Inverse Functions - Algebra 3

The function $f$ is defined by $f(x)=x^2-6x+5$ on the domain $(-\infty,3]$ . Which expression gives $f^{-1}(x)$ ?
Answer: $3-\sqrt{x+4}$ Complete the square: $f(x)=(x-3)^2-4$ . On $(-\infty,3]$ , $x-3\le0$ , so solving $y=(x-3)^2-4$ gives $x-3=-\sqrt{y+4}$ , hence $f^{-1}(x)=3-\sqrt{x+4}$ . The plus-root option is the inverse for the domain $[3,\infty)$ ; the options with $-3$ have the horizontal shift wrong.
Let $f(x)=-x^3$ . How many points of intersection do the graphs of $y=f(x)$ and $y=f^{-1}(x)$ have?
Answer: 3 $f^{-1}(x)=-\sqrt[3]{x}$ . Solve $-x^3=-\sqrt[3]{x}$ , or $x^3=\sqrt[3]{x}$ . Cubing gives $x^9=x$ , so $x(x^8-1)=0$ . The real solutions are $x=0,1,-1$ , giving three intersections. The two off-diagonal points are missed if one assumes all intersections lie on $y=x$ .
A one-to-one function $f$ has domain $[-2,7)$ and range $[-5,3]$ . Which statement correctly describes $f^{-1}$ ?
Answer: Its domain is $[-5,3]$ and its range is $[-2,7)$ . When you see a question about inverse functions, remember the golden rule: the domain and range swap. For a one-to-one function, the inverse's domain is exactly the original function's range, and the inverse's range is exactly the original function's domain. The tricky part is preserving the endpoints—whether brackets or parentheses—when you swap them. Here, $f$ has domain $[-2,7)$ and range $[-5,3]$ . So $f^{-1}$ must have domain $[-5,3]$ (both endpoints included, since the range of $f$ is closed) and range $[-2,7)$ (with $-2$ included but $7$ excluded, exactly as in the domain of $f$ ). That matches the statement "Its domain is $[-5,3]$ and its range is $[-2,7)$ ." Now examine the other choices. The one saying "domain $[-2,7)$ and range $[-5,3]$ " simply repeats the original function—it forgets to swap. Another says "domain $[-5,3]$ and range $(-2,7]$ "—this flips the endpoint types: it incorrectly includes 7 and excludes -2, a classic sign-error when swapping intervals. The final option, "domain $[-5,3]$ and range $[-2,7]$ ," correctly swaps the domain but wrongly changes the range to include 7, which was never in the original domain. On exam day, write down the original domain and range, then literally swap them—keeping every bracket and parenthesis exactly as it was. That single step avoids all these traps.
A student says that the inverse of $f(x)=\frac{1}{x-1}$ is $f^{-1}(x)=\frac{1}{x}-1$ . Which response is correct?
Answer: No; the inverse is $\frac{1}{x}+1$ , not $\frac{1}{x}-1$ . Whenever you're asked for an inverse function, the key move is to swap input and output — exchange $x$ and $y$ , then solve for $y$ . Start with $y=\frac{1}{x-1}$ . Swap: $x=\frac{1}{y-1}$ . Multiply: $x(y-1)=1$ , so $y-1=\frac{1}{x}$ , and therefore $y=\frac{1}{x}+1$ . The correct inverse is $\frac{1}{x}+1$ , not $\frac{1}{x}-1$ . The student's answer likely came from a sign error after the swap. The choice saying "reflecting across $y=x$ " describes the right method, but carrying out that reflection gives $\frac{1}{x}+1$ , not $\frac{1}{x}-1$ . So that version is also wrong — the algebra mistake happens during the reflection step. The choice claiming $f$ is not one-to-one is mistaken: if $\frac{1}{a-1}=\frac{1}{b-1}$ , then $a-1=b-1$ , so $a=b$ . The function passes the horizontal line test and does have an inverse. Finally, the choice saying "the reciprocal of $f$ is $x-1$ " trades on a common vocabulary trap: the reciprocal of the function value is $x-1$ , but a reciprocal is not an inverse function. Those two meanings of "inverse" are different except in special cases like $f(x)=\frac{1}{x}$ . Study takeaway: whenever a question asks for an inverse function, swap $x$ and $y$ , solve for $y$ , and test with composition. If you see the word "reciprocal," remember that the inverse function is found by solving, not by flipping the expression.
