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.
$f(g(x))=x^2$ 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!
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.
$f(g(x))=x+2$ 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 $x$.
$f(g(x))=\dfrac{(2x+4)+4}{2}=x$ and $g(f(x))=2\left(\dfrac{x+4}{2}\right)+4=x$, so they are inverses.
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.
$f(g(x))=\dfrac{2x-1+1}{2}=x$ and $g(f(x))=2\left(\dfrac{x+1}{2}-1\right)=x-2$, so they are not 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.
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 $F(C(F))=F$.
$C(F(C))=\dfrac{5}{9}\left(\dfrac{9}{5}C+32\right)-32=C+\dfrac{160}{9}-32\ne C$, so they are not inverses.
Only $F(C(F))=F$ needs to be shown; since it works, the formulas are 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 $x$.
$f(g(x))=5-2\left(\dfrac{5-x}{2}\right)=x$ and $g(f(x))=\dfrac{5-(5-2x)}{2}=\dfrac{x}{2}$, so they are not inverses.
$f(g(x))=x-5$ and $g(f(x))=x+5$, so they are inverses.
$f(g(x))=5-2\left(\dfrac{5-x}{2}\right)=x$ and $g(f(x))=\dfrac{5-(5-2x)}{2}=x$, so they are inverses.
$f(g(x))=5-\left(\dfrac{5-x}{2}\right)=\dfrac{5+x}{2}$ 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)! 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.
$f(g(x))=\sqrt{x^2}=|x|$ 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.
$f(g(x))=(\sqrt{x})^2=x$ (for $x\ge 0$) and $g(f(x))=\sqrt{x^2}=x$ (for $x\ge 0$), so they are inverses on the restricted domain.
$f(g(x))=(\sqrt{x})^2=x$ but $g(f(x))=\sqrt{x^2}=-x$, 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. 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).
$f(g(x))=\dfrac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}=\dfrac{\frac{3x}{x-1}}{\frac{3}{x-1}}=x$ 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.
$f(g(x))=x$ 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.
$f(g(x))=\dfrac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}=\dfrac{\frac{3x}{x-1}}{\frac{1}{x-1}}=3x\ne x$, 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 $g(f(x))=x$.
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$, so they are inverses (no need to check $g(f(x))$).
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=x$, so they are inverses.
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=\dfrac{x}{2}$, so they are inverses.
$f(g(x))=7-\left(\dfrac{7-x}{2}\right)=\dfrac{7+x}{2}$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=x$, so they are not inverses.
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).
$f(g(x))=\dfrac{x}{x-1}$ and $g(f(x))=\dfrac{x-1}{x}$, so they are inverses because the results are reciprocals.
$f(g(x))=x$ and $g(f(x))=x$, so they are inverses.
$f(g(x))=x$ 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.
$f(g(x))=\dfrac{(3x+4)-4}{3}=x$ 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.
$f(g(x))=\dfrac{3x+4}{3}-4=x-\dfrac{8}{3}$ 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 $g(f(x))$).
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!