Step 1: Replace f(x) with y: y = 2x + 3

Step 2: Swap x and y: x = 2y + 3

Step 3: Solve for y: x - 3 = 2y y = (x - 3)/2

Step 4: Replace y with f⁻¹(x): f⁻¹(x) = (x - 3)/2

Step 5: Verify (check f(f⁻¹(x)) = x): f(f⁻¹(x)) = 2((x - 3)/2) + 3 = (x - 3) + 3 = x ✓

Answer: f⁻¹(x) = (x - 3)/2

Step 1: Write as $y = x + 7$

Step 2: Swap $x$ and $y$ $x = y + 7$

Step 3: Solve for $y$ $y = x - 7$

Step 4: Write inverse $f^{-1}(x) = x - 7$

Answer: $f^{-1}(x) = x - 7$

Step 1: Check f(g(x)) = x: f(g(x)) = f(∛x) = (∛x)³ = x ✓

Step 2: Check g(f(x)) = x: g(f(x)) = g(x³) = ∛(x³) = x ✓

Step 3: Conclusion: Since both compositions equal x, f and g are inverse functions

Answer: Yes, they are inverse functions

Step 1: Replace f(x) with y: y = (x + 1)/(x - 2)

Step 2: Swap x and y: x = (y + 1)/(y - 2)

Step 3: Solve for y (multiply both sides by (y - 2)): x(y - 2) = y + 1 xy - 2x = y + 1

Step 4: Collect y terms: xy - y = 2x + 1 y(x - 1) = 2x + 1

Step 5: Solve for y: y = (2x + 1)/(x - 1)

Step 6: Write inverse: f⁻¹(x) = (2x + 1)/(x - 1)

Answer: f⁻¹(x) = (2x + 1)/(x - 1)

Step 1: Write as $y = \frac{x - 1}{3}$

Step 2: Swap $x$ and $y$ $x = \frac{y - 1}{3}$

Step 3: Solve for $y$ $3x = y - 1$ $y = 3x + 1$

Step 4: Write inverse $f^{-1}(x) = 3x + 1$

Verify: $f(f^{-1}(x)) = \frac{(3x + 1) - 1}{3} = \frac{3x}{3} = x$ ✓

Answer: $f^{-1}(x) = 3x + 1$

Step 1: Write as $y = \frac{2x + 3}{x - 1}$

Step 2: Swap $x$ and $y$ $x = \frac{2y + 3}{y - 1}$

Step 3: Solve for $y$ (multiply both sides by denominator) $x(y - 1) = 2y + 3$ $xy - x = 2y + 3$

Group $y$ terms: $xy - 2y = x + 3$ $y(x - 2) = x + 3$ $y = \frac{x + 3}{x - 2}$

Answer: $f^{-1}(x) = \frac{x + 3}{x - 2}$

Step 1: Replace f(x) with y: y = √(x - 3)

Step 2: Identify domain and range of f: Domain of f: x ≥ 3 Range of f: y ≥ 0

Step 3: Swap x and y: x = √(y - 3)

Step 4: Solve for y: x² = y - 3 y = x² + 3

Step 5: Write inverse: f⁻¹(x) = x² + 3

Step 6: Domain and range of f⁻¹: Domain of f⁻¹ = Range of f: x ≥ 0 Range of f⁻¹ = Domain of f: y ≥ 3

Answer: f⁻¹(x) = x² + 3, Domain: x ≥ 0, Range: y ≥ 3

Step 1: Apply horizontal line test: Does any horizontal line intersect y = x² more than once? Yes - for example, y = 4 intersects at x = 2 and x = -2

Step 2: Conclusion about inverse: f(x) = x² does NOT have an inverse (not one-to-one)

Step 3: Restrict domain to make it one-to-one: Restrict to x ≥ 0 (right half) OR restrict to x ≤ 0 (left half)

Step 4: Find inverse with restriction x ≥ 0: y = x², x ≥ 0 x = y², y ≥ 0 y = √x

Step 5: Verify: f⁻¹(x) = √x, Domain: x ≥ 0

Answer: No inverse without restriction. With x ≥ 0: f⁻¹(x) = √x