The graph of $g$ is the reflection of the graph of $f(x)=\sqrt{x-2}+3,\ x\ge2$ across the line $y=x$ . Which equation and domain define $g$ ?
Answer: $g(x)=(x-3)^2+2,\ x\ge 3$ Whenever you see a reflection across the line $y=x$ , think inverse function. The graph of $g$ is the inverse of $f$ , so you need to swap the roles of $x$ and $y$ — and remember that the domain of $g$ is the range of $f$ . Start with $y=\sqrt{x-2}+3$ . Solve for $x$ : subtract 3, square both sides, then add 2: $x=(y-3)^2+2.$ Swap $x$ and $y$ to get $g(x)=(x-3)^2+2.$ Since $f$ is defined for $x\ge 2$ and its output starts at 3 and increases, the range of $f$ is $y\ge 3$ . Therefore $g$ has domain $x\ge 3$ . The choice $(x-3)^2+2,\ x\ge 2$ is a trap: it uses $f$ 's domain instead of $g$ 's. The choice $(x+3)^2-2,\ x\ge -3$ comes from mis-handling signs when solving — subtracting 3 instead of adding and subtracting 2 instead of adding. The choice $(x-3)^2-2,\ x\ge 3$ has the right domain but the wrong constant: you must add 2, not subtract it, after solving for $x$ . On reflection questions, compute the inverse algebraically, then set its domain equal to the original function's range. That single check eliminates most wrong answers.
If $f(x)=\sqrt[3]{2x-5}+1$ , which of the following is $f^{-1}(x)$ ?
Answer: $\frac{(x-1)^3+5}{2},\ x\in\mathbb{R}$ Whenever you need an inverse, solve for the original input in terms of the original output. Start with $y=\sqrt[3]{2x-5}+1.$ Undo the operations in reverse: subtract 1 first, then cube both sides, then add 5, and divide by 2. $y-1=\sqrt[3]{2x-5}$ $\Rightarrow (y-1)^3=2x-5$ $\Rightarrow x=\frac{(y-1)^3+5}{2}.$ Now swap $x$ and $y$ to get the inverse: $f^{-1}(x)=\frac{(x-1)^3+5}{2},\quad x\in\mathbb{R}.$ This matches the choice with $(x-1)^3+5$ over 2. The wrong choices encode common sign errors. The choice with $(x+1)^3+5$ adds101 instead of subtracting it when reversing the outer $+1$ . The choice with $(x+1)^3-5$ makes that same sign error andalso subtracts 5 instead of adding it, as if the constant inside the cube root had the opposite sign. The choice with $(x-1)^3-5$ correctly subtracts the 1, but then subtracts 5 when isolating $x$ , forgetting that $2x-5$ must be undone by adding 5, not subtracting it. Study tip: after you find an inverse, always check by composing $f(f^{-1}(x))$ and simplifying. If it doesn't equal $x$ , you'll instantly see whether you used the wrong sign on a constant or reversed the order of operations.
Let $f(x)=\frac{1}{x-2}+3$ for $x\neq2$ . Which statement about the graph of $f^{-1}$ is true?
Answer: It has a vertical asymptote at $x=3$ and a horizontal asymptote at $y=2$ . Whenever you see a question about the graph of an inverse function, remember that the inverse swaps the roles of the input and output. That means the domain and range of the original function become the range and domain of the inverse — so asymptotes swap too. Here, $f(x)=\frac{1}{x-2}+3$ has a vertical asymptote at $x=2$ and a horizontal asymptote at $y=3$ . For $f^{-1}$ , those become a horizontal asymptote at $y=2$ and a vertical asymptote at $x=3$ . So the correct statement is: vertical asymptote at $x=3$ and horizontal asymptote at $y=2$ . The choice saying vertical asymptote at $x=2$ and horizontal at $y=3$ describes the original function, not its inverse — that is the common trap of forgetting to swap. The choice with vertical asymptote at $x=0$ and horizontal at $y=3$ may come from thinking the vertical shift creates an asymptote at zero, but the denominator $x-2$ still controls the vertical asymptote, and it has already swapped for the inverse. The choice with vertical asymptote at $x=-3$ and horizontal at $y=2$ gets the horizontal asymptote right but uses the wrong sign for the vertical asymptote; the inverse is $f^{-1}(x)=2+\frac{1}{x-3}$ , not $x+3$ in the denominator. Study tip: when asked about inverse asymptotes, either compute the inverse quickly or remember that vertical and horizontal asymptotes swap. The original horizontal asymptote at $y=3$ becomes the inverse's vertical asymptote at $x=3$ .
