Finding and Verifying Inverse Functions Algebraically - College Algebra

1. Consider the function $h(x) = \sqrt{3x - 2} + 1$. Which of the following correctly represents the inverse function $h^{-1}(x)$?
   Answer: $h^{-1}(x) = \frac{(x - 1)^2 + 2}{3}$ for $x \geq 1$ 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.
2. A student claims that if $f(x) = x^3 - 2$ and $g(x) = \sqrt[3]{x + 2}$, then $g$ is the inverse of $f$. To verify this claim, which composition should equal $x$ for all $x$ in the appropriate domain?
   Answer: Both $f(g(x)) = x$ and $g(f(x)) = x$ must be verified 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.
3. A function $k(x) = \frac{ax + b}{cx + d}$ has inverse $k^{-1}(x) = \frac{dx - b}{-cx + a}$. If $k(2) = 3$ and $k^{-1}(3) = 2$, which statement must be true?
   Answer: This confirms the inverse relationship since $k(2) = 3$ implies $k^{-1}(3) = 2$ 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.
4. Consider $m(x) = \ln(x - 2) + 4$. Which expression correctly represents $m^{-1}(x)$, and what is its domain?
   Answer: $m^{-1}(x) = e^{x-4} + 2$ with domain all real numbers 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.
5. The function $p(x) = \frac{x - 4}{2x + 1}$ has domain $x \neq -\frac{1}{2}$. What is the domain of its inverse function $p^{-1}(x)$?
   Answer: $x \neq \frac{1}{2}$ 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.
6. Let $f(x) = 2^{x-3} + 1$. If $f^{-1}(x)$ represents the inverse function, what is $f^{-1}(9)$?
   Answer: $6$ 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.
7. The function $t(x) = \sqrt{2x - 1} - 3$ is defined for $x \geq \frac{1}{2}$. If the inverse function is $t^{-1}(x) = \frac{(x + 3)^2 + 1}{2}$, what should be the domain restriction for $t^{-1}(x)$?
   Answer: $x \geq -3$ 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.
8. Let $f(x) = \frac{2x + 3}{x - 1}$. 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 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.
9. For the function $w(x) = \frac{x^2 - 1}{x^2 + 1}$ where $x \geq 0$, a student claims the inverse is $w^{-1}(x) = \sqrt{\frac{1 + x}{1 - x}}$. Which verification step would definitively prove or disprove this claim?
   Answer: Compute $w(w^{-1}(x))$ and confirm it equals $x$ for all valid $x$ 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.
10. A student attempts to find the inverse of $r(x) = \frac{3x}{x + 2}$ and gets $r^{-1}(x) = \frac{2x}{x - 3}$. To check this algebraically, they compute $r(r^{-1}(x))$ and obtain $\frac{6x}{2x + 2(x-3)} = \frac{6x}{4x - 6}$. What can be concluded?
    Answer: The inverse is incorrect since $\frac{6x}{4x - 6}$ does not equal $x$ 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.

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