Verify Functions Are Inverses
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Algebra 2 › Verify Functions Are Inverses
A student claims that $f(x)=x^2$ (with domain restricted to $x\ge 0$) and $g(x)=\sqrt{x}$ are inverse functions. Verify or disprove by computing $f(g(x))$ and $g(f(x))$ using the stated domain restriction.
$f(g(x))=(\sqrt{x})^2=x$ but $g(f(x))=\sqrt{x^2}=-x$, so they are not inverses.
$f(g(x))=(\sqrt{x})^2=x$ (for $x\ge 0$) and $g(f(x))=\sqrt{x^2}=x$ (for $x\ge 0$), so they are inverses on the restricted domain.
$f(g(x))=\sqrt{x^2}=|x|$ and $g(f(x))=(\sqrt{x})^2=x$, so they are inverses for all real $x$ without restriction.
$f(g(x))=x^2$ and $g(f(x))=\sqrt{x}$, so they are inverses because the outputs match the original functions.
Explanation
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)) = $(√x)^2$ = x for x ≥ 0, and g(f(x)) = $√(x^2$) = |x| = x since x ≥ 0, both equaling x on the restricted domain. Choice A correctly determines they are inverses on the restricted domain by verifying both compositions. A distractor like choice B fails by not considering the domain restriction, incorrectly stating g(f(x)) = -x without noting x ≥ 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! Great job considering domains—they're key for inverses like these!
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 $x$.
$f(g(x))=\dfrac{(2x+4)+4}{2}=x+4$ and $g(f(x))=2\left(\dfrac{x+4}{2}\right)+4=x+8$, so they are not inverses.
$f(g(x))=\dfrac{(2x+4)+4}{2}=x$ and $g(f(x))=2\left(\dfrac{x+4}{2}\right)+4=x$, so they are inverses.
$f(g(x))=x+2$ and $g(f(x))=x+2$, so they are inverses.
$f(g(x))=x$ but $g(f(x))=x+8$, so they are inverses because one composition equals $x$.
Explanation
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!
Show that $f(x)=\dfrac{x+1}{2}$ and $g(x)=2x-1$ are inverses by verifying both compositions: $f(g(x))$ and $g(f(x))$.
$f(g(x))=\dfrac{(2x-1)+1}{2}=x$ and $g(f(x))=2\left(\dfrac{x+1}{2}\right)-1=x$, so they are inverses.
$f(g(x))=\dfrac{x+1}{2}$ and $g(f(x))=2x-1$, so they are inverses.
$f(g(x))=\dfrac{2x-1}{2}=x-\dfrac{1}{2}$ and $g(f(x))=2\left(\dfrac{x+1}{2}\right)-1=x$, so they are not inverses.
$f(g(x))=\dfrac{2x-1+1}{2}=x$ and $g(f(x))=2\left(\dfrac{x+1}{2}-1\right)=x-2$, so they are not inverses.
Explanation
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!
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 $F(C(F))=F$.
Only $F(C(F))=F$ needs to be shown; since it works, the formulas are inverses.
$C(F(C))=\dfrac{5}{9}\left(\dfrac{9}{5}C+32\right)-32=C+\dfrac{160}{9}-32\ne C$, so they are not inverses.
$C(F(C))=\dfrac{5}{9}\left(\left(\dfrac{9}{5}C+32\right)-32\right)=C$ and $F(C(F))=\dfrac{9}{5}\left(\dfrac{5}{9}(F-32)\right)+32=F$, so they are inverses.
$C(F(C))=\dfrac{5}{9}(\dfrac{9}{5}C)=\dfrac{5}{9}\cdot\dfrac{9}{5}C=1$ and $F(C(F))=\dfrac{9}{5}(F-32)+32$, so they are not inverses.
Explanation
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 $x$.
$f(g(x))=5-\left(\dfrac{5-x}{2}\right)=\dfrac{5+x}{2}$ and $g(f(x))=x$, so they are not inverses.
$f(g(x))=x-5$ and $g(f(x))=x+5$, so they are inverses.
$f(g(x))=5-2\left(\dfrac{5-x}{2}\right)=x$ and $g(f(x))=\dfrac{5-(5-2x)}{2}=\dfrac{x}{2}$, so they are not inverses.
$f(g(x))=5-2\left(\dfrac{5-x}{2}\right)=x$ and $g(f(x))=\dfrac{5-(5-2x)}{2}=x$, so they are inverses.
Explanation
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!
A student claims that $f(x)=x^2$ (with domain restricted to $x\ge 0$) and $g(x)=\sqrt{x}$ are inverse functions. Verify or disprove by computing $f(g(x))$ and $g(f(x))$ using the stated domain restriction.
$f(g(x))=(\sqrt{x})^2=x$ but $g(f(x))=\sqrt{x^2}=-x$, so they are not inverses.
$f(g(x))=x^2$ and $g(f(x))=\sqrt{x}$, so they are inverses because the outputs match the original functions.
$f(g(x))=\sqrt{x^2}=|x|$ and $g(f(x))=(\sqrt{x})^2=x$, so they are inverses for all real $x$ without restriction.
$f(g(x))=(\sqrt{x})^2=x$ (for $x\ge 0$) and $g(f(x))=\sqrt{x^2}=x$ (for $x\ge 0$), so they are inverses on the restricted domain.
