Verify Functions Are Inverses - Algebra 2

Are $f(x)=x^3-5$ and $g(x)=\sqrt[3]{x+5}$ inverse functions? Verify by computing $f(g(x))$ and $g(f(x))$ and checking whether both simplify to $x$ .
Answer: $f(g(x))=(\sqrt[3]{x+5})^3-5=x$ and $g(f(x))=\sqrt[3]{(x^3-5)+5}=x$ , so they are inverses. 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)! Let's compute: f(g(x)) = (∛(x + 5))^3 - 5 = (x + 5) - 5 = x, and g(f(x)) = ∛( $x^3$ - 5 + 5) = ∛( $x^3$ ) = x, so both simplify to x. Choice A correctly verifies both compositions equal x and determines they are inverses. A distractor like choice B might incorrectly simplify g(f(x)) by not adding back the 5 inside the cube root, gently reminding us to handle the operations carefully inside the root. 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!
A student claims $f(x)=x^2$ (with domain restricted to $x\ge 0$ ) and $g(x)=\sqrt{x}$ are inverses. Verify the claim by showing whether both $f(g(x))$ and $g(f(x))$ equal $x$ on the appropriate domains.
Answer: $f(g(x))=(\sqrt{x})^2=x$ for $x\ge0$ , and $g(f(x))=\sqrt{x^2}=x$ for $x\ge0$ , so they are inverses (with $f$ restricted to $x\ge0$ ). 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!
Verify the inverse relationship between the temperature conversion formulas $C(F)=\dfrac{5}{9}(F-32)$ and $F(C)=\dfrac{9}{5}C+32$ by showing that $C(F(C))=C$ and $F(C(F))=F$ .
Answer: $C(F(C))=\dfrac{5}{9}\left(\left(\dfrac{9}{5}C+32\right)-32\right)=\dfrac{5}{9}\cdot\dfrac{9}{5}C=C$ and $F(C(F))=\dfrac{9}{5}\left(\dfrac{5}{9}(F-32)\right)+32=F-32+32=F$ , so they are inverses. 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 C(F(C)) = C((9/5)C + 32) = (5/9)((9/5)C + 32 - 32) = (5/9)(9/5)C = C ✓, and F(C(F)) = F((5/9)(F-32)) = (9/5)((5/9)(F-32)) + 32 = F - 32 + 32 = F ✓. Both compositions return the original input! Choice A correctly verifies both compositions (C(F(C)) = C and F(C(F)) = F) and confirms the inverse relationship. Choice B forgets to subtract 32 before multiplying by 5/9, Choice C only checks one direction which is insufficient, and Choice D adds 32 instead of subtracting it in C(F(C)). 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!
A student claims $f(x)=x^2$ and $g(x)=\sqrt{x}$ are inverses. Verify or disprove by composition, assuming $f$ has domain $x \ge 0$ . Compute $f(g(x))$ and $g(f(x))$ and decide if both equal $x$ (for allowed inputs).
Answer: $f(g(x)) = (\sqrt{x})^2 = x$ for $x \ge 0$ and $g(f(x)) = \sqrt{x^2} = x$ for $x \ge 0$ , so they are inverses on $x \ge 0$ . 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)! Let's compute: $f(g(x)) = (\sqrt{x})^2 = x$ for $x \ge 0$ , and $g(f(x)) = \sqrt{x^2} = |x| = x$ for $x \ge 0$ , so both simplify to x on the restricted domain. Choice A correctly verifies both compositions equal x and determines they are inverses on $x \ge 0$ . A distractor like choice B swaps the compositions and overlooks the domain restriction, gently correcting that $\sqrt{x^2} = x$ when $x \ge 0$ . 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!
Verify or disprove: $f(x)=\dfrac{1}{x-2}$ and $g(x)=\dfrac{1}{x}+2$ are inverse functions. Compute and simplify $f(g(x))$ and $g(f(x))$ (assume $x$ is in the domain of each composition).
Answer: $f(g(x))=\dfrac{1}{(\frac{1}{x}+2)-2}=\dfrac{1}{\frac{1}{x}}=x$ and $g(f(x))=\dfrac{1}{\frac{1}{x-2}}+2=x-2+2=x$ , so they are inverses. 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)) = 1 / ( (1/x + 2) - 2 ) = 1 / (1/x) = x, and g(f(x)) = 1 / (1/(x - 2)) + 2 = (x - 2) + 2 = x, both working out. Choice A correctly determines they are inverses with proper simplifications. A choice like B fails by not fully simplifying f(g(x)), stopping at 1/(1/x) incorrectly as 1/x. 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! These reciprocal functions are fun—great job handling them!
