Answer :
To determine whether a pair of expressions represents inverse functions, we need to check if the given pairs of expressions, say [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], satisfy the conditions:
[tex]\[ f(g(x)) = x \][/tex]
and
[tex]\[ g(f(x)) = x \][/tex]
Let's verify each option step by step.
### Option A:
[tex]\[ f(x) = \frac{4 - 3x}{4x - 2} \][/tex]
[tex]\[ g(x) = \frac{x + 2}{x - 2} \][/tex]
Calculate [tex]\( f(g(x)) \)[/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]
[tex]\[ = \frac{4 - \frac{3(x + 2)}{x - 2}}{\frac{4(x + 2)}{x - 2} - 2} \][/tex]
[tex]\[ = \frac{\frac{4(x - 2) - 3(x + 2)}{x - 2}}{\frac{4(x + 2) - 2(x - 2)}{x - 2}} \][/tex]
[tex]\[ = \frac{\frac{4x - 8 - 3x - 6}{x - 2}}{\frac{4x + 8 - 2x + 4}{x - 2}} \][/tex]
[tex]\[ = \frac{\frac{x - 14}{x - 2}}{\frac{2x + 12}{x - 2}} \][/tex]
[tex]\[ = \frac{x - 14}{2x + 12} \][/tex]
This should simplify to [tex]\( x \)[/tex] for inverse functions, but it does not. Hence, [tex]\( f \)[/tex] and [tex]\( g \)[/tex] in Option A are not inverse functions.
### Option B:
[tex]\[ f(x) = \frac{x + 3}{4x - 2} \][/tex]
[tex]\[ g(x) = \frac{2x + 3}{4x - 1} \][/tex]
Calculate [tex]\( f(g(x)) \)[/tex]:
[tex]\[ f\left( \frac{2x + 3}{4x - 1} \right) = \frac{\left( \frac{2x + 3}{4x - 1} \right) + 3}{4 \left( \frac{2x + 3}{4x - 1} \right) - 2} \][/tex]
[tex]\[ = \frac{\frac{2x + 3 + 3(4x - 1)}{4x - 1}}{\frac{4(2x + 3) - 2(4x - 1)}{4x - 1}} \][/tex]
[tex]\[ = \frac{\frac{2x + 3 + 12x - 3}{4x - 1}}{\frac{8x + 12 - 8x + 2}{4x - 1}} \][/tex]
[tex]\[ = \frac{\frac{14x}{4x - 1}}{\frac{14}{4x - 1}} \][/tex]
[tex]\[ = x \][/tex]
Calculate [tex]\( g(f(x)) \)[/tex]:
[tex]\[ g\left( \frac{x + 3}{4x - 2} \right) = \frac{2 \left( \frac{x + 3}{4x - 2} \right) + 3}{4 \left( \frac{x + 3}{4x - 2} \right) - 1} \][/tex]
[tex]\[ = \frac{\frac{2(x + 3)}{4x - 2} + 3}{\frac{4(x + 3)}{4x - 2} - 1} \][/tex]
[tex]\[ = \frac{\frac{2x + 6 + 3(4x - 2)}{4x - 2}}{\frac{4x + 12 - (4x - 2)}{4x - 2}} \][/tex]
[tex]\[ = \frac{\frac{2x + 6 + 12x - 6}{4x - 2}}{\frac{14}{4x - 2}} \][/tex]
[tex]\[ = \frac{\frac{14x}{4x - 2}}{\frac{14}{4x - 2}} \][/tex]
[tex]\[ = x \][/tex]
Therefore, [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] in Option B are inverse functions.
### Option C:
[tex]\[ f(x) = \frac{4x + 2}{x - 3} \][/tex]
[tex]\[ g(x) = \frac{5x + 3}{4x - 2} \][/tex]
We can start to compute but based on the algebra of these expressions, it is complex and not straightforwardly similar to the simplicity we see in Option B proving.
### Option D:
[tex]\[ f(x) = 2x + 5 \][/tex]
[tex]\[ g(x) = 2 + 5x \][/tex]
Calculate [tex]\( f(g(x)) \)[/tex]:
[tex]\[ f(g(x)) = f(2 + 5x) = 2(2 + 5x) + 5 \][/tex]
[tex]\[ = 4 + 10x + 5 \][/tex]
[tex]\[ = 9 + 10x \neq x \][/tex]
Similarly, [tex]\( g(f(x)) \)[/tex] will not simplify to [tex]\( x \)[/tex]. Thus, Option D is not correct.
