Answer :
To determine whether the function [tex]\( f(x) = \frac{4}{x-5} \)[/tex] is one-to-one, we need to check if each value of [tex]\( f(x) \)[/tex] uniquely corresponds to one value of [tex]\( x \)[/tex].
For the function to be one-to-one, if [tex]\( f(a) = f(b) \)[/tex], then it must be true that [tex]\( a = b \)[/tex].
Let's assume [tex]\( f(a) = f(b) \)[/tex]:
[tex]\[ \frac{4}{a-5} = \frac{4}{b-5} \][/tex]
We can multiply both sides by [tex]\((a-5)(b-5)\)[/tex] to clear the fractions:
[tex]\[ 4(b-5) = 4(a-5) \][/tex]
Divide both sides by 4:
[tex]\[ b-5 = a-5 \][/tex]
Adding 5 to both sides:
[tex]\[ b = a \][/tex]
Thus, we can conclude [tex]\( f \)[/tex] is indeed one-to-one since [tex]\( f(a) = f(b) \)[/tex] leads to [tex]\( a = b \)[/tex].
(a) Write an equation for the inverse function in the form [tex]\( y = f^{-1}(x) \)[/tex].
To find the inverse function, we start by replacing [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = \frac{4}{x-5} \][/tex]
Then we solve for [tex]\( x \)[/tex]. First, we multiply both sides by [tex]\( x-5 \)[/tex]:
[tex]\[ y(x-5) = 4 \][/tex]
Next, we solve for [tex]\( x \)[/tex]:
[tex]\[ yx - 5y = 4 \][/tex]
Add [tex]\( 5y \)[/tex] to both sides:
[tex]\[ yx = 4 + 5y \][/tex]
Divide both sides by [tex]\( y \)[/tex]:
[tex]\[ x = \frac{4 + 5y}{y} \][/tex]
Now, replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to find the inverse function [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}(x) = \frac{4 + 5x}{x} \][/tex]
So,
[tex]\[ f^{-1}(x) = \frac{4 + 5x}{x}, \ x \neq 0 \][/tex]
The correct choice is A:
A. The function is one-to-one. [tex]\( f^{-1}(x) = \frac{4 + 5x}{x}, x \neq 0 \)[/tex].
(b) Graph [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] on the same axes.
1. Graph of [tex]\( f(x) = \frac{4}{x-5} \)[/tex]: This is a hyperbolic function with a vertical asymptote at [tex]\( x = 5 \)[/tex] and a horizontal asymptote at [tex]\( y = 0 \)[/tex].
2. Graph of [tex]\( f^{-1}(x) = \frac{4 + 5x}{x} \)[/tex]: This function will have different behavior due to the terms in the numerator and denominator. It's another rational function but with more complex asymptotes.
(c) Give the domain and range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex].
For [tex]\( f(x) = \frac{4}{x-5} \)[/tex]:
- Domain: All real numbers except [tex]\( x = 5 \)[/tex], as division by zero is undefined.
[tex]\[ \text{Domain of } f: (-\infty, 5) \cup (5, \infty) \][/tex]
- Range: All real numbers except [tex]\( y = 0 \)[/tex], as the function never outputs zero.
[tex]\[ \text{Range of } f: (-\infty, 0) \cup (0, \infty) \][/tex]
For [tex]\( f^{-1}(x) = \frac{4 + 5x}{x} \)[/tex]:
- Domain: All real numbers except [tex]\( x = 0 \)[/tex], as the function involves division by [tex]\( x \)[/tex].
[tex]\[ \text{Domain of } f^{-1}: (-\infty, 0) \cup (0, \infty) \][/tex]
- Range: All real numbers except [tex]\( y = 5 \)[/tex], since it is where the original function’s denominator would be zero.
[tex]\[ \text{Range of } f^{-1}: (-\infty, 5) \cup (5, \infty) \][/tex]
In conclusion:
- The function is one-to-one.
- The inverse function is [tex]\( f^{-1}(x) = \frac{4 + 5x}{x}, x \neq 0 \)[/tex].
- The graph of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] will show both as hyperbolic curves with respective asymptotic behaviors.
