Finding & Verifying Inverse Functions - Math 3

If $p(x) = \frac{x - 2}{3x + 1}$ and $q(x) = \frac{x + 2}{1 - 3x}$ , which statement about these functions is correct?
Answer: $q(x) = p^{-1}(x)$ and they are inverse functions of each other When you encounter questions about inverse functions, you need to verify two key conditions: first, that composing the functions in either order yields the identity function, and second, that their domains and ranges are properly aligned. To check if $q(x) = p^{-1}(x)$ , let's verify that $p(q(x)) = x$ . Starting with $q(x) = \frac{x + 2}{1 - 3x}$ , we substitute this into $p(x)$ : $p(q(x)) = p\left(\frac{x + 2}{1 - 3x}\right) = \frac{\frac{x + 2}{1 - 3x} - 2}{3 \cdot \frac{x + 2}{1 - 3x} + 1}$ Simplifying the numerator: $\frac{x + 2}{1 - 3x} - 2 = \frac{x + 2 - 2(1 - 3x)}{1 - 3x} = \frac{x + 2 - 2 + 6x}{1 - 3x} = \frac{7x}{1 - 3x}$ Simplifying the denominator: $3 \cdot \frac{x + 2}{1 - 3x} + 1 = \frac{3(x + 2) + (1 - 3x)}{1 - 3x} = \frac{3x + 6 + 1 - 3x}{1 - 3x} = \frac{7}{1 - 3x}$ Therefore: $p(q(x)) = \frac{\frac{7x}{1 - 3x}}{\frac{7}{1 - 3x}} = x$ Similarly, you can verify that $q(p(x)) = x$ . Since both compositions yield the identity function, $q(x) = p^{-1}(x)$ . Answer choice A incorrectly suggests domain misalignment, but the domains work properly for the inverse relationship. Choice B wrongly claims domain restrictions are needed beyond what's already present. Choice C uses faulty reasoning with a specific value test that doesn't properly verify the inverse relationship. Remember: to prove two functions are inverses, show that both $f(g(x)) = x$ and $g(f(x)) = x$ through algebraic manipulation, not just specific value checks.
If $f(x) = \sqrt{x + 3} - 1$ where $x \geq -3$ , then $f^{-1}(2)$ equals which value?
Answer: $6$ When you encounter inverse function problems, remember that $f^{-1}(2)$ asks "what input value gives an output of 2?" Rather than finding the inverse function formula, you can solve this directly by setting $f(x) = 2$ . Starting with $f(x) = \sqrt{x + 3} - 1 = 2$ , add 1 to both sides: $\sqrt{x + 3} = 3$ . Square both sides to eliminate the square root: $x + 3 = 9$ . Solving for $x$ : $x = 6$ . Let's verify: $f(6) = \sqrt{6 + 3} - 1 = \sqrt{9} - 1 = 3 - 1 = 2$ ✓ Now examining the wrong answers: Choice A ( $12$ ) likely comes from incorrectly setting up $\sqrt{x + 3} = 2$ instead of $\sqrt{x + 3} = 3$ , leading to $x + 3 = 4$ , so $x = 1$ . But this gives $f(1) = 1$ , not 2. Some students might then add more steps incorrectly. Choice B ( $9$ ) represents stopping at $x + 3 = 9$ without subtracting 3. Choice C ( $\sqrt{6}$ ) might result from confusing the inverse process or not fully simplifying $\sqrt{9}$ . Study tip: For inverse function evaluation problems, don't waste time finding the complete inverse function. Instead, set $f(x)$ equal to your target output and solve for $x$ . Always verify your answer by plugging it back into the original function.
Let $p(x) = 2x + 1$ and $q(x) = x^2$ for $x \geq 0$ . If $r(x) = p(q(x))$ , what is $r^{-1}(19)$ ? Give your answer in simplest form.
