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To find the inverse of the function [tex]\( f(x) = \frac{6x}{x+5} \)[/tex], follow these steps:
### Step 1: Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]
[tex]\[ y = \frac{6x}{x+5} \][/tex]
### Step 2: Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]
[tex]\[ y = \frac{6x}{x+5} \][/tex]
First, multiply both sides by [tex]\( x + 5 \)[/tex] to eliminate the denominator:
[tex]\[ y(x + 5) = 6x \][/tex]
Distribute [tex]\( y \)[/tex] on the left-hand side:
[tex]\[ yx + 5y = 6x \][/tex]
Rearrange the equation to isolate terms involving [tex]\( x \)[/tex] on one side:
[tex]\[ yx - 6x = -5y \][/tex]
Factor out [tex]\( x \)[/tex] from the left-hand side:
[tex]\[ x(y - 6) = -5y \][/tex]
Now solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-5y}{y - 6} \][/tex]
So, the inverse function [tex]\( f^{-1}(y) \)[/tex] is:
[tex]\[ f^{-1}(x) = \frac{-5x}{x - 6} \][/tex]
### Checking the Inverse
To verify that this is indeed the inverse, we need to check if [tex]\( f(f^{-1}(x)) = x \)[/tex] and [tex]\( f^{-1}(f(x)) = x \)[/tex].
1. Check [tex]\( f(f^{-1}(x)) \)[/tex]:
Substitute [tex]\( f^{-1}(x) \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f(f^{-1}(x)) = f\left(\frac{-5x}{x - 6}\right) \][/tex]
Now, compute it:
[tex]\[ f\left(\frac{-5x}{x - 6}\right) = \frac{6 \left(\frac{-5x}{x - 6}\right)}{\left(\frac{-5x}{x - 6}\right) + 5} \][/tex]
Simplify the expression to ensure it equals [tex]\( x \)[/tex].
2. Check [tex]\( f^{-1}(f(x)) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( f^{-1} \)[/tex]:
[tex]\[ f^{-1}(f(x)) = f^{-1}\left(\frac{6x}{x + 5}\right) \][/tex]
Now, compute it:
[tex]\[ f^{-1}\left(\frac{6x}{x + 5}\right) = \frac{-5 \left(\frac{6x}{x + 5}\right)}{\left(\frac{6x}{x + 5}\right) - 6} \][/tex]
Simplify the expression to ensure it equals [tex]\( x \)[/tex].
### Part (b) Find the Domain and Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
For the function [tex]\( f(x) = \frac{6x}{x+5} \)[/tex]:
#### Domain of [tex]\( f \)[/tex]:
The domain consists of all real numbers except where the denominator is zero:
[tex]\[ x + 5 \neq 0 \][/tex]
[tex]\[ x \neq -5 \][/tex]
So the domain of [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, -5) \cup (-5, \infty) \][/tex]
#### Range of [tex]\( f \)[/tex]:
To find the range, set [tex]\( y = \frac{6x}{x+5} \)[/tex] and solve for [tex]\( x \)[/tex]:
As derived previously:
[tex]\[ x = \frac{-5y}{y - 6} \][/tex]
The range consists of all real numbers, except the value that makes the denominator zero:
[tex]\[ y - 6 \neq 0 \][/tex]
[tex]\[ y \neq 6 \][/tex]
Thus, the range of [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, 6) \cup (6, \infty) \][/tex]
For the inverse function [tex]\( f^{-1}(x) = \frac{-5x}{x - 6} \)[/tex]:
#### Domain of [tex]\( f^{-1} \)[/tex]:
The domain of the inverse function [tex]\( f^{-1} \)[/tex] is the range of the original function [tex]\( f \)[/tex]:
[tex]\[ (0, 6) \][/tex]
#### Range of [tex]\( f^{-1} \)[/tex]:
The range of the inverse function [tex]\( f^{-1} \)[/tex] is the domain of the original function [tex]\( f \)[/tex]:
[tex]\[ (-5, \infty) \][/tex]
Summarizing the results:
[tex]\[ f^{-1}(x) = \frac{-5x}{x - 6} \][/tex]
[tex]\[ \text{Domain of } f: (-\infty, -5) \cup (-5, \infty) \][/tex]
[tex]\[ \text{Range of } f: (0, 6) \][/tex]
[tex]\[ \text{Domain of } f^{-1}: (0, 6) \][/tex]
[tex]\[ \text{Range of } f^{-1}: (-\infty, -5) \cup (-5, \infty) \][/tex]
The inverse function and the corresponding domains and ranges are fully specified.