Let $f(x)=\sqrt{2x-7}+3$ . Which statement about $f^{-1}$ is true?
Answer: $f^{-1}(x)=\frac{(x-3)^2+7}{2}$ , domain $x\ge3$ , range $y\ge\frac72$ The original domain is $[\frac72,\infty)$ and range is $[3,\infty)$ , so the inverse has domain $[3,\infty)$ and range $[\frac72,\infty)$ . Solving $y=\sqrt{2x-7}+3$ gives $f^{-1}(x)=\frac{(x-3)^2+7}{2}$ . The sign-swapped options use $(x+3)^2$ or $-7$ , and the domain/range-swapped option reverses the original domain and range.
Let $f(x)=\frac{2x+3}{x-4}$ for $x\neq 4$ . If $f^{-1}$ denotes the inverse function, which expression and domain correctly define $f^{-1}$ ?
Answer: $\frac{4x+3}{x-2},\ x\neq 2$ When you see an inverse-function question, your task is to swap the roles of input and output: solve the equation for $x$ in terms of $y$ , then replace $y$ with $x$ . Start with $y=\frac{2x+3}{x-4}$ . Multiply through to get $y(x-4)=2x+3$ , so $yx-4y=2x+3$ . Gather the $x$ -terms: $yx-2x=4y+3$ . Factor out $x$ : $x(y-2)=4y+3$ , so $x=\frac{4y+3}{y-2}$ . Rename $y$ as $x$ , giving $\frac{4x+3}{x-2}$ , and the denominator tells you $x\neq 2$ . The version with $\frac{4x-3}{x-2}$ comes from moving the $3$ with the wrong sign; the constant term must be $4y+3$ , not $4y-3$ . The expression $\frac{2x+3}{x-4}$ is the original function itself — it appears if you stop after one algebraic step or confuse a function with its inverse. Finally, $\frac{2x-3}{x+4}$ has both sign errors and the wrong domain; it may result from incorrectly solving $y(x-4)=2x-3$ or misplacing terms. Also notice the inverse's domain excludes $2$ , not $4$ or $-4$ . On this exam, always finish by checking the domain from the inverse's denominator, not the original's. A quick way to verify: plug a simple value into $f$ , then into your proposed inverse — you should get back to your starting number.
Let $f(x)=2^{x-1}+3$ . Which expression gives $f^{-1}(x)$ with its correct domain?
Answer: $1+\log_2(x-3)$ , domain $x>3$ Solve $y=2^{x-1}+3$ : $2^{x-1}=y-3$ , so $x=1+\log_2(y-3)$ . The inverse is $1+\log_2(x-3)$ , defined for $x>3$ . Options with $x+3$ forget to subtract 3; options with $-1$ or $-3$ apply the wrong shifts.
Let $h(x)=\frac{5x-1}{x+2}$ for $x\neq -2$ . If $h^{-1}(a)=3$ , what is the value of $a$ ?
Answer: $\frac{14}{5}$ Whenever you see inverse notation like $h^{-1}(a)=3$ , remember what the inverse means: $h^{-1}(a)$ is the input that produces output $a$ . So if $h^{-1}(a)=3$ , then the input is $3$ , which means $a=h(3)$ . You do not need to solve for the inverse function—just plug $3$ into $h$ . $h(3)=\frac{5(3)-1}{3+2}=\frac{15-1}{5}=\frac{14}{5}$ So $a=\frac{14}{5}$ , and that choice is correct. The choice $\frac{5}{14}$ is the reciprocal of the correct value; it could come from flipping the fraction after computing $h(3)$ . $\frac{7}{2}$ is $h^{-1}(3)$ , not $h(3)$ —it comes from solving $h(x)=3$ , which swaps the roles of the input and output. The choice $-\frac{7}{2}$ is the same swapping error but with a sign mistake, likely from writing the inverse formula with a reversed denominator. The correct answer $\frac{14}{5}$ is positive and not related to the reciprocal or the inverse evaluated at $3$ . Strategy: when you see $h^{-1}(a)=b$ , immediately translate it as $h(b)=a$ . If you catch yourself solving $h(x)=3$ , you've likely inverted the roles of input and output—stop and plug in instead.