Answer: $3$ First find $r(x) = p(q(x)) = p(x^2) = 2x^2 + 1$ . To find $r^{-1}(19)$ , solve $r(x) = 19$ : $2x^2 + 1 = 19$ , so $2x^2 = 18$ , giving $x^2 = 9$ , and $x = \pm 3$ . However, since $q(x) = x^2$ is restricted to $x \geq 0$ , the composition $r(x)$ inherits this domain restriction, so $r^{-1}(19) = 3$ . Choice B gives the value of $x^2$ rather than $x$ . Choice C is equivalent to choice A but unnecessarily complex. Choice D ignores the domain restriction that makes $r(x)$ one-to-one.
The function $f(x) = 3x - 7$ has domain $x \geq 2$ . What is the range of the inverse function $f^{-1}(x)$ ?
Answer: $x \geq 2$ When you encounter questions about inverse functions and their domains and ranges, remember this key relationship: the domain of a function becomes the range of its inverse, and the range of a function becomes the domain of its inverse. Let's work through this systematically. First, we need to find the range of the original function $f(x) = 3x - 7$ with domain $x \geq 2$ . Since this is a linear function with a positive slope, it's strictly increasing. When $x = 2$ (the minimum input), we get $f(2) = 3(2) - 7 = -1$ . As $x$ increases beyond 2, $f(x)$ increases without bound. Therefore, the range of $f(x)$ is $y \geq -1$ . Now, since the range of $f(x)$ is $y \geq -1$ , this becomes the domain of $f^{-1}(x)$ . The domain of $f(x)$ is $x \geq 2$ , so this becomes the range of $f^{-1}(x)$ . The answer is D. Let's examine why the other options are incorrect. Choice A ( $x \geq 7$ ) likely comes from incorrectly plugging in the domain value: $f(2) = 3(2) - 7 = -1$ , but someone might have calculated $3 + 7 - 3 = 7$ . Choice B ( $x \geq -1$ ) confuses the range of the original function with the range of the inverse—this is actually the domain of $f^{-1}(x)$ . Choice C ( $x \geq \frac{7}{3}$ ) might result from solving $3x - 7 = 0$ incorrectly. Remember: domain and range swap places when you move from a function to its inverse. Always identify both for the original function first.
Let $f(x) = x^2 - 4x + 5$ for $x \geq 2$ . If $f^{-1}$ exists on this restricted domain, what is $f^{-1}(1) + f^{-1}(5)$ ?
Answer: $6$ First, let's complete the square: $f(x) = (x - 2)^2 + 1$ . For $x \geq 2$ , this function is one-to-one with range $y \geq 1$ . To find $f^{-1}(1)$ , solve $f(x) = 1$ : $(x - 2)^2 + 1 = 1$ , so $(x - 2)^2 = 0$ , giving $x = 2$ . To find $f^{-1}(5)$ , solve $f(x) = 5$ : $(x - 2)^2 + 1 = 5$ , so $(x - 2)^2 = 4$ , giving $x - 2 = \pm 2$ . Since $x \geq 2$ , we take $x - 2 = 2$ , so $x = 4$ . Therefore $f^{-1}(1) + f^{-1}(5) = 2 + 4 = 6$ . Choice A uses only the positive solution from the first equation. Choice C would result from taking both solutions $x = 0$ and $x = 4$ for the second equation, ignoring the domain restriction. Choice D represents just $f^{-1}(1)$ .
Functions $u(x) = \frac{2x - 1}{3}$ and $v(x) = \frac{3x + 1}{2}$ are tested to see if they are inverses. What is the value of $u(v(x)) + v(u(x))$ ?
Answer: $2x$ When you encounter questions about function composition, you're testing whether two functions "undo" each other. If functions are true inverses, then $u(v(x)) = x$ and $v(u(x)) = x$ , so their sum would equal $2x$ . Let's calculate each composition step by step. For $u(v(x))$ , substitute $v(x) = \frac{3x + 1}{2}$ into $u(x) = \frac{2x - 1}{3}$ : $u(v(x)) = u\left(\frac{3x + 1}{2}\right) = \frac{2 \cdot \frac{3x + 1}{2} - 1}{3} = \frac{3x + 1 - 1}{3} = \frac{3x}{3} = x$ For $v(u(x))$ , substitute $u(x) = \frac{2x - 1}{3}$ into $v(x) = \frac{3x + 1}{2}$ : $v(u(x)) = v\left(\frac{2x - 1}{3}\right) = \frac{3 \cdot \frac{2x - 1}{3} + 1}{2} = \frac{2x - 1 + 1}{2} = \frac{2x}{2} = x$ Therefore, $u(v(x)) + v(u(x)) = x + x = 2x$ , confirming these functions are indeed inverses. Choice A, $\frac{4x + 1}{3}$ , results from incorrectly combining terms during composition. Choice B, $x$ , comes from mistakenly thinking the sum equals just one composition instead of both. Choice D, $\frac{6x + 2}{6}$ , appears to stem from algebraic errors in the composition process, though it does simplify to $x + \frac{1}{3}$ . Remember: when checking if functions are inverses, compute both compositions carefully. If each equals $x$ , they're inverses, and their sum will always be $2x$ .
