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Sagot :
To determine whether the function [tex]\( f(x) = 3x - 5 \)[/tex] is one-to-one, follow these steps:
Step 1: Verify if the function is one-to-one:
A function [tex]\( f(x) \)[/tex] is one-to-one if for any two different inputs [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex], the outputs [tex]\( f(x_1) \)[/tex] and [tex]\( f(x_2) \)[/tex] are different. Mathematically, [tex]\( f(x_1) = f(x_2) \)[/tex] implies [tex]\( x_1 = x_2 \)[/tex].
For [tex]\( f(x) = 3x - 5 \)[/tex]:
Suppose [tex]\( f(x_1) = f(x_2) \)[/tex].
[tex]\[ 3x_1 - 5 = 3x_2 - 5 \][/tex]
Adding 5 to both sides:
[tex]\[ 3x_1 = 3x_2 \][/tex]
Dividing both sides by 3:
[tex]\[ x_1 = x_2 \][/tex]
Since [tex]\( x_1 = x_2 \)[/tex] is always true for any different inputs [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex], [tex]\( f(x) = 3x - 5 \)[/tex] is a one-to-one function.
Step 2: Find the inverse function:
To find the inverse function [tex]\( f^{-1}(x) \)[/tex], we need to solve the equation [tex]\( y = 3x - 5 \)[/tex] for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
Starting from the equation:
[tex]\[ y = 3x - 5 \][/tex]
Add 5 to both sides:
[tex]\[ y + 5 = 3x \][/tex]
Divide both sides by 3:
[tex]\[ x = \frac{y + 5}{3} \][/tex]
Thus, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ y = f^{-1}(x) = \frac{x + 5}{3} \][/tex]
Step 3: Graph [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex]:
To graph [tex]\( f(x) = 3x - 5 \)[/tex] and its inverse [tex]\( f^{-1}(x) = \frac{x + 5}{3} \)[/tex] on the same axes:
1. Plot the function [tex]\( f(x) = 3x - 5 \)[/tex].
2. Plot the inverse function [tex]\( f^{-1}(x) = \frac{x + 5}{3} \)[/tex].
3. The graphs of these functions should be reflections of each other across the line [tex]\( y = x \)[/tex].
Step 4: Determine the domain and range:
1. Domain of [tex]\( f(x) \)[/tex]:
Since [tex]\( f(x) \)[/tex] is a linear function and defined for all real numbers, its domain is all real numbers:
[tex]\[ \text{Domain of } f(x) = (-\infty, \infty) \][/tex]
2. Range of [tex]\( f(x) \)[/tex]:
As a linear function, [tex]\( f(x) \)[/tex] can produce any real number output, so its range is also all real numbers:
[tex]\[ \text{Range of } f(x) = (-\infty, \infty) \][/tex]
3. Domain of [tex]\( f^{-1}(x) \)[/tex]:
Since [tex]\( f^{-1}(x) \)[/tex] is also a linear function and defined for all real numbers, its domain is all real numbers:
[tex]\[ \text{Domain of } f^{-1}(x) = (-\infty, \infty) \][/tex]
4. Range of [tex]\( f^{-1}(x) \)[/tex]:
As a linear function, [tex]\( f^{-1}(x) \)[/tex] can produce any real number output, so its range is:
[tex]\[ \text{Range of } f^{-1}(x) = (-\infty, \infty) \][/tex]
Conclusion for part (a):
The function [tex]\( f(x) \)[/tex] is one-to-one.
The equation for the inverse function is:
[tex]\[ y = f^{-1}(x) = \frac{x + 5}{3} \][/tex]
So the correct choice is:
A. The function [tex]\( f(x) \)[/tex] is one-to-one and [tex]\( f^{-1}(x) = \frac{x + 5}{3} \)[/tex].
Graphing:
While I cannot graph directly here, you can plot the lines [tex]\( f(x) = 3x - 5 \)[/tex] and [tex]\( f^{-1}(x) = \frac{x + 5}{3} \)[/tex] using graphing software or graph paper. Remember, the two should be symmetric with respect to the line [tex]\( y = x \)[/tex].
Domain and Range Summary:
- Domain of [tex]\( f(x) \)[/tex]: All real numbers [tex]\( (-\infty, \infty) \)[/tex]
- Range of [tex]\( f(x) \)[/tex]: All real numbers [tex]\( (-\infty, \infty) \)[/tex]
- Domain of [tex]\( f^{-1}(x) \)[/tex]: All real numbers [tex]\( (-\infty, \infty) \)[/tex]
- Range of [tex]\( f^{-1}(x) \)[/tex]: All real numbers [tex]\( (-\infty, \infty) \)[/tex]
Step 1: Verify if the function is one-to-one:
A function [tex]\( f(x) \)[/tex] is one-to-one if for any two different inputs [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex], the outputs [tex]\( f(x_1) \)[/tex] and [tex]\( f(x_2) \)[/tex] are different. Mathematically, [tex]\( f(x_1) = f(x_2) \)[/tex] implies [tex]\( x_1 = x_2 \)[/tex].
