Mathematics 10 — Exercise 4.1 · Q8 (iv)

Full question

Verify f(f⁻¹(x))=f⁻¹(f(x))=x on the valid domains.

Step-by-step solution

1. Write the output as y. To find the inverse, solve this equation for the original input x.

   $$y=\frac{x-4}{x+2}$$

2. Multiply both sides by the denominator. The original denominator must remain nonzero.

   $$y\times \left(x+2\right)=x-4$$

3. Expand the left bracket.

   $$x\times y+2\times y=x-4$$

4. Subtract the x term on the right and the y-only term on the left.

   $$x\times y-x=-2\times y-4$$

5. Factor out the original input x.

   $$x\times \left(y-1\right)=-2\times y-4$$

6. Divide both sides by the bracket, which must be nonzero.

   $$x=\frac{-2\times y-4}{y-1}$$

7. Interchange the names x and y: the old output is now the inverse input.

   $$\mathrm{f⁻¹}\left(x\right)=\frac{-2\times x-4}{x-1}$$

8. The inverse domain is the original range. Keep any exclusions or nonnegative-input restriction.

   $$(-\infty ,1)\cup (1,\infty )$$

9. Compute f(f⁻¹(x)).

10. Composition means apply the inner function first, then use its entire output as the input of the outer function.

    $$\frac{-2\times x-4}{x-1}$$

11. Replace every x of the outer function by this full bracketed inner expression.

    $$\frac{\frac{-2\times x-4}{x-1}-4}{\frac{-2\times x-4}{x-1}+2}$$

12. Substitute the function expressions, keeping the brackets.

    $$\frac{\frac{-2\times x-4}{x-1}-4}{\frac{-2\times x-4}{x-1}+2}$$

13. Distribute each bracket product: each term multiplies every term in the other bracket.

    $$-\frac{2\times x}{-\frac{2\times x^2}{x-1}+2\times x-\frac{2\times x}{x-1}-2+\frac{4}{x-1}}-\frac{4}{-\frac{2\times x^2}{x-1}+2\times x-\frac{2\times x}{x-1}-2+\frac{4}{x-1}}-\frac{4}{-\frac{2\times x}{x-1}+2-\frac{4}{x-1}}$$

14. Write EVERY distributed product before collecting terms.

    $$\frac{\frac{-4}{x-1}}{\frac{-2\times x-4}{x-1}+2}+\frac{\frac{-2\times x}{x-1}}{\frac{-2\times x-4}{x-1}+2}+\frac{-4}{\frac{-2\times x-4}{x-1}+2}$$