Consider the piecewise function $f(x) = \begin{cases} 2x + 1 & \text{if } x < 0 \\ x^2 + 1 & \text{if } x \geq 0 \end{cases}$ . For what values of $x$ does $f$ have an inverse function when restricted to the domain $x \geq -2$ ?
Answer: The function has an inverse only when further restricted to $x \geq 0$ For a function to have an inverse, it must be one-to-one (injective). On $[-2, 0)$ , $f(x) = 2x + 1$ gives values from $f(-2) = -3$ to $f(0^-) = 1$ . On $[0, \infty)$ , $f(x) = x^2 + 1$ gives values from $f(0) = 1$ to $\infty$ . The issue is that $f(x) = x^2 + 1$ for $x \geq 0$ is not one-to-one since $f(a) = f(-a)$ for any $a > 0$ . However, when we restrict to $x \geq 0$ , the quadratic piece $x^2 + 1$ is one-to-one (increasing). Also, we need to check if there's overlap in range values. The linear piece on $[-2, 0)$ has range $[-3, 1)$ and the quadratic piece on $[0, \infty)$ has range $[1, \infty)$ . At $x = 0$ , both pieces would give $f(0) = 1$ , but since $x = 0$ belongs to the quadratic piece, there's no issue. The function is one-to-one on $[0, \infty)$ . Choice A ignores that the quadratic piece needs restriction. Choice C over-restricts. Choice D incorrectly concludes no inverse exists.
Given that $g(x) = \frac{ax + b}{cx + d}$ where $ad - bc \neq 0$ , and $g(1) = 2$ , $g(2) = 3$ , if $g^{-1}(2) = 1$ , what additional relationship must hold?
Answer: No additional relationship is required since $g^{-1}(2) = 1$ is automatically satisfied when $g(1) = 2$ Since g and $g^{-1}$ are inverse functions, if $g(1) = 2$ , then automatically $g^{-1}(2) = 1$ by the definition of inverse functions. This is not an additional constraint but rather a consequence of the given information. The condition $g^{-1}(2) = 1$ doesn't provide new information beyond $g(1) = 2$ . Choice A incorrectly suggests we can deduce $g^{-1}(3) = 2$ , but while $g(2) = 3$ does imply $g^{-1}(3) = 2$ , the question asks what MUST hold given $g^{-1}(2) = 1$ . Choice B proposes an unrelated constraint. Choice C gives a true but irrelevant formula for $g(0)$ .
Let $f(x) = \frac{2x + 3}{x - 1}$ where $x \neq 1$ . If $g(x)$ is the inverse function of $f(x)$ , what is $g(5)$ ?
Answer: $\frac{8}{3}$ To find g(5), we need to solve f(x) = 5. Setting $\frac{2x + 3}{x - 1} = 5$ , we get $2x + 3 = 5(x - 1) = 5x - 5$ . This gives us $2x + 3 = 5x - 5$ , so $8 = 3x$ , and $x = \frac{8}{3}$ . Therefore $g(5) = \frac{8}{3}$ . Choice B results from incorrectly solving $2x + 3 = 5x + 5$ . Choice C comes from setting up $\frac{2x + 3}{x - 1} = 5$ but solving $2x + 3 = 5x - 1$ . Choice D results from the error $2x + 3 = 5(x - 1)$ becoming $2x + 3 = 5x - 3$ .
