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Sagot :
To determine which pair of expressions represents inverse functions, we need to verify that for a given pair of functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], the following conditions are met:
1. [tex]\( f(g(x)) = x \)[/tex]
2. [tex]\( g(f(x)) = x \)[/tex]
Let's analyze each option step-by-step:
Option A:
[tex]\[ f(x) = \frac{4 - 3x}{4x - 2}, \quad g(x) = \frac{x + 2}{x - 2} \][/tex]
1. [tex]\( f(g(x)) = f\left(\frac{x + 2}{x - 2}\right) \)[/tex]:
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{x + 2}{x - 2}\right) = \frac{4 - 3\left(\frac{x + 2}{x - 2}\right)}{4\left(\frac{x + 2}{x - 2}\right) - 2} \][/tex]
2. [tex]\( g(f(x)) = g\left(\frac{4 - 3x}{4x - 2}\right) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left(\frac{4 - 3x}{4x - 2}\right) = \frac{\left(\frac{4 - 3x}{4x - 2}\right) + 2}{\left(\frac{4 - 3x}{4x - 2}\right) - 2} \][/tex]
After simplifying, we find that [tex]\( f(g(x)) \neq x \)[/tex] and [tex]\( g(f(x)) \neq x \)[/tex]. Therefore, these functions are not inverses.
Option B:
[tex]\[ f(x) = \frac{x + 3}{4x - 2}, \quad g(x) = \frac{2x + 3}{4x - 1} \][/tex]
1. [tex]\( f(g(x)) = f\left(\frac{2x + 3}{4x - 1}\right) \)[/tex]:
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{2x + 3}{4x - 1}\right) = \frac{\frac{2x + 3}{4x - 1} + 3}{4\left(\frac{2x + 3}{4x - 1}\right) - 2} = x \][/tex]
2. [tex]\( g(f(x)) = g\left(\frac{x + 3}{4x - 2}\right) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left(\frac{x + 3}{4x - 2}\right) = \frac{2\left(\frac{x + 3}{4x - 2}\right) + 3}{4\left(\frac{x + 3}{4x - 2}\right) - 1} = x \][/tex]
These simplifications confirm that [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are indeed inverse functions.
Option C:
[tex]\[ f(x) = \frac{4x + 2}{x - 3}, \quad g(x) = \frac{5x + 3}{4x - 2} \][/tex]
1. [tex]\( f(g(x)) = f\left(\frac{5x + 3}{4x - 2}\right) \)[/tex]:
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{5x + 3}{4x - 2}\right) = \frac{4\left(\frac{5x + 3}{4x - 2}\right) + 2}{\left(\frac{5x + 3}{4x - 2}\right) - 3} \][/tex]
2. [tex]\( g(f(x)) = g\left(\frac{4x + 2}{x - 3}\right) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left(\frac{4x + 2}{x - 3}\right) = \frac{5\left(\frac{4x + 2}{x - 3}\right) + 3}{4\left(\frac{4x + 2}{x - 3}\right) - 2} \][/tex]
After simplification, it is concluded that these expressions do not satisfy the inverse conditions.
Option D:
[tex]\[ f(x) = 2x + 5, \quad g(x) = 2 + 5x \][/tex]
1. [tex]\( f(g(x)) = f(2 + 5x) \)[/tex]:
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(2 + 5x) = 2(2 + 5x) + 5 = 4 + 10x + 5 = 9 + 10x \neq x \][/tex]
2. [tex]\( g(f(x)) = g(2x + 5) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(2x + 5) = 2 + 5(2x + 5) = 2 + 10x + 25 = 27 + 10x \neq x \][/tex]
Therefore, these functions are also not inverses.
After reviewing all options:
The pair of expressions that represents inverse functions are those in Option B:
[tex]\[ \boxed{2} \][/tex]
1. [tex]\( f(g(x)) = x \)[/tex]
2. [tex]\( g(f(x)) = x \)[/tex]
Let's analyze each option step-by-step:
Option A:
[tex]\[ f(x) = \frac{4 - 3x}{4x - 2}, \quad g(x) = \frac{x + 2}{x - 2} \][/tex]
1. [tex]\( f(g(x)) = f\left(\frac{x + 2}{x - 2}\right) \)[/tex]:
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{x + 2}{x - 2}\right) = \frac{4 - 3\left(\frac{x + 2}{x - 2}\right)}{4\left(\frac{x + 2}{x - 2}\right) - 2} \][/tex]
2. [tex]\( g(f(x)) = g\left(\frac{4 - 3x}{4x - 2}\right) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left(\frac{4 - 3x}{4x - 2}\right) = \frac{\left(\frac{4 - 3x}{4x - 2}\right) + 2}{\left(\frac{4 - 3x}{4x - 2}\right) - 2} \][/tex]
After simplifying, we find that [tex]\( f(g(x)) \neq x \)[/tex] and [tex]\( g(f(x)) \neq x \)[/tex]. Therefore, these functions are not inverses.
Option B:
[tex]\[ f(x) = \frac{x + 3}{4x - 2}, \quad g(x) = \frac{2x + 3}{4x - 1} \][/tex]
1. [tex]\( f(g(x)) = f\left(\frac{2x + 3}{4x - 1}\right) \)[/tex]:
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{2x + 3}{4x - 1}\right) = \frac{\frac{2x + 3}{4x - 1} + 3}{4\left(\frac{2x + 3}{4x - 1}\right) - 2} = x \][/tex]
2. [tex]\( g(f(x)) = g\left(\frac{x + 3}{4x - 2}\right) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left(\frac{x + 3}{4x - 2}\right) = \frac{2\left(\frac{x + 3}{4x - 2}\right) + 3}{4\left(\frac{x + 3}{4x - 2}\right) - 1} = x \][/tex]
These simplifications confirm that [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex]. Therefore, [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are indeed inverse functions.
Option C:
[tex]\[ f(x) = \frac{4x + 2}{x - 3}, \quad g(x) = \frac{5x + 3}{4x - 2} \][/tex]
1. [tex]\( f(g(x)) = f\left(\frac{5x + 3}{4x - 2}\right) \)[/tex]:
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{5x + 3}{4x - 2}\right) = \frac{4\left(\frac{5x + 3}{4x - 2}\right) + 2}{\left(\frac{5x + 3}{4x - 2}\right) - 3} \][/tex]
2. [tex]\( g(f(x)) = g\left(\frac{4x + 2}{x - 3}\right) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left(\frac{4x + 2}{x - 3}\right) = \frac{5\left(\frac{4x + 2}{x - 3}\right) + 3}{4\left(\frac{4x + 2}{x - 3}\right) - 2} \][/tex]
After simplification, it is concluded that these expressions do not satisfy the inverse conditions.
Option D:
[tex]\[ f(x) = 2x + 5, \quad g(x) = 2 + 5x \][/tex]
1. [tex]\( f(g(x)) = f(2 + 5x) \)[/tex]:
Substitute [tex]\( g(x) \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[ f(2 + 5x) = 2(2 + 5x) + 5 = 4 + 10x + 5 = 9 + 10x \neq x \][/tex]
2. [tex]\( g(f(x)) = g(2x + 5) \)[/tex]:
Substitute [tex]\( f(x) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(2x + 5) = 2 + 5(2x + 5) = 2 + 10x + 25 = 27 + 10x \neq x \][/tex]
Therefore, these functions are also not inverses.
After reviewing all options:
The pair of expressions that represents inverse functions are those in Option B:
[tex]\[ \boxed{2} \][/tex]
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