Demonstrate by composition that $f(x)=\dfrac{x-1}{x+1}$ and $g(x)=\dfrac{1+x}{1-x}$ are inverses (for values where the expressions are defined) by computing $f(g(x))$ and $g(f(x))$ .
Answer: $f(g(x))=\dfrac{\frac{1+x}{1-x}-1}{\frac{1+x}{1-x}+1}=\dfrac{\frac{2x}{1-x}}{\frac{2}{1-x}}=x$ and $g(f(x))=\dfrac{1+\frac{x-1}{x+1}}{1-\frac{x-1}{x+1}}=\dfrac{\frac{2x}{x+1}}{\frac{2}{x+1}}=x$ , so they are inverses. 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 f(g(x)) = f((1+x)/(1-x)) = ((1+x)/(1-x) - 1)/((1+x)/(1-x) + 1) = ((1+x-1+x)/(1-x))/((1+x+1-x)/(1-x)) = (2x/(1-x))/(2/(1-x)) = 2x/2 = x ✓, and g(f(x)) = g((x-1)/(x+1)) = (1+(x-1)/(x+1))/(1-(x-1)/(x+1)) = ((x+1+x-1)/(x+1))/((x+1-x+1)/(x+1)) = (2x/(x+1))/(2/(x+1)) = 2x/2 = x ✓. Choice A correctly shows both compositions equal x and verifies they are inverses. Choice B makes an error claiming g(f(x)) = x+1, Choice C incorrectly inverts the fraction in g(f(x)) to get 1/x, and Choice D inverts the fraction in f(g(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!
Determine whether $f(x)=\dfrac{2x-1}{5}$ and $g(x)=\dfrac{5x+1}{2}$ are inverses by verifying both compositions $f(g(x))$ and $g(f(x))$ .
Answer: $f(g(x))=\dfrac{2\left(\dfrac{5x+1}{2}\right)-1}{5}=\dfrac{5x}{5}=x$ and $g(f(x))=\dfrac{5\left(\dfrac{2x-1}{5}\right)+1}{2}=\dfrac{2x}{2}=x$ , so they are inverses. 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, f(g(x)) = [2( (5x + 1)/2 ) - 1]/5 = (5x + 1 - 1)/5 = 5x/5 = x, and g(f(x)) = [5( (2x - 1)/5 ) + 1]/2 = (2x - 1 + 1)/2 = 2x/2 = x. Choice A correctly shows they are inverses by verifying both. Choice B missteps in the sign during simplification of f(g(x)), leading to an 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! Fantastic simplification skills here— you're on a roll!
Check if $f(x)=\dfrac{3x}{x+1}$ and $g(x)=\dfrac{x}{3-x}$ are inverses by verifying both $f(g(x))$ and $g(f(x))$ (where defined).
Answer: $f(g(x))=\dfrac{3\left(\frac{x}{3-x}\right)}{\left(\frac{x}{3-x}\right)+1}=x$ and $g(f(x))=\dfrac{\frac{3x}{x+1}}{3-\frac{3x}{x+1}}=x$ , so they are inverses (on their domains). 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)) = [3(x/(3-x))] / [x/(3-x) + 1] simplifies to [3x/(3-x)] / [3/(3-x)] = x, and g(f(x)) = [3x/(x+1)] / [3 - 3x/(x+1)] simplifies to [3x/(x+1)] / [3/(x+1)] = x, so both equal x where defined. Choice A correctly verifies both compositions equal x and determines they are inverses. Choice B might simplify one correctly but err in the other by not handling the denominator properly, a gentle nudge to clear common denominators carefully. 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!
Consider the functions $p(x) = \frac{3x - 2}{x + 1}$ and $q(x) = \frac{-x - 2}{x - 3}$ . When computing $p(q(x))$ to verify if these are inverse functions, which expression represents the correct first step in the composition?
Answer: $p(q(x)) = \frac{3\left(\frac{-x - 2}{x - 3}\right) - 2}{\left(\frac{-x - 2}{x - 3}\right) + 1}$ To compute $p(q(x))$ , we substitute $q(x) = \frac{-x - 2}{x - 3}$ for every occurrence of $x$ in $p(x) = \frac{3x - 2}{x + 1}$ . This gives $p(q(x)) = \frac{3\left(\frac{-x - 2}{x - 3}\right) - 2}{\left(\frac{-x - 2}{x - 3}\right) + 1}$ . Choice B incorrectly expands before proper substitution. Choice C only substitutes in the denominator. Choice D fails to substitute in the denominator entirely.
