algebraically find the inverse of a function​

9514 1404 393

Answer:

  1)  f⁻¹(x) = 6 ± 2√(x -1)

  3)  y = (x +4)² -2

  5)  y = (x -4)³ -4

Step-by-step explanation:

In general, swap x and y, then solve for y. Quadratics, as in the first problem, do not have an inverse function: the inverse relation is double-valued, unless the domain is restricted. Here, we're just going to consider these to be "solve for ..." problems, without too much concern for domain or range.

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1) x = f(y)

  x = (1/4)(y -6)² +1

  4(x -1) = (y-6)² . . . . . . subtract 1, multiply by 4

  ±2√(x -1) = y -6 . . . . square root

  y = 6 ± 2√(x -1) . . . . inverse relation

  f⁻¹(x) = 6 ± 2√(x -1) . . . . in functional form

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3) x = √(y +2) -4

  x +4 = √(y +2) . . . . add 4

  (x +4)² = y +2 . . . . square both sides

  y = (x +4)² -2 . . . . . subtract 2

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5) x = ∛(y +4) +4

  x -4 = ∛(y +4) . . . . . subtract 4

  (x -4)³ = y +4 . . . . . cube both sides

  y = (x -4)³ -4 . . . . . . subtract 4

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оооо
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