For [tex]\( f(x) = 3x - 5 \)[/tex]:
Suppose [tex]\( f(x_1) = f(x_2) \)[/tex].
[tex]\[ 3x_1 - 5 = 3x_2 - 5 \][/tex]
Adding 5 to both sides:
[tex]\[ 3x_1 = 3x_2 \][/tex]
Dividing both sides by 3:
[tex]\[ x_1 = x_2 \][/tex]
Since [tex]\( x_1 = x_2 \)[/tex] is always true for any different inputs [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex], [tex]\( f(x) = 3x - 5 \)[/tex] is a one-to-one function.
Step 2: Find the inverse function:
To find the inverse function [tex]\( f^{-1}(x) \)[/tex], we need to solve the equation [tex]\( y = 3x - 5 \)[/tex] for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex].
Starting from the equation:
[tex]\[ y = 3x - 5 \][/tex]
Add 5 to both sides:
[tex]\[ y + 5 = 3x \][/tex]
Divide both sides by 3:
[tex]\[ x = \frac{y + 5}{3} \][/tex]
Thus, the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ y = f^{-1}(x) = \frac{x + 5}{3} \][/tex]
Step 3: Graph [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex]:
To graph [tex]\( f(x) = 3x - 5 \)[/tex] and its inverse [tex]\( f^{-1}(x) = \frac{x + 5}{3} \)[/tex] on the same axes:
1. Plot the function [tex]\( f(x) = 3x - 5 \)[/tex].
2. Plot the inverse function [tex]\( f^{-1}(x) = \frac{x + 5}{3} \)[/tex].
3. The graphs of these functions should be reflections of each other across the line [tex]\( y = x \)[/tex].
Step 4: Determine the domain and range:
1. Domain of [tex]\( f(x) \)[/tex]:
Since [tex]\( f(x) \)[/tex] is a linear function and defined for all real numbers, its domain is all real numbers:
[tex]\[ \text{Domain of } f(x) = (-\infty, \infty) \][/tex]
2. Range of [tex]\( f(x) \)[/tex]:
As a linear function, [tex]\( f(x) \)[/tex] can produce any real number output, so its range is also all real numbers:
[tex]\[ \text{Range of } f(x) = (-\infty, \infty) \][/tex]
3. Domain of [tex]\( f^{-1}(x) \)[/tex]:
Since [tex]\( f^{-1}(x) \)[/tex] is also a linear function and defined for all real numbers, its domain is all real numbers:
[tex]\[ \text{Domain of } f^{-1}(x) = (-\infty, \infty) \][/tex]
4. Range of [tex]\( f^{-1}(x) \)[/tex]:
As a linear function, [tex]\( f^{-1}(x) \)[/tex] can produce any real number output, so its range is:
[tex]\[ \text{Range of } f^{-1}(x) = (-\infty, \infty) \][/tex]
Conclusion for part (a):
The function [tex]\( f(x) \)[/tex] is one-to-one.
The equation for the inverse function is:
[tex]\[ y = f^{-1}(x) = \frac{x + 5}{3} \][/tex]
So the correct choice is:
A. The function [tex]\( f(x) \)[/tex] is one-to-one and [tex]\( f^{-1}(x) = \frac{x + 5}{3} \)[/tex].
Graphing:
While I cannot graph directly here, you can plot the lines [tex]\( f(x) = 3x - 5 \)[/tex] and [tex]\( f^{-1}(x) = \frac{x + 5}{3} \)[/tex] using graphing software or graph paper. Remember, the two should be symmetric with respect to the line [tex]\( y = x \)[/tex].
Domain and Range Summary:
- Domain of [tex]\( f(x) \)[/tex]: All real numbers [tex]\( (-\infty, \infty) \)[/tex]
- Range of [tex]\( f(x) \)[/tex]: All real numbers [tex]\( (-\infty, \infty) \)[/tex]
- Domain of [tex]\( f^{-1}(x) \)[/tex]: All real numbers [tex]\( (-\infty, \infty) \)[/tex]
- Range of [tex]\( f^{-1}(x) \)[/tex]: All real numbers [tex]\( (-\infty, \infty) \)[/tex]
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