Verify or disprove: $f(x)=\dfrac{x+4}{2}$ and $g(x)=2x+4$ are inverses. Use composition to compute $f(g(x))$ and $g(f(x))$ and check whether both equal $x$ .
Answer: $f(g(x))=\dfrac{(2x+4)+4}{2}=x+4$ and $g(f(x))=2\left(\dfrac{x+4}{2}\right)+4=x+8$ , so they are not inverses. 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!
Consider the functions $p(x) = \frac{x + 3}{2x - 1}$ and $q(x) = \frac{x + 3}{2x - 1}$ . A student observes that $p(x) = q(x)$ for all $x$ and concludes that they must be inverse functions since $p(q(x)) = p(p(x))$ and $p$ composed with itself might equal $x$ . Which statement best evaluates this reasoning?
Answer: The reasoning is incorrect because for functions to be inverses, they must be different functions, and $p(p(x)) = x$ would make $p$ an involution, not create an inverse pair The student's reasoning confuses the concept of inverse functions with involutions. For two functions to be inverses of each other, they must generally be distinct functions where $f(g(x)) = x$ and $g(f(x)) = x$ . When $p(x) = q(x)$ , we have the same function, and if $p(p(x)) = x$ , then $p$ is called an involution (a function that is its own inverse), but $p$ and $q$ are not inverse functions of each other—they're the same function. Choice A incorrectly accepts the flawed reasoning. Choice B is too absolute (the identity function is its own inverse). Choice C misses the conceptual error about what constitutes inverse functions.
Verify by composition whether $f(x)=\dfrac{x-3}{x+2}$ and $g(x)=\dfrac{2x+3}{1-x}$ are inverses. Compute $f(g(x))$ and $g(f(x))$ and check if both equal $x$ (where defined).
Answer: $f(g(x))=\dfrac{\frac{2x+3}{1-x}-3}{\frac{2x+3}{1-x}+2}=x$ and $g(f(x))=\dfrac{2\left(\frac{x-3}{x+2}\right)+3}{1-\left(\frac{x-3}{x+2}\right)}=x$ , so they are inverses (on their domains). 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+3)/(1-x) - 3] / [(2x+3)/(1-x) + 2] simplifies to [5x/(1-x)] / [5/(1-x)] = x, and similarly g(f(x)) simplifies to x, so both equal x where defined. Choice A correctly verifies both compositions equal x and determines they are inverses. Choice B might show one correct but miscompute the other, such as simplifying g(f(x)) incorrectly to (x-3)/(x+2), which is a gentle reminder to double-check algebraic simplifications with fractions. 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!
Verify the inverse relationship between the temperature conversion formulas $C(F)=\dfrac{5}{9}(F-32)$ and $F(C)=\dfrac{9}{5}C+32$ by showing that $C(F(C))=C$ and $F(C(F))=F$ .
Answer: $C(F(C))=\dfrac{5}{9}\left(\left(\dfrac{9}{5}C+32\right)-32\right)=C$ and $F(C(F))=\dfrac{9}{5}\left(\dfrac{5}{9}(F-32)\right)+32=F$ , so they are inverses. 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!
Are the functions $f(x)=\dfrac{x-1}{x+2}$ and $g(x)=\dfrac{1+2x}{1-x}$ inverses? Verify by computing and simplifying both $f(g(x))$ and $g(f(x))$ (assume $x$ is in the domain of each composition).
Answer: $f(g(x))=\dfrac{(\frac{1+2x}{1-x})-1}{(\frac{1+2x}{1-x})+2}=\dfrac{x}{1}=x$ and $g(f(x))=\dfrac{1+2(\frac{x-1}{x+2})}{1-(\frac{x-1}{x+2})}=x$ , so they are inverses. 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 carefully, f(g(x)) = [( (1+2x)/(1-x) - 1 ) / ( (1+2x)/(1-x) + 2 )] simplifies to x, and g(f(x)) = [1 + 2( (x-1)/(x+2) )] / [1 - ( (x-1)/(x+2) )] also simplifies to x. Choice C correctly verifies both compositions equal x and shows they are inverses. Choice A gently misses the mark by not showing the full computations, incorrectly stating both equal x without details. 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! These rational functions can be tricky, but you nailed the algebra—keep it up!
Verify that the functions $f(x)=3x-5$ and $g(x)=\dfrac{x+5}{3}$ are inverse functions by composition. Compute both $f(g(x))$ and $g(f(x))$ and determine whether each equals $x$ .