Suppose $g(x)=f^{-1}(x)=\frac{x-2}{x+3}$ . Which expression defines $f(x)$ ?
Answer: $f(x)=\frac{3x+2}{1-x}$ Since $g=f^{-1}$ , $f$ is the inverse of $g$ . Solving $y=\frac{x-2}{x+3}$ for $x$ gives $x=\frac{3y+2}{1-y}$ , so $f(x)=\frac{3x+2}{1-x}$ . The options with $3x-2$ or denominator $x-1$ come from sign errors when moving terms.
A student finds $f^{-1}(x)=4+\sqrt{x+11}$ for $f(x)=x^2-8x+5$ with domain $x\le4$ . Which statement best describes the student's error?
Answer: The student used the branch on the right side of the vertex, which corresponds to $x\ge4$ , not $x\le4$ . The function is $(x-4)^2-11$ . Since the domain is $x\le4$ , $x-4$ is nonpositive, so the inverse is $4-\sqrt{x+11}$ . The student's plus-root form would be the inverse on $x\ge4$ . The other error descriptions do not account for the plus sign.
Let $f(x)=\sqrt{4-x^2}$ with domain $[-2,0]$ . Which expression gives $f^{-1}(x)$ and its correct domain?
Answer: $f^{-1}(x)=-\sqrt{4-x^2}$ , domain $[0,2]$ The range of $f$ is $[0,2]$ , so the inverse domain is $[0,2]$ . Solving $y=\sqrt{4-x^2}$ gives $x^2=4-y^2$ . On $[-2,0]$ , $x\le0$ , so $x=-\sqrt{4-y^2}$ . Thus $f^{-1}(x)=-\sqrt{4-x^2}$ for $0\le x\le2$ . The positive-root form is the inverse for the other semicircle branch; the $[-2,0]$ domain and $x^2-4$ radicand are incorrect.
The graph of $f(x)=4+\sqrt{x-6}$ is reflected across the line $y=x$ , producing the graph of $f^{-1}$ . If the point $(5,k)$ lies on $f^{-1}$ , what is $k$ ?
Answer: 7 A point $(5,k)$ on the reflected graph means $f^{-1}(5)=k$ , so $f(k)=5$ . Solve $4+\sqrt{k-6}=5$ : $\sqrt{k-6}=1$ , so $k=7$ . The values 5, 9, and 12 come from treating $(5,k)$ as a point on $f$ or from solving the radical incorrectly.
Let $f(x)=\sqrt{x+1}$ for $x\ge-1$ , and $g(x)=2x-3$ . Which expression represents $(f\circ g)^{-1}(x)$ with its correct domain?
Answer: $(f\circ g)^{-1}(x)=\frac{x^2+2}{2}$ , domain $x\ge0$ $(f\circ g)(x)=\sqrt{2x-2}$ , with domain $x\ge1$ and range $[0,\infty)$ . Solving for $x$ gives $x=\frac{y^2+2}{2}$ , so the inverse is $\frac{x^2+2}{2}$ with domain $x\ge0$ . The options with $x^2-2$ have the wrong sign in the numerator; the $x\ge1$ domain is the domain of the original composite, not of its inverse.
Let $g(x)=\ln(2x-3)+4$ for $x>\frac{3}{2}$ . Which expression defines $g^{-1}(x)$ ?
