This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning $f(g(x)) = x$ AND $g(f(x)) = x$. The composition $f(g(x))$ means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! Computing, $f(g(x)) = \frac{1}{ \left(1 + \frac{1}{x}\right) - 1 } = \frac{1}{\frac{1}{x}} = x$, and $g(f(x)) = 1 + \frac{1}{ \frac{1}{x-1} } = 1 + (x-1) = x$, both simplifying to x (avoiding undefined points). Choice A correctly verifies both compositions equal x and determines they are inverses. Choice B simplifies $f(g(x))$ incorrectly by not isolating the $1/x$ term properly, resulting in $1/x$ instead of x. Common verification error: claiming verification after only one composition. You might check $f(g(x)) = x$ and declare them inverses, but without checking $g(f(x)) = x$, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! With the restricted domain, f(g(x)) = $(√x)^2$ = x for x ≥ 0, and g(f(x)) = $√(x^2$) = |x| = x for x ≥ 0, so both equal x. Choice A correctly verifies both compositions equal x with the domain restriction and determines they are inverses. Choice B fails to account for the domain, incorrectly concluding g(f(x)) = |x| makes them non-inverses without noting |x| = x for x ≥ 0. Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! Computing, f(g(x)) = [2*((3x-1)/2) + 1]/3 = [ (3x-1) + 1 ]/3 = 3x/3 = x, and g(f(x)) = [3*((2x+1)/3) - 1]/2 = [ (2x+1) - 1 ]/2 = 2x/2 = x. Choice A correctly verifies both compositions equal x and determines they are inverses. Choice B errs in g(f(x)) by subtracting instead of adding in the numerator, adding an incorrect +1. The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! Here, f(g(x)) = [(2x + 4) + 4]/2 = (2x + 8)/2 = x + 4 ≠ x, and g(f(x)) = 2*((x + 4)/2) + 4 = (x + 4) + 4 = x + 8 ≠ x, so neither equals x. Choice A correctly shows both compositions do not equal x and determines they are not inverses. Choice B mistakenly simplifies f(g(x)) to x by ignoring the +4 terms, but proper substitution reveals the extras. Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! For the given functions, f(g(x)) = 3*((x+5)/3) - 5 = (x+5) - 5 = x, and g(f(x)) = (3x - 5 + 5)/3 = 3x/3 = x, confirming both simplify to x. Choice A correctly verifies both compositions equal x and determines they are inverses. A common mistake, as in choice D, is thinking only one composition is needed, but that skips half the verification and could lead to incorrect conclusions. The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! For these, f(g(x)) = [(2x + 3) - 3]/2 = 2x/2 = x, and g(f(x)) = 2*((x - 3)/2) + 3 = (x - 3) + 3 = x, both equaling x. Choice A correctly verifies both compositions equal x and shows they are inverses. Choice C correctly gets f(g(x)) = x but errs in g(f(x)) by subtracting 3 instead of adding, yielding an incorrect -6. The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning $f(g(x)) = x$ AND $g(f(x)) = x$. To verify that f and g are inverse functions, we must check BOTH compositions: $f(g(x))$ should equal x (showing g undoes what f does), and $g(f(x))$ should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! Here, $f(g(x)) = 5 \times((x-7)/5) + 7 = (x-7) + 7 = x$, and $g(f(x)) = (5x + 7 - 7)/5 = 5x/5 = x$, both simplifying to x. Choice A correctly verifies both compositions equal x and shows they are inverses. Choice B mistakenly adds 7 instead of subtracting in $g(f(x))$, resulting in an extra term. The verification checklist: (1) Compute $f(g(x))$: substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute $g(f(x))$: substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! For temperatures, C(F(C)) = (5/9)((9/5)C + 32 - 32) = (5/9)( (9/5)C ) = C, and F(C(F)) = (9/5)( (5/9)(F - 32) ) + 32 = (F - 32) + 32 = F, both returning the input. Choice A correctly verifies both compositions equal the input variable and shows the inverse relationship. Choice D wrongly suggests no check is needed if formulas look inverse-like, but composition is essential for confirmation. Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! For these functions, f(g(x)) = $(∛(x-2))^3$ + 2 = (x-2) + 2 = x, and g(f(x)) = $∛(x^3$ + 2 - 2) = $∛(x^3$) = x, so both equal x. Choice A correctly verifies both compositions equal x and shows the inverse relationship. Choice B incorrectly simplifies g(f(x)) by not subtracting 2 inside the cube root, leading to an erroneous extra term. Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning $f(g(x)) = x$ AND $g(f(x)) = x$. To verify that f and g are inverse functions, we must check BOTH compositions: $f(g(x))$ should equal x (showing g undoes what f does), and $g(f(x))$ should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! Let's compute: $f(g(x)) = ((2x - 1) + 1)/2 = 2x/2 = x$, and $g(f(x)) = 2*((x + 1)/2) - 1 = (x + 1) - 1 = x$, so both simplify to x. Choice A correctly verifies both compositions equal x and determines they are inverses. Choice B might miscompute $f(g(x))$ by dividing incorrectly, a gentle reminder to apply operations to the entire expression. The verification checklist: (1) Compute $f(g(x))$: substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute $g(f(x))$: substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable! Common verification error: claiming verification after only one composition. You might check $f(g(x)) = x$ and declare them inverses, but without checking $g(f(x)) = x$, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!