### Step 1: Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]
[tex]\[ y = \frac{6x}{x+5} \][/tex]
### Step 2: Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]
[tex]\[ y = \frac{6x}{x+5} \][/tex]
First, multiply both sides by [tex]\( x + 5 \)[/tex] to eliminate the denominator:
[tex]\[ y(x + 5) = 6x \][/tex]
Distribute [tex]\( y \)[/tex] on the left-hand side:
[tex]\[ yx + 5y = 6x \][/tex]
Rearrange the equation to isolate terms involving [tex]\( x \)[/tex] on one side:
[tex]\[ yx - 6x = -5y \][/tex]
Factor out [tex]\( x \)[/tex] from the left-hand side:
[tex]\[ x(y - 6) = -5y \][/tex]
Now solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-5y}{y - 6} \][/tex]
So, the inverse function [tex]\( f^{-1}(y) \)[/tex] is:
[tex]\[ f^{-1}(x) = \frac{-5x}{x - 6} \][/tex]
### Checking the Inverse
To verify that this is indeed the inverse, we need to check if [tex]\( f(f^{-1}(x)) = x \)[/tex] and [tex]\( f^{-1}(f(x)) = x \)[/tex].
1. Check [tex]\( f(f^{-1}(x)) \)[/tex]:
Substitute [tex]\( f^{-1}(x) \)[/tex] into [tex]\( f \)[/tex]:
[tex]\[ f(f^{-1}(x)) = f\left(\frac{-5x}{x - 6}\right) \][/tex]
Now, compute it:
[tex]\[ f\left(\frac{-5x}{x - 6}\right) = \frac{6 \left(\frac{-5x}{x - 6}\right)}{\left(\frac{-5x}{x - 6}\right) + 5} \][/tex]
Simplify the expression to ensure it equals [tex]\( x \)[/tex].
2. Check [tex]\( f^{-1}(f(x)) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( f^{-1} \)[/tex]:
[tex]\[ f^{-1}(f(x)) = f^{-1}\left(\frac{6x}{x + 5}\right) \][/tex]
Now, compute it:
[tex]\[ f^{-1}\left(\frac{6x}{x + 5}\right) = \frac{-5 \left(\frac{6x}{x + 5}\right)}{\left(\frac{6x}{x + 5}\right) - 6} \][/tex]
Simplify the expression to ensure it equals [tex]\( x \)[/tex].
### Part (b) Find the Domain and Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
For the function [tex]\( f(x) = \frac{6x}{x+5} \)[/tex]:
#### Domain of [tex]\( f \)[/tex]:
The domain consists of all real numbers except where the denominator is zero:
[tex]\[ x + 5 \neq 0 \][/tex]
[tex]\[ x \neq -5 \][/tex]
So the domain of [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, -5) \cup (-5, \infty) \][/tex]
#### Range of [tex]\( f \)[/tex]:
To find the range, set [tex]\( y = \frac{6x}{x+5} \)[/tex] and solve for [tex]\( x \)[/tex]:
As derived previously:
[tex]\[ x = \frac{-5y}{y - 6} \][/tex]
The range consists of all real numbers, except the value that makes the denominator zero:
[tex]\[ y - 6 \neq 0 \][/tex]
[tex]\[ y \neq 6 \][/tex]
Thus, the range of [tex]\( f \)[/tex] is:
[tex]\[ (-\infty, 6) \cup (6, \infty) \][/tex]
For the inverse function [tex]\( f^{-1}(x) = \frac{-5x}{x - 6} \)[/tex]:
#### Domain of [tex]\( f^{-1} \)[/tex]:
The domain of the inverse function [tex]\( f^{-1} \)[/tex] is the range of the original function [tex]\( f \)[/tex]:
[tex]\[ (0, 6) \][/tex]
#### Range of [tex]\( f^{-1} \)[/tex]:
The range of the inverse function [tex]\( f^{-1} \)[/tex] is the domain of the original function [tex]\( f \)[/tex]:
[tex]\[ (-5, \infty) \][/tex]
Summarizing the results:
[tex]\[ f^{-1}(x) = \frac{-5x}{x - 6} \][/tex]
[tex]\[ \text{Domain of } f: (-\infty, -5) \cup (-5, \infty) \][/tex]
[tex]\[ \text{Range of } f: (0, 6) \][/tex]
[tex]\[ \text{Domain of } f^{-1}: (0, 6) \][/tex]
[tex]\[ \text{Range of } f^{-1}: (-\infty, -5) \cup (-5, \infty) \][/tex]
The inverse function and the corresponding domains and ranges are fully specified.
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