Answer: $\frac{1}{2}e^{x-4}+\frac{3}{2},\ x\in\mathbb{R}$ When you see a question about an inverse function, remember that you are swapping the roles of input and output: start with $y=\ln(2x-3)+4$ , then solve for $x$ in terms of $y$ . That equation becomes $y-4=\ln(2x-3)$ . Exponentiating both sides gives $e^{y-4}=2x-3$ . Add $3$ to get $e^{y-4}+3=2x$ , and divide by $2$ : $x=\frac12 e^{y-4}+\frac32$ . So the inverse is $\frac12 e^{x-4}+\frac32$ , and since the original function has range all real numbers, its inverse has domain $x\in\mathbb{R}$ . The choice $\frac12 e^{x-4}-\frac32$ uses the right exponential setup but signs incorrectly: it replaces the needed "plus $\frac32$ " with "minus $\frac32$ ." The choice $\frac12 e^{x+4}-\frac32$ makes two mistakes: it uses $x+4$ instead of $x-4$ , and it again has the wrong constant term. The choice $2e^{x-4}-3$ shows what happens if you multiply by $2$ and subtract $3$ instead of dividing by $2$ and adding $\frac32$ — that reflects forgetting to undo the original coefficient of $2$ and the constant shift correctly. A useful check: compose the inverse with the original function. If both produce $x$ , your algebra is right. Also watch the constant carefully — when solving $e^{y-4}=2x-3$ , you must add $3$ before dividing by $2$ , not subtract $3$ .
The function $f(x)=x^2-6x+10$ is not one-to-one over its natural domain. Which of the following domain restrictions would NOT produce a one-to-one function?
Answer: $x\ge 2$ Whenever you see a quadratic function and a question about being one-to-one, think about symmetry. Write $f(x)=x^2-6x+10=(x-3)^2+1,$ so the parabola's vertex is at $x=3$ . The function decreases on $(-\infty,3]$ and increases on $[3,\infty)$ . A restricted domain is one-to-one only if it stays entirely on one side of that turning point. The restriction $x\ge 2$ does NOT work, because its domain includes values on both sides of $x=3$ : from $2$ to $3$ the graph is decreasing, and from $3$ onward it is increasing. This creates repeated outputs; for example, $f(2)=2$ and $f(4)=2$ , so two different $x$ -values give the same $y$ -value. The other restrictions do keep the domain on one side: $x\ge 5$ lies completely to the right of $3$ ; $x\le 3$ is the left side plus the vertex; and $x\le 0$ is also entirely left of $3$ . Each passes the horizontal line test on that restricted domain. The trap here is assuming that any restriction starting below the vertex will work. It won't if it also includes values above the vertex. Study tip: complete the square to find the axis $x=h$ , then make sure the restricted domain is a subset of $(-\infty,h]$ or $[h,\infty)$ . If it straddles $h$ , check symmetric pairs of inputs.
Let $f(x)=\frac{3x-2}{x+k}$ , where $x\ne -k$ . If $f^{-1}(2)=5$ , what is $k$ ?
Answer: $\frac32$ $f^{-1}(2)=5$ is equivalent to $f(5)=2$ . Thus $\frac{15-2}{5+k}=2$ , so $13=10+2k$ and $k=\frac32$ . Using $f(2)=5$ would reverse the definition of inverse; sign slips produce the other fractions.
Suppose $f$ and $g$ are inverse functions. Which statement is NOT necessarily true?
Answer: For every $x$ in the domain of $f$ , $f(g(x))=x$ . Whenever you see inverse functions, immediately think about two things: the domain/range swap and the composition cancellation properties. Inverses "undo" each other, but only within the correct domains. The false statement is the one claiming that for every $x$ in the domain of $f$ , $f(g(x))=x$ . This is not necessarily true. If $x$ is in the domain of $f$ , then $g(x)$ is only defined when $x$ also lies in the domain of $g$ , which equals the range of $f$ . Those sets need not be the same. The guaranteed cancellation is $f(g(x))=x$ for every $x$ in the domain of $g$ , and $g(f(x))=x$ for every $x$ in the domain of $f$ . The other choices are true by definition: the range of $g$ equals the domain of $f$ , since inverse functions swap domains and ranges. The graph of $g$ is indeed the reflection of the graph of $f$ across the line $y=x$ . And the statement that $f(g(x))=x$ for every $x$ in the domain of $g$ is exactly the cancellation property for inverses. The trap here is mixing up the two composition directions. On the exam, write down both cancellation rules with their correct domains before answering. That makes it easy to spot which statement has the domain in the wrong place.

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