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Functions, MET1-NHT 2018 VCAA 5

Let  `h: R^+ ∪ {0} → R, \ h(x) = ( 7)/(x + 2) - 3`.

  1.  State the range of  `h`.  (1 mark)
  2.  Find the rule for  `h^-1`.  (2 marks)
Show Answers Only
  1. `(-3, (1)/(2))`
  2. `(7)/(x + 3) – 2`
Show Worked Solution

a.    `y_text(max)\  text(occurs when) \ x = 0`

`y_text(max) = (7)/(2) – 3 = (1)/(2)`

`text(As) \ \ x → ∞ , \ (7)/(x + 2) \ → \ 0^+`

`:. \ text(Range) \ \ h(x) = (-3, (1)/(2))`
 

b.    `y = (7)/(x + 2)`
 

`text(Inverse: swap) \ x ↔ y`

`x` `= (7)/(y + 2) – 3`
`x + 3` `= (7)/(y + 2)`
`y + 2` `= (7)/(x + 3)`
`y` `= (7)/(x + 3) – 2`

  
`:. \ h^-1 = (7)/(x + 3) – 2`

Algebra, MET1 2019 VCAA 2

  1. Let  `f: R\{1/3} -> R,\ f(x) = 1/(3x - 1)`.

     

     

    Find the rule of  `f^(-1)`.  (2 marks) 

  2. State the domain of  `f^(-1)`.  (1 mark)
  3. Let  `g`  be the function obtained by applying the transformation  `T`  to the function  `f`, where
  4. `qquad qquad qquad T ([(x), (y)]) = [(x), (y)] + [(c), (d)]`
  5. and  `c, d in R`.
  6. Find the values of  `c`  and  `d`  given that  `g = f^(-1)`.  (1 mark)
Show Answers Only
  1. `y = 1/(3x) + 1/3`
  2. `x in R\ text(\){0}`
  3. `c = -1/3,\ d = 1/3`
Show Worked Solution

a.   `text(Let)\ \ y = 1/(3x – 1)`

`text(Inverse:  swap)\ \ x↔ y`

`x` `= 1/(3y – 1)`
`3y – 1` `= 1/x`
`3y` `= 1/x + 1`
`y` `= 1/(3x) + 1/3`

 

b.   `x in R\ text(\){0}`

 

c.   `f(x) = 1/(3x – 1) \ -> \ f(x) = 1/(3x) + 1/3`

`f(x + 1/3) = 1/(3(x + 1/3) – 1) = 1/(3x)`

`:. c = -1/3,\ \ d = 1/3`

Algebra, MET1 2018 VCAA 5

Let  `f: (2, ∞) -> R`, where  `f(x) = 1/((x - 2)^2)`.

State the rule and domain of  `f^(−1)`.  (3 marks)

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`f^(−1)(x) = 1/sqrtx + 2`

`text(Domain of)\ \ f^(−1)(x):\ (0, ∞)`

Show Worked Solution

`y = 1/((x – 2)^2)`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= 1/((y – 2)^2)`
`(y – 2)^2` `= 1/x`
`y – 2` `= 1/sqrtx`
`y` `= 1/sqrtx + 2`

 
`:. f^(−1)(x) = 1/sqrtx + 2`

`text(Domain of)\ \ f^(−1)(x):\ (0, ∞)\ \ text(or)\ \ R^+`

Functions, MET1 2009 VCAA 3

Let  `f: R\ text(\)\ {0} -> R`  where  `f(x) = 3/x - 4.`  Find  `f^-1`, the inverse function of  `f.`  (3 marks)

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`:. f^-1 (x) = 3/(x + 4),\ \ x in R\ text(\)\ {– 4}`

Show Worked Solution

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

MARKER’S COMMENT: “Few students” realised that both the equation and the domain had to be specified.

`text(For inverse swap)\ \ x harr y`

`x` `= 3/y – 4`
`3/y` `= x+4`
`y` `= 3/(x + 4)`

 

`:. f^-1 (x) = 3/(x + 4),\ \ x in R\ text(\)\ {– 4}`

Algebra, MET2 2013 VCAA 7 MC

The function  `g: text{[−a, a]} -> R, \ g(x) = sin (2(x - pi/6))`  has an inverse function.

The maximum possible value of  `a`  is

  1. `pi/12`
  2. `1`
  3. `pi/6`
  4. `pi/4`
  5. `pi/2`
Show Answers Only

`A`

Show Worked Solution

`text(If)\ \ y=sin x,\ \ text(the inverse function exists in)`

♦ Mean mark 37%.

`text(the domain)\ \ \ -pi/2<= x <= pi/2`

`text(Find)\ x\ text(such that:)`

`2(x – pi/6) = +- pi/2`

 

`:. g^-1\ \ text(exists in the domain)\ \ [-pi/12, (5pi)/12]`

 

`text(S)text(ince the form of the domain is)\ \ [a,-a],`

`a = pi/12`

`=>   A`