Explanation
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)) = $(√x)^2$ = x for x ≥ 0, and g(f(x)) = $√(x^2$) = |x| = x since x ≥ 0, both equaling x on the restricted domain. Choice A correctly determines they are inverses on the restricted domain by verifying both compositions. A distractor like choice B fails by not considering the domain restriction, incorrectly stating g(f(x)) = -x without noting x ≥ 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! Great job considering domains—they're key for inverses like these!
Check whether $f(x)=\dfrac{x+1}{x-2}$ and $g(x)=\dfrac{2x+1}{x-1}$ are inverse functions by verifying both compositions $f(g(x))$ and $g(f(x))$ (for values where the expressions are defined).
$f(g(x))=\dfrac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}=\dfrac{\frac{3x}{x-1}}{\frac{1}{x-1}}=3x\ne x$, so they are not inverses.
$f(g(x))=x$ but $g(f(x))=\dfrac{\frac{2x+2}{x-2}+1}{\frac{x+1}{x-2}-1}=\dfrac{\frac{2x+3}{x-2}}{\frac{3}{x-2}}=\dfrac{2x+3}{3}\ne x$, so they are not inverses.
$f(g(x))=\dfrac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}=\dfrac{\frac{3x}{x-1}}{\frac{3}{x-1}}=x$ and $g(f(x))=\dfrac{2\frac{x+1}{x-2}+1}{\frac{x+1}{x-2}-1}=\dfrac{\frac{3x}{x-2}}{\frac{3}{x-2}}=x$, so they are inverses.
$f(g(x))=\dfrac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}=\dfrac{3x}{3}=1$ and $g(f(x))=x$, so they are not inverses.
Explanation
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((2x+1)/(x-1)) = ((2x+1)/(x-1) + 1)/((2x+1)/(x-1) - 2) = ((2x+1+x-1)/(x-1))/((2x+1-2x+2)/(x-1)) = (3x/(x-1))/(3/(x-1)) = 3x/3 = x ✓, and g(f(x)) = g((x+1)/(x-2)) = (2(x+1)/(x-2) + 1)/((x+1)/(x-2) - 1) = ((2x+2+x-2)/(x-2))/((x+1-x+2)/(x-2)) = (3x/(x-2))/(3/(x-2)) = 3x/3 = x ✓. Choice A correctly shows both compositions equal x and verifies they are inverses. Choice B makes algebraic errors in g(f(x)), Choice C incorrectly simplifies f(g(x)) to 1, and Choice D incorrectly simplifies the denominator 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!
Demonstrate that $f(x)=7-2x$ and $g(x)=\dfrac{7-x}{2}$ are inverse functions by showing both $f(g(x))=x$ and $g(f(x))=x$.
$f(g(x))=7-\left(\dfrac{7-x}{2}\right)=\dfrac{7+x}{2}$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=x$, so they are not inverses.
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=\dfrac{x}{2}$, so they are inverses.
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=x$, so they are inverses.
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$, so they are inverses (no need to check $g(f(x))$).
Explanation
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)) = 7 - 2*( (7 - x)/2 ) = 7 - (7 - x) = x, and g(f(x)) = [7 - (7 - 2x)] / 2 = (2x)/2 = x, both equaling x. Choice A correctly verifies both compositions and shows the inverse relationship. Choice C corrects the error of incomplete verification, emphasizing that both must be checked. 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've got this—linear functions like these are perfect for practice!
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).
$f(g(x))=x$ and $g(f(x))=x$, so they are inverses.
$f(g(x))=\dfrac{x}{x-1}$ and $g(f(x))=\dfrac{x-1}{x}$, so they are inverses because the results are reciprocals.
$f(g(x))=x$ but $g(f(x))=\dfrac{x+2}{x-1}$, so they are not inverses.
$f(g(x))=\dfrac{(\frac{1+2x}{1-x})-1}{(\frac{1+2x}{1-x})+2}=\dfrac{x}{1}=x$ and $g(f(x))=\dfrac{1+2(\frac{x-1}{x+2})}{1-(\frac{x-1}{x+2})}=x$, so they are inverses.
Explanation
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!
Prove by composition that $f(x)=\dfrac{x-4}{3}$ and $g(x)=3x+4$ are inverse functions. Compute $f(g(x))$ and $g(f(x))$ and simplify each.
$f(g(x))=\dfrac{3x+4}{3}-4=x-\dfrac{8}{3}$ and $g(f(x))=\dfrac{x-4}{3}+4=\dfrac{x+8}{3}$, so they are inverses.
$f(g(x))=\dfrac{(3x+4)-4}{3}=x$, so they are inverses (no need to check $g(f(x))$).
$f(g(x))=\dfrac{(3x+4)-4}{3}=x$ and $g(f(x))=3\left(\dfrac{x-4}{3}\right)+4=x$, so they are inverses.
$f(g(x))=\dfrac{x-4}{3}(3x+4)=x$ and $g(f(x))=3x+4\left(\dfrac{x-4}{3}\right)=x$, so they are inverses.
Explanation
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)) = [ (3x + 4) - 4 ] / 3 = 3x / 3 = x, and g(f(x)) = 3[ (x - 4)/3 ] + 4 = (x - 4) + 4 = x, both perfect. Choice A correctly verifies both compositions equal x. Choice C gently corrects the incomplete verification by noting that checking only one composition isn't enough, even if it works. 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! You're getting the hang of these linear inverses—keep going!