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Sagot :
To determine whether the function [tex]\(f(x) = 5x - 8\)[/tex] is one-to-one, we need to check if it satisfies the requirements of a one-to-one function. A function is one-to-one if and only if every y-value corresponds to exactly one x-value.
### Step-by-Step Solution
#### Step 1: Verify if the function is one-to-one
One way to check if a function is one-to-one is to use the Horizontal Line Test: if no horizontal line intersects the graph of the function more than once, then the function is one-to-one.
For [tex]\(f(x) = 5x - 8\)[/tex]:
- The function is a linear equation with a non-zero slope (in this case, 5).
- Linear functions with non-zero slopes are always one-to-one because the slope ensures that the function is always increasing or always decreasing, which means no horizontal line will intersect the graph more than once.
Therefore, [tex]\(f(x) = 5x - 8\)[/tex] is one-to-one.
---
#### Step 2: Find the inverse function
To find the inverse of the function [tex]\(f(x) = 5x - 8\)[/tex], we need to express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]. Follow these algebraic steps:
1. Start with the equation [tex]\(y = 5x - 8\)[/tex].
2. Switch the roles of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] to find the inverse:
[tex]\[ x = 5y - 8 \][/tex]
3. Solve for [tex]\(y\)[/tex]:
[tex]\[ x + 8 = 5y \][/tex]
[tex]\[ y = \frac{x + 8}{5} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x + 8}{5} \][/tex]
Thus, the correct choice is:
A. The function [tex]\( f(x) \)[/tex] is one-to-one and [tex]\( f^{-1}(x) = \frac{x + 8}{5} \)[/tex].
---
#### Step 3: Graph [tex]\(f(x)\)[/tex] and [tex]\(f^{-1}(x)\)[/tex]
To graph both [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] on the same axes:
- The graph of [tex]\( f(x) = 5x - 8 \)[/tex] is a straight line with slope 5 and y-intercept -8.
- The graph of [tex]\( f^{-1}(x) = \frac{x + 8}{5} \)[/tex] is a straight line with slope [tex]\( \frac{1}{5} \)[/tex] and y-intercept [tex]\( \frac{8}{5} \approx 1.6 \)[/tex].
The line [tex]\( y = x \)[/tex] should be included as it is the line of reflection for the function and its inverse.
---
#### Step 4: Determine the domain and range of [tex]\(f\)[/tex] and [tex]\(f^{-1}\)[/tex]
For the function [tex]\( f(x) = 5x - 8 \)[/tex]:
- The domain of [tex]\( f \)[/tex]: All real numbers ([tex]\( -\infty, \infty \)[/tex])
- The range of [tex]\( f \)[/tex]: All real numbers ([tex]\( -\infty, \infty \)[/tex])
For the inverse function [tex]\( f^{-1}(x) = \frac{x + 8}{5} \)[/tex]:
- The domain of [tex]\( f^{-1} \)[/tex]: All real numbers ([tex]\( -\infty, \infty \)[/tex])
- The range of [tex]\( f^{-1} \)[/tex]: All real numbers ([tex]\( -\infty, \infty \)[/tex])
### Summary:
- The function [tex]\( f(x) = 5x - 8 \)[/tex] is one-to-one.
- Inverse function: [tex]\( f^{-1}(x) = \frac{x + 8}{5} \)[/tex]
- Domain and Range:
- [tex]\( f(x): \)[/tex] Domain = All real numbers, Range = All real numbers
- [tex]\( f^{-1}(x): \)[/tex] Domain = All real numbers, Range = All real numbers
### Step-by-Step Solution
#### Step 1: Verify if the function is one-to-one
One way to check if a function is one-to-one is to use the Horizontal Line Test: if no horizontal line intersects the graph of the function more than once, then the function is one-to-one.
For [tex]\(f(x) = 5x - 8\)[/tex]:
- The function is a linear equation with a non-zero slope (in this case, 5).
- Linear functions with non-zero slopes are always one-to-one because the slope ensures that the function is always increasing or always decreasing, which means no horizontal line will intersect the graph more than once.
Therefore, [tex]\(f(x) = 5x - 8\)[/tex] is one-to-one.
---
#### Step 2: Find the inverse function
To find the inverse of the function [tex]\(f(x) = 5x - 8\)[/tex], we need to express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]. Follow these algebraic steps:
1. Start with the equation [tex]\(y = 5x - 8\)[/tex].
2. Switch the roles of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] to find the inverse:
[tex]\[ x = 5y - 8 \][/tex]
3. Solve for [tex]\(y\)[/tex]:
[tex]\[ x + 8 = 5y \][/tex]
[tex]\[ y = \frac{x + 8}{5} \][/tex]
So, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{x + 8}{5} \][/tex]
Thus, the correct choice is:
A. The function [tex]\( f(x) \)[/tex] is one-to-one and [tex]\( f^{-1}(x) = \frac{x + 8}{5} \)[/tex].
---
#### Step 3: Graph [tex]\(f(x)\)[/tex] and [tex]\(f^{-1}(x)\)[/tex]
To graph both [tex]\( f(x) \)[/tex] and [tex]\( f^{-1}(x) \)[/tex] on the same axes:
- The graph of [tex]\( f(x) = 5x - 8 \)[/tex] is a straight line with slope 5 and y-intercept -8.
- The graph of [tex]\( f^{-1}(x) = \frac{x + 8}{5} \)[/tex] is a straight line with slope [tex]\( \frac{1}{5} \)[/tex] and y-intercept [tex]\( \frac{8}{5} \approx 1.6 \)[/tex].
The line [tex]\( y = x \)[/tex] should be included as it is the line of reflection for the function and its inverse.
---
#### Step 4: Determine the domain and range of [tex]\(f\)[/tex] and [tex]\(f^{-1}\)[/tex]
For the function [tex]\( f(x) = 5x - 8 \)[/tex]:
- The domain of [tex]\( f \)[/tex]: All real numbers ([tex]\( -\infty, \infty \)[/tex])
- The range of [tex]\( f \)[/tex]: All real numbers ([tex]\( -\infty, \infty \)[/tex])
For the inverse function [tex]\( f^{-1}(x) = \frac{x + 8}{5} \)[/tex]:
- The domain of [tex]\( f^{-1} \)[/tex]: All real numbers ([tex]\( -\infty, \infty \)[/tex])
- The range of [tex]\( f^{-1} \)[/tex]: All real numbers ([tex]\( -\infty, \infty \)[/tex])
### Summary:
- The function [tex]\( f(x) = 5x - 8 \)[/tex] is one-to-one.
- Inverse function: [tex]\( f^{-1}(x) = \frac{x + 8}{5} \)[/tex]
- Domain and Range:
- [tex]\( f(x): \)[/tex] Domain = All real numbers, Range = All real numbers
- [tex]\( f^{-1}(x): \)[/tex] Domain = All real numbers, Range = All real numbers
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