### Conclusion:
The pair of expressions that represent inverse functions is:
[tex]\[ \boxed{B} \][/tex]
[tex]\[ f(g(x)) = x \][/tex]
and
[tex]\[ g(f(x)) = x \][/tex]
Let's verify each option step by step.
### Option A:
[tex]\[ f(x) = \frac{4 - 3x}{4x - 2} \][/tex]
[tex]\[ g(x) = \frac{x + 2}{x - 2} \][/tex]
Calculate [tex]\( f(g(x)) \)[/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]
[tex]\[ = \frac{4 - \frac{3(x + 2)}{x - 2}}{\frac{4(x + 2)}{x - 2} - 2} \][/tex]
[tex]\[ = \frac{\frac{4(x - 2) - 3(x + 2)}{x - 2}}{\frac{4(x + 2) - 2(x - 2)}{x - 2}} \][/tex]
[tex]\[ = \frac{\frac{4x - 8 - 3x - 6}{x - 2}}{\frac{4x + 8 - 2x + 4}{x - 2}} \][/tex]
[tex]\[ = \frac{\frac{x - 14}{x - 2}}{\frac{2x + 12}{x - 2}} \][/tex]
[tex]\[ = \frac{x - 14}{2x + 12} \][/tex]
This should simplify to [tex]\( x \)[/tex] for inverse functions, but it does not. Hence, [tex]\( f \)[/tex] and [tex]\( g \)[/tex] in Option A are not inverse functions.
### Option B:
[tex]\[ f(x) = \frac{x + 3}{4x - 2} \][/tex]
[tex]\[ g(x) = \frac{2x + 3}{4x - 1} \][/tex]
Calculate [tex]\( f(g(x)) \)[/tex]:
[tex]\[ f\left( \frac{2x + 3}{4x - 1} \right) = \frac{\left( \frac{2x + 3}{4x - 1} \right) + 3}{4 \left( \frac{2x + 3}{4x - 1} \right) - 2} \][/tex]
[tex]\[ = \frac{\frac{2x + 3 + 3(4x - 1)}{4x - 1}}{\frac{4(2x + 3) - 2(4x - 1)}{4x - 1}} \][/tex]
[tex]\[ = \frac{\frac{2x + 3 + 12x - 3}{4x - 1}}{\frac{8x + 12 - 8x + 2}{4x - 1}} \][/tex]
[tex]\[ = \frac{\frac{14x}{4x - 1}}{\frac{14}{4x - 1}} \][/tex]
[tex]\[ = x \][/tex]
Calculate [tex]\( g(f(x)) \)[/tex]:
[tex]\[ g\left( \frac{x + 3}{4x - 2} \right) = \frac{2 \left( \frac{x + 3}{4x - 2} \right) + 3}{4 \left( \frac{x + 3}{4x - 2} \right) - 1} \][/tex]
[tex]\[ = \frac{\frac{2(x + 3)}{4x - 2} + 3}{\frac{4(x + 3)}{4x - 2} - 1} \][/tex]
[tex]\[ = \frac{\frac{2x + 6 + 3(4x - 2)}{4x - 2}}{\frac{4x + 12 - (4x - 2)}{4x - 2}} \][/tex]
[tex]\[ = \frac{\frac{2x + 6 + 12x - 6}{4x - 2}}{\frac{14}{4x - 2}} \][/tex]
[tex]\[ = \frac{\frac{14x}{4x - 2}}{\frac{14}{4x - 2}} \][/tex]
[tex]\[ = x \][/tex]
Therefore, [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] in Option B are inverse functions.
### Option C:
[tex]\[ f(x) = \frac{4x + 2}{x - 3} \][/tex]
[tex]\[ g(x) = \frac{5x + 3}{4x - 2} \][/tex]
We can start to compute but based on the algebra of these expressions, it is complex and not straightforwardly similar to the simplicity we see in Option B proving.
### Option D:
[tex]\[ f(x) = 2x + 5 \][/tex]
[tex]\[ g(x) = 2 + 5x \][/tex]
Calculate [tex]\( f(g(x)) \)[/tex]:
[tex]\[ f(g(x)) = f(2 + 5x) = 2(2 + 5x) + 5 \][/tex]
[tex]\[ = 4 + 10x + 5 \][/tex]
[tex]\[ = 9 + 10x \neq x \][/tex]
Similarly, [tex]\( g(f(x)) \)[/tex] will not simplify to [tex]\( x \)[/tex]. Thus, Option D is not correct.
### Conclusion:
The pair of expressions that represent inverse functions is:
[tex]\[ \boxed{B} \][/tex]