The function $f(x) = \frac{ax + b}{cx + d}$ has inverse $f^{-1}(x) = \frac{dx - b}{-cx + a}$ . If $f(2) = 3$ and $f^{-1}(3) = 2$ , which relationship must hold?
Answer: $ad - bc = -1$ and $f(2) = 3$ implies $f^{-1}(3) = 2$ by the definition of inverse functions For a rational function to have an inverse, the determinant $ad - bc \neq 0$ . The given inverse formula is correct when $ad - bc = -1$ (or we could multiply by -1 to get the standard form). Since f and $f^{-1}$ are inverses, if $f(2) = 3$ , then automatically $f^{-1}(3) = 2$ by definition - these aren't independent conditions. Choice A has the wrong determinant value. Choice B incorrectly suggests the conditions are independent. Choice D gives an impossible determinant condition (this would make the function non-invertible).
A function $g(x) = ax + b$ has an inverse $g^{-1}(x) = \frac{x - 4}{3}$ . If $g(g^{-1}(7)) + g^{-1}(g(-2)) = k$ , what is the value of $k$ ?
Answer: $5$ When you encounter inverse function problems, remember that inverse functions "undo" each other. If $g$ and $g^{-1}$ are inverses, then $g(g^{-1}(x)) = x$ and $g^{-1}(g(x)) = x$ for all values in their domains. Let's find the original function $g(x)$ first. Since $g^{-1}(x) = \frac{x - 4}{3}$ , we can find $g(x)$ by swapping variables and solving. Let $y = \frac{x - 4}{3}$ , so $3y = x - 4$ , which gives us $x = 3y + 4$ . Therefore, $g(x) = 3x + 4$ . Now we can evaluate $g(g^{-1}(7)) + g^{-1}(g(-2))$ . Using the inverse function property, $g(g^{-1}(7)) = 7$ . Similarly, $g^{-1}(g(-2)) = -2$ . Therefore, $k = 7 + (-2) = 5$ . Looking at the wrong answers: Choice A ( $3$ ) might result from incorrectly calculating $g^{-1}(7) = \frac{7-4}{3} = 1$ and then adding $1 + 2 = 3$ . Choice B ( $9$ ) could come from adding $g^{-1}(7) = 1$ and $g(-2) = 3(-2) + 4 = -2$ incorrectly, getting $1 + 8 = 9$ . Choice D ( $1$ ) might result from finding only $g^{-1}(7) = 1$ and stopping there. Remember: when you see $f(f^{-1}(a))$ or $f^{-1}(f(a))$ , these always simplify to just $a$ . This inverse function property is your shortcut to avoid unnecessary calculations.
Consider the function $h(x) = \sqrt{x + 4} - 2$ where $x \geq -4$ . If $h^{-1}(x)$ represents the inverse function, which statement correctly verifies that $h^{-1}(x) = x^2 + 4x$ for $x \geq 0$ ?
Answer: $h(h^{-1}(3)) = h(21) = \sqrt{25} - 2 = 3$ and $h^{-1}(h(5)) = h^{-1}(1) = 5$ To verify inverse functions, we need both $h(h^{-1}(x)) = x$ and $h^{-1}(h(x)) = x$ . First, $h^{-1}(3) = 3^2 + 4(3) = 9 + 12 = 21$ , so $h(h^{-1}(3)) = h(21) = \sqrt{21 + 4} - 2 = \sqrt{25} - 2 = 5 - 2 = 3$ ✓. Next, $h(5) = \sqrt{5 + 4} - 2 = \sqrt{9} - 2 = 3 - 2 = 1$ , so $h^{-1}(h(5)) = h^{-1}(1) = 1^2 + 4(1) = 5$ ✓. Choice B incorrectly computes $\sqrt{21 + 4}$ as $\sqrt{21}$ . Choice C uses $h^{-1}(3) = 9$ instead of 21. Choice D uses an invalid test point since $h^{-1}(-2)$ is undefined for $x \geq 0$ .
Consider $f(x) = \frac{1}{x + 2}$ for $x \neq -2$ . Which expression represents $f^{-1}(x)$ and what is its domain?