15. Calculate this individual product, including its sign and powers.

    $$\frac{-4}{\frac{-2\times x-4}{x-1}+2}=\frac{2\times x}{3}-\frac{2}{3}$$

16. Calculate this individual product, including its sign and powers.

    $$\frac{\frac{-4}{x-1}}{\frac{-2\times x-4}{x-1}+2}=\frac{2}{3}$$

17. Calculate this individual product, including its sign and powers.

    $$\frac{\frac{-2\times x}{x-1}}{\frac{-2\times x-4}{x-1}+2}=\frac{x}{3}$$

18. For multiplication of fractions, multiply the numerators together and the denominators together.

    $$\frac{\left(-2\right)\times 1}{1\times 3}$$

19. Calculate these two products separately.

    $$-\frac{2}{3}$$

20. Calculate this operation before combining the remaining terms.

    $$\left(-2\right)\times \frac{1}{3}=-\frac{2}{3}$$

21. For multiplication of fractions, multiply the numerators together and the denominators together.

    $$\frac{2\times 1}{1\times 3}$$

22. Calculate these two products separately.

    $$\frac{2}{3}$$

23. Calculate this operation before combining the remaining terms.

    $$2\times \frac{1}{3}=\frac{2}{3}$$

24. To add these fractions, first use a common denominator: the least common multiple of their denominators.

    $$\mathrm{LCD}=3$$

25. Multiply this numerator AND denominator by the same factor. Its value stays unchanged.

    $$-\frac{2}{3}=\frac{\left(-2\right)\times 1}{3\times 1}$$

26. Multiply this numerator AND denominator by the same factor. Its value stays unchanged.

    $$\frac{2}{3}=\frac{2\times 1}{3\times 1}$$

27. Keep the common denominator and add the signed numerators.

    $$\frac{-2+2}{3}=\frac{0}{3}$$

28. The numerator and denominator have this greatest common divisor. Divide BOTH by it.

    $$\mathrm{GCD}=3$$

29. Show the two divisions explicitly.

    $$\frac{\frac{0}{3}}{\frac{3}{3}}$$

30. Evaluate the numerator division.

    $$\frac{0}{3}=0$$

31. Evaluate the denominator division.

    $$\frac{3}{3}=1$$

32. Write the reduced fraction.

    $$0\times \frac{1}{1}$$

33. These terms have the same variable part. Add ONLY their coefficients.

    $$-\frac{2}{3}+\frac{2}{3}=0$$

34. For multiplication of fractions, multiply the numerators together and the denominators together.

    $$\frac{1\times 1}{1\times 3}$$

35. Calculate these two products separately.

    $$\frac{1}{3}$$

36. Calculate this operation before combining the remaining terms.

    $$1\times \frac{1}{3}=\frac{1}{3}$$

37. Multiply this numerator AND denominator by the same factor. Its value stays unchanged.

    $$\frac{1}{3}=\frac{1\times 1}{3\times 1}$$

38. Keep the common denominator and add the signed numerators.

    $$\frac{2+1}{3}=\frac{3}{3}$$

39. Show the two divisions explicitly.

    $$\frac{\frac{3}{3}}{\frac{3}{3}}$$

40. Write the reduced fraction.

    $$\frac{1}{1}$$

41. These terms have the same variable part. Add ONLY their coefficients.

    $$\frac{2}{3}+\frac{1}{3}=1$$

42. Put the simplified groups together.

    $$0+x$$

43. Collect terms with the same power and simplify their coefficients.

    $$x$$

44. Retain inputs where the inner function is defined and where its output makes the outer function defined. Neither denominator may be zero.

    $$(x-1\ne 0,x-1\ne 0)$$

45. Compute f⁻¹(f(x)) independently.

46. Composition means apply the inner function first, then use its entire output as the input of the outer function.

    $$\frac{x-4}{x+2}$$

47. Replace every x of the outer function by this full bracketed inner expression.

    $$\frac{-2\times \left(x-4\right)\times \frac{1}{x+2}-4}{\frac{x-4}{x+2}-1}$$

48. Substitute the function expressions, keeping the brackets.

    $$\frac{-2\times \left(x-4\right)\times \frac{1}{x+2}-4}{\frac{x-4}{x+2}-1}$$

49. Distribute each bracket product: each term multiplies every term in the other bracket.

    $$-\frac{2\times x}{\frac{x^2}{x+2}-x-\frac{2\times x}{x+2}-2-\frac{8}{x+2}}+\frac{8}{\frac{x^2}{x+2}-x-\frac{2\times x}{x+2}-2-\frac{8}{x+2}}-\frac{4}{\frac{x}{x+2}-1-\frac{4}{x+2}}$$

50. Write EVERY distributed product before collecting terms.

    $$\frac{-2\times x\times \frac{1}{x+2}}{\frac{x-4}{x+2}-1}+\frac{-2\times \left(-4\right)\times \frac{1}{x+2}}{\frac{x-4}{x+2}-1}+\frac{-4}{\frac{x-4}{x+2}-1}$$

51. Calculate this individual product, including its sign and powers.

    $$\frac{-4}{\frac{x-4}{x+2}-1}=\frac{2\times x}{3}+\frac{4}{3}$$

52. Calculate this individual product, including its sign and powers.

    $$\frac{-2\times \left(-4\right)\times \frac{1}{x+2}}{\frac{x-4}{x+2}-1}=-\frac{4}{3}$$

53. Calculate this individual product, including its sign and powers.

    $$\frac{-2\times x\times \frac{1}{x+2}}{\frac{x-4}{x+2}-1}=\frac{x}{3}$$

54. For multiplication of fractions, multiply the numerators together and the denominators together.

    $$\frac{4\times 1}{1\times 3}$$

55. Calculate these two products separately.

    $$\frac{4}{3}$$

56. Calculate this operation before combining the remaining terms.

    $$4\times \frac{1}{3}=\frac{4}{3}$$

57. For multiplication of fractions, multiply the numerators together and the denominators together.

    $$\frac{\left(-1\right)\times 4}{1\times 3}$$

58. Calculate these two products separately.

    $$-\frac{4}{3}$$

59. Calculate this operation before combining the remaining terms.

    $$\left(-1\right)\times \frac{4}{3}=-\frac{4}{3}$$

60. Multiply this numerator AND denominator by the same factor. Its value stays unchanged.

    $$\frac{4}{3}=\frac{4\times 1}{3\times 1}$$

61. Multiply this numerator AND denominator by the same factor. Its value stays unchanged.

    $$-\frac{4}{3}=\frac{\left(-4\right)\times 1}{3\times 1}$$

62. Keep the common denominator and add the signed numerators.

    $$\frac{4-4}{3}=\frac{0}{3}$$

63. These terms have the same variable part. Add ONLY their coefficients.

    $$\frac{4}{3}-\frac{4}{3}=0$$

64. These terms have the same variable part. Add ONLY their coefficients.

    $$\frac{2}{3}+\frac{1}{3}=1$$

65. Put the simplified groups together.

    $$0+x$$

66. Collect terms with the same power and simplify their coefficients.

    $$x$$

67. Retain inputs where the inner function is defined and where its output makes the outer function defined. Neither denominator may be zero.

    $$(x+2\ne 0,x+2\ne 0)$$

Answer

The final answer is shown below.