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Algebra, MET1-NHT 2019 VCAA 5b

Let  `h:[-3/2, oo) -> R,\ h(x) = sqrt(2x + 3) - 2.`

Find the domain and the rule of the inverse function  `h^(-1)`.  (3 marks)

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`[–2, oo)`

Show Worked Solution

 `y = sqrt (2x + 3) – 2`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= sqrt(2y + 3) – 2`
`sqrt(2y + 3)` `= x + 2`
`2y + 3` `= (x + 2)^2`
`y` `= 1/2(x + 2)^2 – 3/2`
`:. h^(-1)` `= 1/2(x + 2)^2 – 3/2`

 

`text(Domain)\ \ h^(-1)(x)` `= text(Range)\ h(x)`
  `= [–2, oo)`

Algebra, MET2 2008 VCAA 7 MC

The inverse of the function  `f: R^+ -> R,\ f(x) = 1/sqrt x - 3`  is

  1. `{:f^-1: R^+ -> R, qquad qquad qquad qquad f^-1(x) = (x + 3)^2:}`
  2. `{:f^-1: R^+ -> R, qquad qquad qquad qquad f^-1(x) = 1/x^2 + 3:}`
  3. `{:f^-1: (3, oo) -> R, qquad qquad qquad f^-1 (x) = (-1)/(x - 3)^2:}`
  4. `{:f^-1: text{(−3, ∞)} -> R, qquad qquad f^-1 (x) = 1/(x + 3)^2:}`
  5. `{:f^-1: text{(−3, ∞)} -> R, qquad qquad f^-1 (x) = -1/x^2 - 3:}`
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`D`

Show Worked Solution

`text(Let)\ \ y = f(x)`

`text(Inverse:  swap)\ \ x harr y`

`x` `= 1/sqrt y – 3`
`x + 3` `= 1/sqrt y`
`y` `= 1/(x + 3)^2 = f^-1(x)`

 

`text(Domain)\ (f^-1(x))` `= text(Range)\ (f)`
  `= (– 3, oo)`

`=>   D`

Algebra, MET2 2016 VCAA 5 MC

Which one of the following is the inverse function of  `g: [3, oo) -> R,\ g(x) = sqrt (2x - 6)?`

  1. `g^(-1): [3, oo) -> R,\ g^(-1) (x) = (x^2 + 6)/2`
  2. `g^(-1): [0, oo) -> R,\ g^(-1) (x) = (2x - 6)^2`
  3. `g^(-1): [0, oo) -> R,\ g^(-1) (x) = sqrt (x/2 + 6)`
  4. `g^(-1): [0, oo) -> R,\ g^(-1) (x) = (x^2 + 6)/2`
  5. `g^(-1): R -> R,\ g^(-1) (x) = (x^2 + 6)/2`
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`D`

Show Worked Solution

`text(Let)\ \ y = g(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= sqrt (2y – 6)`
`x^2` `= 2y – 6`
`y` `= (x^2 + 6)/2`

 

`text(Domain)\ (g^(-1)) = text(Range)\ (g) = [0, oo)`

`=>   D`

Algebra, MET2 2011 VCAA 5 MC

The inverse function of  `g: [2,∞) -> R, g(x) = sqrt(2x - 4)`  is

  1. `g^(−1): [2,∞) -> R, g^(−1)(x) = (x^2 + 4)/2`
  2. `g^(−1): [0,∞) -> R, g^(−1)(x) = (2x - 4)^2`
  3. `g^(−1): [0,∞) -> R, g^(−1)(x) = sqrt(x/2 + 4)`
  4. `g^(−1): [0,∞) -> R, g^(−1)(x) = (x^2 + 4)/2`
  5. `g^(−1): R -> R, g^(−1)(x) = (x^2 + 4)/2`
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`=> D`

Show Worked Solution

`text(Let)\ \ y = g(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= sqrt(2y – 4)`
`x^2` `= 2y-4`
`2y` `=x^2+4`
`:. y` `= (x^2 +4)/2`

 

`g^(−1)(x) = (x^2 + 4)/2`

 

`text(Domain)\ (g^(−1)) = text(Range)\ g(x) = [0,∞)`

`=> D`

Algebra, MET2 2015 VCAA 2 MC

The inverse function of  `f:\ text{(−2, ∞)} -> R,\ f(x) = 1/sqrt(x + 2)` is

A.   `f^-1:\ R^+ -> R` `f^-1(x) = 1/x^2 - 2`
B.   `f^-1: R text(\{0}) -> R` `f^-1(x) = 1/x^2 - 2`
C.   `f^-1: R^+ -> R` `f^-1(x) = 1/x^2 + 2`
D.   `f^-1:\ text{(−2, ∞)} -> R` `f^-1(x) = x^2 + 2`
E.   `f^-1:\ (2, oo) -> R` `f^-1(x) = 1/(x^2 - 2)`
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`A`

Show Worked Solution

`text(Let)\ y = f(x)`

♦ Mean mark 50%.

`text(Inverse: swap)\ x ↔ y`

`x` `= 1/(sqrt(y + 2))`
`sqrt(y+2)` `=1/x`
`:.y` `=1/(x^2) – 2`

 

`text(Domain of)\ \ f^(−1)` `= text(Range)\ f(x)`
  `= R^+`

`=>   A`

Algebra, MET2 2014 VCAA 9 MC

The inverse of the function  `f: R^+ -> R,\ f(x) = 1/sqrt x + 4`  is

A.    `f^-1: (4, oo) -> R` `f^-1(x) = 1/(x - 4)^2`
B.    `f^-1: R^+ -> R` `f^-1(x) = 1/x^2 + 4`
C.    `f^-1: R^+ -> R` `f^-1(x) = (x + 4)^2`
D.    `f^-1:\ text{(−4, ∞)} -> R`        `f^-1(x) = 1/(x + 4)^2`
E.    `f^-1:\ text{(−∞, 4)} -> R` `f^-1(x) = 1/(x - 4)^2`
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`A`

Show Worked Solution

`text(Let)\ \ y = f(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= 1/sqrty + 4`
`x – 4` `= 1/sqrty`
`sqrty` `= 1/(x – 4)`
`y` `= 1/((x – 4)^2) = f^(−1)(x)`

 

`text(Domain)(f^(−1)) = text(Range)\ (f) = (4,∞)`

`=>   A`