Answer :
To determine if the function [tex]\( f(x) = \frac{2}{x-7} \)[/tex] is one-to-one, we need to check if it passes the Horizontal Line Test. A function is one-to-one if and only if every horizontal line intersects its graph at most once.
We start by analyzing the function:
[tex]\[ f(x) = \frac{2}{x-7} \][/tex]
For this function to be one-to-one, each [tex]\( y \)[/tex]-value must correspond to exactly one [tex]\( x \)[/tex]-value. If we were to graph the function [tex]\( f(x) \)[/tex], we would see that it takes on a unique [tex]\( y \)[/tex]-value for each [tex]\( x \)[/tex]-value in its domain ([tex]\( x \ne 7 \)[/tex]). Therefore, [tex]\( f(x) \)[/tex] is one-to-one.
Since the function is one-to-one, we can proceed with finding its inverse.
### a. Solve for the Inverse Function
1. Write the function with [tex]\( y \)[/tex] instead of [tex]\( f(x) \)[/tex]:
[tex]\[ y = \frac{2}{x-7} \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to solve for the inverse:
[tex]\[ x = \frac{2}{y-7} \][/tex]
3. Solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ x(y-7) = 2 \][/tex]
[tex]\[ xy - 7x = 2 \][/tex]
[tex]\[ xy = 2 + 7x \][/tex]
[tex]\[ y = \frac{2 + 7x}{x} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{2 + 7x}{x} \][/tex]
We also note that for [tex]\( f^{-1}(x) \)[/tex], [tex]\( x \neq 0 \)[/tex] since division by zero is undefined.
### b. Graph of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
Both functions [tex]\( f(x) = \frac{2}{x-7} \)[/tex] and [tex]\( f^{-1}(x) = \frac{2 + 7x}{x} \)[/tex] should be graphed on the same set of axes.
1. The graph of [tex]\( f(x) \)[/tex] has a vertical asymptote at [tex]\( x = 7 \)[/tex] and a horizontal asymptote at [tex]\( y = 0 \)[/tex].
2. The graph of [tex]\( f^{-1}(x) \)[/tex] has a vertical asymptote at [tex]\( x = 0 \)[/tex] and a horizontal asymptote at [tex]\( y = 7 \)[/tex].
These graphs intersect the line [tex]\( y = x \)[/tex].
### c. Domain and Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
- For [tex]\( f(x) = \frac{2}{x-7} \)[/tex]:
- Domain: All real numbers except [tex]\( x = 7 \)[/tex] (since the function is undefined at [tex]\( x = 7 \)[/tex]).
- Range: All real numbers except [tex]\( y = 0 \)[/tex].
- For [tex]\( f^{-1}(x) = \frac{2 + 7x}{x} \)[/tex]:
- Domain: All real numbers except [tex]\( x = 0 \)[/tex] (since the function is undefined at [tex]\( x = 0 \)[/tex]).
- Range: All real numbers except [tex]\( y = 7 \)[/tex].
### Summary
- The function [tex]\( f(x) = \frac{2}{x-7} \)[/tex] is one-to-one.
- The inverse function is: [tex]\( f^{-1}(x) = \frac{2 + 7x}{x} \)[/tex], where [tex]\( x \neq 0 \)[/tex].
- The graphs of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] should be plotted, taking note of their respective asymptotes.
- The domain of [tex]\( f \)[/tex] is [tex]\( (-\infty, 7) \cup (7, \infty) \)[/tex] and its range is [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex].
- The domain of [tex]\( f^{-1} \)[/tex] is [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex] and its range is [tex]\( (-\infty, 7) \cup (7, \infty) \)[/tex].
We start by analyzing the function:
[tex]\[ f(x) = \frac{2}{x-7} \][/tex]
For this function to be one-to-one, each [tex]\( y \)[/tex]-value must correspond to exactly one [tex]\( x \)[/tex]-value. If we were to graph the function [tex]\( f(x) \)[/tex], we would see that it takes on a unique [tex]\( y \)[/tex]-value for each [tex]\( x \)[/tex]-value in its domain ([tex]\( x \ne 7 \)[/tex]). Therefore, [tex]\( f(x) \)[/tex] is one-to-one.
Since the function is one-to-one, we can proceed with finding its inverse.
### a. Solve for the Inverse Function
1. Write the function with [tex]\( y \)[/tex] instead of [tex]\( f(x) \)[/tex]:
[tex]\[ y = \frac{2}{x-7} \][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to solve for the inverse:
[tex]\[ x = \frac{2}{y-7} \][/tex]
3. Solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ x(y-7) = 2 \][/tex]
[tex]\[ xy - 7x = 2 \][/tex]
[tex]\[ xy = 2 + 7x \][/tex]
[tex]\[ y = \frac{2 + 7x}{x} \][/tex]
Thus, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{2 + 7x}{x} \][/tex]
We also note that for [tex]\( f^{-1}(x) \)[/tex], [tex]\( x \neq 0 \)[/tex] since division by zero is undefined.
### b. Graph of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
Both functions [tex]\( f(x) = \frac{2}{x-7} \)[/tex] and [tex]\( f^{-1}(x) = \frac{2 + 7x}{x} \)[/tex] should be graphed on the same set of axes.
1. The graph of [tex]\( f(x) \)[/tex] has a vertical asymptote at [tex]\( x = 7 \)[/tex] and a horizontal asymptote at [tex]\( y = 0 \)[/tex].
2. The graph of [tex]\( f^{-1}(x) \)[/tex] has a vertical asymptote at [tex]\( x = 0 \)[/tex] and a horizontal asymptote at [tex]\( y = 7 \)[/tex].
These graphs intersect the line [tex]\( y = x \)[/tex].
### c. Domain and Range of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex]
- For [tex]\( f(x) = \frac{2}{x-7} \)[/tex]:
- Domain: All real numbers except [tex]\( x = 7 \)[/tex] (since the function is undefined at [tex]\( x = 7 \)[/tex]).
- Range: All real numbers except [tex]\( y = 0 \)[/tex].
- For [tex]\( f^{-1}(x) = \frac{2 + 7x}{x} \)[/tex]:
- Domain: All real numbers except [tex]\( x = 0 \)[/tex] (since the function is undefined at [tex]\( x = 0 \)[/tex]).
- Range: All real numbers except [tex]\( y = 7 \)[/tex].
### Summary
- The function [tex]\( f(x) = \frac{2}{x-7} \)[/tex] is one-to-one.
- The inverse function is: [tex]\( f^{-1}(x) = \frac{2 + 7x}{x} \)[/tex], where [tex]\( x \neq 0 \)[/tex].
- The graphs of [tex]\( f \)[/tex] and [tex]\( f^{-1} \)[/tex] should be plotted, taking note of their respective asymptotes.
- The domain of [tex]\( f \)[/tex] is [tex]\( (-\infty, 7) \cup (7, \infty) \)[/tex] and its range is [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex].
- The domain of [tex]\( f^{-1} \)[/tex] is [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex] and its range is [tex]\( (-\infty, 7) \cup (7, \infty) \)[/tex].
