College Algebra Quiz: Finding And Verifying Inverse Functions Algebraically

What this quiz covers

This quiz focuses on Finding And Verifying Inverse Functions Algebraically, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The inverse is correct since the expression simplifies to $x$
2. The inverse is incorrect since $\frac{6x}{4x - 6}$ does not equal $x$ (correct answer)
3. The calculation contains an error in the denominator simplification
4. More work is needed since $\frac{6x}{4x - 6}$ can be simplified further

Explanation: To verify an inverse, r(r⁻¹(x)) must equal x. The student correctly computed r(r⁻¹(x)) = 6x/(4x-6). This can be simplified: 6x/(4x-6) = 6x/[2(2x-3)] = 3x/(2x-3), which does NOT equal x. Therefore, the proposed inverse is incorrect. The correct inverse should be r⁻¹(x) = 2x/(x-3). Choice A incorrectly claims it simplifies to x. Choice C suggests an error that doesn't exist. Choice D misses that even when simplified, the result doesn't equal x.

Question 2

1. $f(g(x)) = (\sqrt[3]{x + 2})^3 - 2 = x + 2 - 2 = x$
2. $g(f(x)) = \sqrt[3]{(x^3 - 2) + 2} = \sqrt[3]{x^3} = x$
3. Both $f(g(x)) = x$ and $g(f(x)) = x$ must be verified (correct answer)
4. $f(g(x)) = (\sqrt[3]{x + 2})^3 - 2 = x - 2$, so they are not inverses

Explanation: To verify that two functions are inverses, BOTH compositions f(g(x)) = x AND g(f(x)) = x must equal x. While choices A and B each show one correct composition, complete verification requires checking both. A complete verification would show: f(g(x)) = (∛(x+2))³ - 2 = (x+2) - 2 = x, and g(f(x)) = ∛((x³-2)+2) = ∛(x³) = x. Choice D makes an error in the cube root calculation.

Question 3

1. This confirms the inverse relationship since $k(2) = 3$ implies $k^{-1}(3) = 2$ (correct answer)
2. The functions cannot be inverses because $k(2) \neq k^{-1}(2)$
3. Additional verification is needed by checking $k^{-1}(2) = 3$
4. The inverse formula is incorrect based on these values

Explanation: The definition of inverse functions requires that if f(a) = b, then f⁻¹(b) = a. Since k(2) = 3 and k⁻¹(3) = 2, this is exactly the relationship that confirms they are inverses. Choice B misunderstands inverse functions—there's no requirement that k(2) = k⁻¹(2). Choice C suggests unnecessary additional checking when the given information already confirms the inverse relationship. Choice D incorrectly concludes the formula is wrong when the values actually support it.

Question 4

1. $m^{-1}(x) = e^{x-4} + 2$ with domain $x > 4$
2. $m^{-1}(x) = e^{x-4} + 2$ with domain all real numbers (correct answer)
3. $m^{-1}(x) = e^{x+4} + 2$ with domain all real numbers
4. $m^{-1}(x) = e^{x-4} - 2$ with domain all real numbers

Explanation: To find the inverse: Start with y = ln(x-2) + 4. Subtract 4: y - 4 = ln(x-2). Exponentiate: e^(y-4) = x-2. Solve for x: x = e^(y-4) + 2. So m⁻¹(x) = e^(x-4) + 2. The domain of m⁻¹(x) equals the range of m(x). Since ln(x-2) can output any real number, m(x) = ln(x-2) + 4 has range (-∞,∞). Therefore m⁻¹(x) has domain all real numbers. Choice A has the correct formula but wrong domain. Choice C has incorrect exponent. Choice D has wrong constant term.

Question 5

1. $h^{-1}(x) = \frac{(x - 1)^2 + 2}{3}$ for $x \geq 1$ (correct answer)
2. $h^{-1}(x) = \frac{(x + 1)^2 + 2}{3}$ for $x \geq -1$
3. $h^{-1}(x) = \frac{x^2 - 2x + 3}{3}$ for $x \geq 1$
4. $h^{-1}(x) = \frac{(x - 1)^2 + 2}{3}$ for $x \geq 0$

Explanation: To find the inverse: Start with y = √(3x - 2) + 1. Subtract 1: y - 1 = √(3x - 2). Square both sides: (y - 1)² = 3x - 2. Solve for x: x = [(y - 1)² + 2]/3. So h⁻¹(x) = [(x - 1)² + 2]/3. The domain restriction comes from h(x) ≥ 1 (since √(3x - 2) ≥ 0), so h⁻¹ has domain x ≥ 1. Choice B has wrong signs. Choice C expands incorrectly. Choice D has wrong domain restriction.