Answer: $f^{-1}(x) = \frac{1}{x} - 2$ with domain $x \neq 0$ To find the inverse, let $y = \frac{1}{x + 2}$ . Solving for x: $y(x + 2) = 1$ , so $yx + 2y = 1$ , giving $yx = 1 - 2y$ , and $x = \frac{1 - 2y}{y} = \frac{1}{y} - 2$ . Therefore $f^{-1}(x) = \frac{1}{x} - 2$ . The domain excludes $x = 0$ since we can't divide by zero. We can verify: $f(f^{-1}(x)) = f\left(\frac{1}{x} - 2\right) = \frac{1}{\frac{1}{x} - 2 + 2} = \frac{1}{\frac{1}{x}} = x$ . Choice B shows the intermediate step before simplification. Choice C incorrectly restricts to positive values only. Choice D has algebraic errors in the numerator.
If $f(x) = \frac{2x + 3}{x - 1}$ where $x \neq 1$ , and $g(x)$ is the inverse function of $f(x)$ , what is $g(5)$ ?
Answer: $\frac{8}{3}$ To find g(5), we need to find the value of x such that f(x) = 5. Setting up the equation: $\frac{2x + 3}{x - 1} = 5$ . Cross-multiplying: $2x + 3 = 5(x - 1) = 5x - 5$ . Solving: $2x + 3 = 5x - 5$ , so $8 = 3x$ , giving $x = \frac{8}{3}$ . Therefore, $g(5) = \frac{8}{3}$ . Choice B results from incorrectly setting up $2x + 3 = 5x + 5$ . Choice C comes from the error $2x + 3 = 5x - 1$ . Choice D results from solving $2x + 3 = 4(x - 1)$ instead of $5(x - 1)$ .
Function $f(x) = \frac{x}{x - 3}$ where $x \neq 3$ has an inverse. If we compute $(f^{-1} \circ f)(5)$ , which step correctly identifies a potential error?
Answer: The composition should equal 5, but we must verify 5 is in the domain of $f^{-1} \circ f$ For inverse functions, $(f^{-1} \circ f)(x) = x$ for all $x$ in the domain of $f$ . Since $5 \neq 3$ , it's in the domain of $f$ , so $(f^{-1} \circ f)(5) = 5$ . However, we should verify that 5 is indeed in the domain where this composition is defined. The domain of $f^{-1} \circ f$ is the set of $x$ values where $f(x)$ is defined AND $f(x)$ is in the domain of $f^{-1}$ . Choice A focuses on the wrong condition. Choice C gives an incorrect value and reasoning. Choice D misunderstands how function composition works.
Let $f(x) = x^2 + 6x + 5$ for $x \geq -3$ . To find $f^{-1}(x)$ , a student completes the square and gets $f(x) = (x + 3)^2 - 4$ . What is $f^{-1}(12)$ ?
Answer: $1$ When finding inverse functions, you're essentially solving for the input that produces a given output. The completed square form $f(x) = (x + 3)^2 - 4$ makes this process much cleaner than working with the original quadratic. To find $f^{-1}(12)$ , you need to solve $f(x) = 12$ . Using the completed square form: $(x + 3)^2 - 4 = 12$ $(x + 3)^2 = 16$ $x + 3 = \pm 4$ $x = -3 \pm 4$ This gives you $x = 1$ or $x = -7$ . However, the domain restriction $x \geq -3$ is crucial here. Since $-7 < -3$ , this solution is outside the allowed domain. Therefore, $x = 1$ is the only valid solution, making $f^{-1}(12) = 1$ . Answer C) $1$ is correct. Answer A) $4$ likely comes from solving $(x + 3)^2 = 16$ and mistakenly thinking $x + 3 = 4$ gives $x = 4$ instead of $x = 1$ . Answer B) $-7$ represents the other solution to the equation that gets eliminated by the domain restriction. Answer D) $-1$ might result from arithmetic errors in the solving process or confusion about the domain boundary. Study tip: Always check domain restrictions when finding inverse functions. Quadratic functions are only one-to-one (and thus invertible) when their domain is restricted, so you'll often need to eliminate one of the two solutions you get from solving the quadratic equation.

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