- The domain and range of both functions were specified.
For the function to be one-to-one, if [tex]\( f(a) = f(b) \)[/tex], then it must be true that [tex]\( a = b \)[/tex].
Let's assume [tex]\( f(a) = f(b) \)[/tex]:
[tex]\[ \frac{4}{a-5} = \frac{4}{b-5} \][/tex]
We can multiply both sides by [tex]\((a-5)(b-5)\)[/tex] to clear the fractions:
[tex]\[ 4(b-5) = 4(a-5) \][/tex]
Divide both sides by 4:
[tex]\[ b-5 = a-5 \][/tex]
Adding 5 to both sides:
[tex]\[ b = a \][/tex]
Thus, we can conclude [tex]\( f \)[/tex] is indeed one-to-one since [tex]\( f(a) = f(b) \)[/tex] leads to [tex]\( a = b \)[/tex].
(a) Write an equation for the inverse function in the form [tex]\( y = f^{-1}(x) \)[/tex].
To find the inverse function, we start by replacing [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = \frac{4}{x-5} \][/tex]
Then we solve for [tex]\( x \)[/tex]. First, we multiply both sides by [tex]\( x-5 \)[/tex]:
[tex]\[ y(x-5) = 4 \][/tex]
Next, we solve for [tex]\( x \)[/tex]:
[tex]\[ yx - 5y = 4 \][/tex]
Add [tex]\( 5y \)[/tex] to both sides:
[tex]\[ yx = 4 + 5y \][/tex]
Divide both sides by [tex]\( y \)[/tex]:
[tex]\[ x = \frac{4 + 5y}{y} \][/tex]
Now, replace [tex]\( y \)[/tex] with [tex]\( x \)[/tex] to find the inverse function [tex]\( f^{-1}(x) \)[/tex]:
[tex]\[ f^{-1}(x) = \frac{4 + 5x}{x} \][/tex]
So,
[tex]\[ f^{-1}(x) = \frac{4 + 5x}{x}, \ x \neq 0 \][/tex]
The correct choice is A:
A. The function is one-to-one. [tex]\( f^{-1}(x) = \frac{4 + 5x}{x}, x \neq 0 \)[/tex].
(b) Graph [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] on the same axes.
1. Graph of [tex]\( f(x) = \frac{4}{x-5} \)[/tex]: This is a hyperbolic function with a vertical asymptote at [tex]\( x = 5 \)[/tex] and a horizontal asymptote at [tex]\( y = 0 \)[/tex].
2. Graph of [tex]\( f^{-1}(x) = \frac{4 + 5x}{x} \)[/tex]: This function will have different behavior due to the terms in the numerator and denominator. It's another rational function but with more complex asymptotes.
(c) Give the domain and range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex].
For [tex]\( f(x) = \frac{4}{x-5} \)[/tex]:
- Domain: All real numbers except [tex]\( x = 5 \)[/tex], as division by zero is undefined.
[tex]\[ \text{Domain of } f: (-\infty, 5) \cup (5, \infty) \][/tex]
- Range: All real numbers except [tex]\( y = 0 \)[/tex], as the function never outputs zero.
[tex]\[ \text{Range of } f: (-\infty, 0) \cup (0, \infty) \][/tex]
For [tex]\( f^{-1}(x) = \frac{4 + 5x}{x} \)[/tex]:
- Domain: All real numbers except [tex]\( x = 0 \)[/tex], as the function involves division by [tex]\( x \)[/tex].
[tex]\[ \text{Domain of } f^{-1}: (-\infty, 0) \cup (0, \infty) \][/tex]
- Range: All real numbers except [tex]\( y = 5 \)[/tex], since it is where the original function’s denominator would be zero.
[tex]\[ \text{Range of } f^{-1}: (-\infty, 5) \cup (5, \infty) \][/tex]
In conclusion:
- The function is one-to-one.
- The inverse function is [tex]\( f^{-1}(x) = \frac{4 + 5x}{x}, x \neq 0 \)[/tex].
- The graph of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] will show both as hyperbolic curves with respective asymptotic behaviors.
- The domain and range of both functions were specified.