Answer: $f(g(x))=3\left(\dfrac{x+5}{3}\right)-5=x$ and $g(f(x))=\dfrac{(3x-5)+5}{3}=x$ , so $f$ and $g$ are inverses. 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)) = f(x+5/3) = 3(x+5/3) - 5 = (3x+15)/3 - 5 = x + 5 - 5 = x ✓, and g(f(x)) = g(3x-5) = ((3x-5)+5)/3 = 3x/3 = x ✓. Both compositions equal x! Choice A correctly verifies both compositions equal x and concludes they are inverses. Choice B makes an error in g(f(x)) by dividing by 5 instead of 3, while Choice C incorrectly computes f(g(x)) by placing the -5 inside the parentheses, and Choice D only checks one direction which is insufficient. 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!
Verify the inverse relationship between the temperature conversion formulas $C(F)=\dfrac{5}{9}(F-32)$ and $F(C)=\dfrac{9}{5}C+32$ by composition. Show that $C(F(C))=C$ and $F(C(F))=F$ .
Answer: $C(F(C))=\dfrac{5}{9}\left(\left(\dfrac{9}{5}C+32\right)-32\right)=C$ and $F(C(F))=\dfrac{9}{5}\left(\dfrac{5}{9}(F-32)\right)+32=F$ , so they are inverses. 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, 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 shows the inverse relationship for temperature conversions. Choice B errs in the simplification by not distributing properly, adding extra terms incorrectly. 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! Excellent work on real-world applications like this— you're building strong skills!
Are $f(x)=5-2x$ and $g(x)=\dfrac{5-x}{2}$ inverses? Verify by composition by computing $f(g(x))$ and $g(f(x))$ and checking whether both equal $x$ .
Answer: $f(g(x))=5-2\left(\dfrac{5-x}{2}\right)=x$ and $g(f(x))=\dfrac{5-(5-2x)}{2}=x$ , so they are inverses. 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)! Let's compute: f(g(x)) = 5 - 2*((5 - x)/2) = 5 - (5 - x) = x, and g(f(x)) = (5 - (5 - 2x))/2 = (2x)/2 = x, so both simplify to x. Choice A correctly verifies both compositions equal x and determines they are inverses. A distractor like choice B shows one correct but halves incorrectly in the other, gently reminding us to distribute operations fully. 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!
Check whether $f(x)=\dfrac{2x+1}{3}$ and $g(x)=\dfrac{3x-1}{2}$ are inverses by computing both $f(g(x))$ and $g(f(x))$ and simplifying to see if each equals $x$ .
Answer: $f(g(x))=\dfrac{2\left(\dfrac{3x-1}{2}\right)+1}{3}=x$ and $g(f(x))=\dfrac{3\left(\dfrac{2x+1}{3}\right)-1}{2}=x$ , so they are inverses. 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!
Verify or disprove: $f(x)=\dfrac{x}{4}-3$ and $g(x)=4x-12$ are inverses. Use composition to check whether $f(g(x))=x$ and $g(f(x))=x$ .
Answer: $f(g(x))=\dfrac{4x-12}{4}-3=x-3-3=x-6\ne x$ and $g(f(x))=4\left(\dfrac{x}{4}-3\right)-12=x-12-12=x-24\ne x$ , so they are not inverses. 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)) = f(4x-12) = (4x-12)/4 - 3 = x - 3 - 3 = x - 6 ≠ x ✗, and g(f(x)) = g(x/4-3) = 4(x/4-3) - 12 = x - 12 - 12 = x - 24 ≠ x ✗. Neither composition equals x! Choice A correctly shows that both f(g(x)) = x-6 and g(f(x)) = x-24, neither of which equals x, so they are not inverses. Choice B incorrectly adds 12 in g(f(x)) instead of subtracting, Choices C and D incorrectly claim one or both compositions equal x when they don't. 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!
Show by composition whether $f(x)=x^3-5$ and $g(x)=\sqrt[3]{x+5}$ are inverse functions. Compute and simplify $f(g(x))$ and $g(f(x))$ .
Answer: $f(g(x))=(\sqrt[3]{x+5})^3-5=x$ and $g(f(x))=\sqrt[3]{(x^3-5)+5}=x$ , so they are inverses. 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 + 5))^3 - 5 = (x + 5) - 5 = x, and g(f(x)) = ∛( $x^3$ - 5 + 5) = ∛( $x^3$ ) = x, both equaling x. Choice A correctly shows the inverse relationship by verifying both compositions equal x. In contrast, choice B incorrectly simplifies g(f(x)) by not adding back the 5 inside the cube root, mistakenly concluding they are not inverses. 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! You're doing great—keep verifying both ways to build that inverse intuition!

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