
(b) State the domain and range of [tex]\( f \)[/tex].
For the function [tex]\( f(x) = 4x + 3 \)[/tex]:
- The domain of [tex]\( f \)[/tex]: Since [tex]\( f(x) \)[/tex] is a linear function, [tex]\( x \)[/tex] can be any real number. Thus, the domain is:
[tex]\[ \text{Domain of } f: \text{all real numbers} \][/tex]
- The range of [tex]\( f \)[/tex]: Because the function is a linear polynomial with a non-zero slope, it can take any real value as output. Thus, the range is:
[tex]\[ \text{Range of } f: \text{all real numbers} \][/tex]
(c) State the domain and range of [tex]\( f^{-1} \)[/tex].
For the inverse function [tex]\( f^{-1}(x) = \frac{x - 3}{4} \)[/tex]:
- The domain of [tex]\( f^{-1} \)[/tex]: The domain of the inverse function is the range of the original function [tex]\( f \)[/tex]. Since the range of [tex]\( f \)[/tex] is all real numbers, the domain of [tex]\( f^{-1} \)[/tex] is:
[tex]\[ \text{Domain of } f^{-1}: \text{all real numbers} \][/tex]
- The range of [tex]\( f^{-1} \)[/tex]: The range of the inverse function is the domain of the original function [tex]\( f \)[/tex]. Since the domain of [tex]\( f \)[/tex] is all real numbers, the range of [tex]\( f^{-1} \)[/tex] is:
[tex]\[ \text{Range of } f^{-1}: \text{all real numbers} \][/tex]
(d) Graph [tex]\( f \)[/tex], [tex]\( f^{-1} \)[/tex], and [tex]\( y = x \)[/tex] on the same set of axes.
To graph these functions:
1. Graph of [tex]\( f(x) = 4x + 3 \)[/tex]:
- It's a straight line with a slope of 4 and y-intercept of 3.
- Passes through points like [tex]\( (0, 3) \)[/tex] and [tex]\( (1, 7) \)[/tex].
2. Graph of [tex]\( f^{-1}(x) = \frac{x - 3}{4} \)[/tex]:
- It's a straight line with a slope of [tex]\( \frac{1}{4} \)[/tex] and y-intercept of [tex]\( -\frac{3}{4} \)[/tex].
- Passes through points like [tex]\( (3, 0) \)[/tex] and [tex]\( (7, 1) \)[/tex].
3. Graph of [tex]\( y = x \)[/tex]:
- It's the identity line that passes through points like [tex]\( (0, 0) \)[/tex] and [tex]\( (1, 1) \)[/tex].
These three lines should intersect at the points where the original function f and its inverse reflect across the line [tex]\( y = x \)[/tex].
So, the solution to the initial question is:
[tex]\[ f^{-1}(x) = \frac{x - 3}{4} \][/tex]
(a) Find the inverse of [tex]\( f \)[/tex].
To find the inverse of the function [tex]\( f(x) = 4x + 3 \)[/tex]:
1. Start with the equation [tex]\( y = 4x + 3 \)[/tex].
2. Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]: [tex]\( y = 4x + 3 \)[/tex].
3. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y - 3 = 4x \][/tex]
[tex]\[ x = \frac{y - 3}{4} \][/tex]
4. Replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] and [tex]\( x \)[/tex] with [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}(x) = \frac{x - 3}{4} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x - 3}{4} \][/tex]
(b) State the domain and range of [tex]\( f \)[/tex].
For the function [tex]\( f(x) = 4x + 3 \)[/tex]:
- The domain of [tex]\( f \)[/tex]: Since [tex]\( f(x) \)[/tex] is a linear function, [tex]\( x \)[/tex] can be any real number. Thus, the domain is:
[tex]\[ \text{Domain of } f: \text{all real numbers} \][/tex]
- The range of [tex]\( f \)[/tex]: Because the function is a linear polynomial with a non-zero slope, it can take any real value as output. Thus, the range is:
[tex]\[ \text{Range of } f: \text{all real numbers} \][/tex]
(c) State the domain and range of [tex]\( f^{-1} \)[/tex].
For the inverse function [tex]\( f^{-1}(x) = \frac{x - 3}{4} \)[/tex]:
- The domain of [tex]\( f^{-1} \)[/tex]: The domain of the inverse function is the range of the original function [tex]\( f \)[/tex]. Since the range of [tex]\( f \)[/tex] is all real numbers, the domain of [tex]\( f^{-1} \)[/tex] is:
[tex]\[ \text{Domain of } f^{-1}: \text{all real numbers} \][/tex]
- The range of [tex]\( f^{-1} \)[/tex]: The range of the inverse function is the domain of the original function [tex]\( f \)[/tex]. Since the domain of [tex]\( f \)[/tex] is all real numbers, the range of [tex]\( f^{-1} \)[/tex] is:
[tex]\[ \text{Range of } f^{-1}: \text{all real numbers} \][/tex]
(d) Graph [tex]\( f \)[/tex], [tex]\( f^{-1} \)[/tex], and [tex]\( y = x \)[/tex] on the same set of axes.
To graph these functions:
1. Graph of [tex]\( f(x) = 4x + 3 \)[/tex]:
- It's a straight line with a slope of 4 and y-intercept of 3.
- Passes through points like [tex]\( (0, 3) \)[/tex] and [tex]\( (1, 7) \)[/tex].
2. Graph of [tex]\( f^{-1}(x) = \frac{x - 3}{4} \)[/tex]:
- It's a straight line with a slope of [tex]\( \frac{1}{4} \)[/tex] and y-intercept of [tex]\( -\frac{3}{4} \)[/tex].
- Passes through points like [tex]\( (3, 0) \)[/tex] and [tex]\( (7, 1) \)[/tex].
3. Graph of [tex]\( y = x \)[/tex]:
- It's the identity line that passes through points like [tex]\( (0, 0) \)[/tex] and [tex]\( (1, 1) \)[/tex].
These three lines should intersect at the points where the original function f and its inverse reflect across the line [tex]\( y = x \)[/tex].
So, the solution to the initial question is:
[tex]\[ f^{-1}(x) = \frac{x - 3}{4} \][/tex]
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