Question 6

1. Check that $w(0) = -1$ and $w^{-1}(-1) = 0$
2. Verify the domain restrictions match appropriately between functions
3. Compute $w(w^{-1}(x))$ and confirm it equals $x$ for all valid $x$ (correct answer)
4. Check that both functions are one-to-one on their domains

Explanation: The most definitive verification of inverse functions is to show that w(w⁻¹(x)) = x (and ideally w⁻¹(w(x)) = x as well). This directly tests the fundamental property of inverse functions. Computing w(w⁻¹(x)) = w(√[(1+x)/(1-x)]) and checking if it simplifies to x would definitively prove or disprove the claim. Choice A only checks one point, which is insufficient. Choice B is important but not definitive. Choice D verifies a necessary condition but doesn't confirm the specific inverse relationship.

Question 7

1. $2$ (from evaluating $f(5)$ incorrectly)
2. $\frac{13}{3}$
3. $\frac{8}{3}$ (correct answer)
4. $\frac{5}{2}$

Explanation: To find g(5), we need to solve f(x) = 5. Setting up: $\frac{2x + 3}{x - 1} = 5$. 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 A results from incorrectly evaluating the original function at 5. Choice B results from algebraic error in cross-multiplication. Choice D comes from incorrectly switching numerator and denominator.

Question 8

1. $x \geq \frac{1}{2}$
2. $x \geq 1$
3. $x \geq 0$
4. $x \geq -3$ (correct answer)

Explanation: When working with inverse functions, remember that the domain of the inverse function equals the range of the original function. To find the correct domain restriction for $t^{-1}(x)$, you need to determine what output values the original function $t(x) = \sqrt{2x - 1} - 3$ can produce. Since $t(x)$ is defined for $x \geq \frac{1}{2}$, the expression under the square root starts at $2(\frac{1}{2}) - 1 = 0$, so $\sqrt{2x - 1} \geq 0$. This means $t(x) = \sqrt{2x - 1} - 3 \geq 0 - 3 = -3$. As $x$ increases beyond $\frac{1}{2}$, the square root grows without bound, so $t(x)$ can take any value greater than or equal to $-3$. Therefore, the range of $t(x)$ is $y \geq -3$, making the domain of $t^{-1}(x)$ also $x \geq -3$. Answer choice A ($x \geq \frac{1}{2}$) incorrectly uses the domain of the original function rather than its range. Choice B ($x \geq 1$) might come from mistakenly adding the constant terms $\frac{1}{2} + \frac{1}{2} = 1$. Choice C ($x \geq 0$) could result from forgetting about the $-3$ shift and assuming the square root function starts at zero. Study tip: Always remember that domain and range switch places between a function and its inverse. To find the domain of an inverse function, determine the range of the original function by analyzing its minimum and maximum possible output values.

Question 9

1. $x \neq -\frac{1}{2}$
2. $x \neq \frac{1}{2}$ (correct answer)
3. $x \neq 4$
4. $x \neq -4$

Explanation: The domain of p⁻¹(x) equals the range of p(x). To find the range, we solve y = (x-4)/(2x+1) for x: y(2x+1) = x-4, so 2xy + y = x - 4, giving 2xy - x = -4 - y, so x(2y-1) = -4-y, and x = (-4-y)/(2y-1). This is undefined when 2y-1 = 0, i.e., when y = 1/2. Therefore, the range of p(x) is all real numbers except 1/2, making the domain of p⁻¹(x) all real numbers except x ≠ 1/2. Choice A confuses domain with range. Choices C and D result from misunderstanding the relationship between domains and ranges of inverse functions.

Question 10

1. $6$ (correct answer)
2. $3 + 3 = 6$
3. $\log_2(10) + 3$
4. $\log_2(9) + 3$

Explanation: To find f⁻¹(9), solve f(x) = 9: 2^(x-3) + 1 = 9, so 2^(x-3) = 8. Since 8 = 2³, we have x - 3 = 3, giving x = 6. Therefore f⁻¹(9) = 6. Choice B shows correct working but with unnecessary steps. Choice C results from forgetting to subtract 1 before taking the logarithm. Choice D results from taking log₂(9) directly without properly isolating the exponential term.