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Sagot :
To determine for which function [tex]\( f(x) \)[/tex] is equal to its inverse [tex]\( f^{-1}(x) \)[/tex], let's analyze each of the given functions step by step.
### Function A: [tex]\( f(x) = \frac{x+6}{x-6} \)[/tex]
To find the inverse [tex]\( f^{-1}(x) \)[/tex], we start by switching [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the function and then solve for [tex]\( y \)[/tex].
1. [tex]\( y = \frac{x + 6}{x - 6} \)[/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: [tex]\( x = \frac{y + 6}{y - 6} \)[/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x(y - 6) = y + 6 \\ xy - 6x = y + 6 \\ xy - y = 6 + 6x \\ y(x - 1) = 6(x + 1) \\ y = \frac{6(x + 1)}{x - 1} \][/tex]
So, [tex]\( f^{-1}(x) = \frac{6(x + 1)}{x - 1} \)[/tex].
Clearly, [tex]\( f(x) \neq f^{-1}(x) \)[/tex].
### Function B: [tex]\( f(x) = \frac{x+2}{x-2} \)[/tex]
Similarly, find the inverse [tex]\( f^{-1}(x) \)[/tex].
1. [tex]\( y = \frac{x + 2}{x - 2} \)[/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: [tex]\( x = \frac{y + 2}{y - 2} \)[/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x(y - 2) = y + 2 \\ xy - 2x = y + 2 \\ xy - y = 2 + 2x \\ y(x - 1) = 2(x + 1) \\ y = \frac{2(x + 1)}{x - 1} \][/tex]
So, [tex]\( f^{-1}(x) = \frac{2(x + 1)}{x - 1} \)[/tex].
Clearly, [tex]\( f(x) \neq f^{-1}(x) \)[/tex].
### Function C: [tex]\( f(x) = \frac{x + 1}{x - 1} \)[/tex]
Similarly, find the inverse [tex]\( f^{-1}(x) \)[/tex].
1. [tex]\( y = \frac{x + 1}{x - 1} \)[/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: [tex]\( x = \frac{y + 1}{y - 1} \)[/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x(y - 1) = y + 1 \\ xy - x = y + 1 \\ xy - y = x + 1 \\ y(x - 1) = x + 1 \\ y = \frac{x + 1}{x - 1} \][/tex]
We find that [tex]\( f^{-1}(x) = \frac{x + 1}{x - 1} \)[/tex].
Clearly, [tex]\( f(x) = f^{-1}(x) \)[/tex].
### Function D: [tex]\( f(x) = \frac{x + 5}{x - 5} \)[/tex]
Similarly, find the inverse [tex]\( f^{-1}(x) \)[/tex].
1. [tex]\( y = \frac{x + 5}{x - 5} \)[/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: [tex]\( x = \frac{y + 5}{y - 5} \)[/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x(y - 5) = y + 5 \\ xy - 5x = y + 5 \\ xy - y = 5 + 5x \\ y(x - 1) = 5(x + 1) \\ y = \frac{5(x + 1)}{x - 1} \][/tex]
So, [tex]\( f^{-1}(x) = \frac{5(x + 1)}{x - 1} \)[/tex].
Clearly, [tex]\( f(x) \neq f^{-1}(x) \)[/tex].
Therefore, the function for which [tex]\( f(x) \)[/tex] is equal to its inverse [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ \boxed{f(x) = \frac{x + 1}{x - 1}} \][/tex] (Option C)
### Function A: [tex]\( f(x) = \frac{x+6}{x-6} \)[/tex]
To find the inverse [tex]\( f^{-1}(x) \)[/tex], we start by switching [tex]\( x \)[/tex] and [tex]\( y \)[/tex] in the function and then solve for [tex]\( y \)[/tex].
1. [tex]\( y = \frac{x + 6}{x - 6} \)[/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: [tex]\( x = \frac{y + 6}{y - 6} \)[/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x(y - 6) = y + 6 \\ xy - 6x = y + 6 \\ xy - y = 6 + 6x \\ y(x - 1) = 6(x + 1) \\ y = \frac{6(x + 1)}{x - 1} \][/tex]
So, [tex]\( f^{-1}(x) = \frac{6(x + 1)}{x - 1} \)[/tex].
Clearly, [tex]\( f(x) \neq f^{-1}(x) \)[/tex].
### Function B: [tex]\( f(x) = \frac{x+2}{x-2} \)[/tex]
Similarly, find the inverse [tex]\( f^{-1}(x) \)[/tex].
1. [tex]\( y = \frac{x + 2}{x - 2} \)[/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: [tex]\( x = \frac{y + 2}{y - 2} \)[/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x(y - 2) = y + 2 \\ xy - 2x = y + 2 \\ xy - y = 2 + 2x \\ y(x - 1) = 2(x + 1) \\ y = \frac{2(x + 1)}{x - 1} \][/tex]
So, [tex]\( f^{-1}(x) = \frac{2(x + 1)}{x - 1} \)[/tex].
Clearly, [tex]\( f(x) \neq f^{-1}(x) \)[/tex].
### Function C: [tex]\( f(x) = \frac{x + 1}{x - 1} \)[/tex]
Similarly, find the inverse [tex]\( f^{-1}(x) \)[/tex].
1. [tex]\( y = \frac{x + 1}{x - 1} \)[/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: [tex]\( x = \frac{y + 1}{y - 1} \)[/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x(y - 1) = y + 1 \\ xy - x = y + 1 \\ xy - y = x + 1 \\ y(x - 1) = x + 1 \\ y = \frac{x + 1}{x - 1} \][/tex]
We find that [tex]\( f^{-1}(x) = \frac{x + 1}{x - 1} \)[/tex].
Clearly, [tex]\( f(x) = f^{-1}(x) \)[/tex].
### Function D: [tex]\( f(x) = \frac{x + 5}{x - 5} \)[/tex]
Similarly, find the inverse [tex]\( f^{-1}(x) \)[/tex].
1. [tex]\( y = \frac{x + 5}{x - 5} \)[/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]: [tex]\( x = \frac{y + 5}{y - 5} \)[/tex]
3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x(y - 5) = y + 5 \\ xy - 5x = y + 5 \\ xy - y = 5 + 5x \\ y(x - 1) = 5(x + 1) \\ y = \frac{5(x + 1)}{x - 1} \][/tex]
So, [tex]\( f^{-1}(x) = \frac{5(x + 1)}{x - 1} \)[/tex].
Clearly, [tex]\( f(x) \neq f^{-1}(x) \)[/tex].
Therefore, the function for which [tex]\( f(x) \)[/tex] is equal to its inverse [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ \boxed{f(x) = \frac{x + 1}{x - 1}} \][/tex] (Option C)
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