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Functions, MET2 2022 VCAA 6 MC

Which of the pairs of functions below are not inverse functions?

  1. \( \begin {cases}f(x)=5x+3         &\ \ x\in R \\ g(x)=\dfrac{x-3}{5}         &\ \ x \in R \\ \end{cases}\)
     
  2. \( \begin {cases}f(x)=\frac{2}{3}x+2         &\ \ x\in R \\ g(x)=\frac{3}{2}x-3         &\ \ x \in R \\ \end{cases}\)
     
  3. \( \begin {cases}f(x)=x^2         &\ \ x<0 \\ g(x)=\sqrt{x}         &\ \ x >0 \\ \end{cases}\)
     
  4. \( \begin {cases}f(x)=\dfrac{1}{x}        &\ \ x\neq 0 \\ g(x)=\dfrac{1}{x}         &\ \ x \neq 0 \\ \end{cases}\)
     
  5. \( \begin {cases}f(x)=\log_e(x)+1        &\ \ x>0 \\ g(x)=e^{x-1}        &\ \ x \in R \\ \end{cases}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Graphically, it can be seen that } f(x)=x^2\ \text{and }g(x)=\sqrt{x}\ \text{are not}\)

\(\text{inverse functions as }g(x)\ \text{is not a reflection of }f(x)\ \text{in the line }y=x.\)

\(g(x)\ \text{should be the curve drawn as } h(x)=-\sqrt{x}\ \text{below}.\)

\(\Rightarrow C\)


♦ Mean mark 47%.

Calculus, MET1 2020 VCAA 6

Let  `f:[0,2] -> R`, where  `f(x) = 1/sqrt2 sqrtx`.

  1. Find the domain and the rule for  `f^(-1)`, the inverse function of  `f`.   (2 marks)

The graph of  `y = f(x)`, where  `x ∈ [0, 2]`, is shown on the axes below.
 

     
 

  1. On the axes above, sketch the graph of  `f^(-1)`  over its domain. Label the endpoints and point(s) of intersection with the function  `f`, giving their coordinates.   (2 marks)

  2. Find the total area of the two regions: one region bounded by the functions  `f` and `f^(-1)`, and the other region bounded by  `f, f^(-1)`  and the line  `x = 1`. Give your answer in the form  `(a-bsqrtb)/6`, where  `a, b ∈ ZZ^+`.   (4 marks)

Show Answers Only
  1. `text(Domain) = [0, 1]`

     

    `f^(-1)(x) = 2x^2`

  2.  
       
  3.  `(5 + 2sqrt2)/6\ \ text(u²)`
Show Worked Solution
a.    `text(Domain)\ \ f^(-1)(x)` `= text(Range)\ \ f(x)=[0,1]`

 
`y = 1/sqrt2 x`

`text(Inverse: swap)\ \ x ↔ y`

`x` `= 1/sqrt2 sqrty`
`sqrty` `= sqrt2 x`
`y` `= 2x^2`

 
`:. f^(-1)(x) = 2x^2`

 

b.     

 

c.     
`A` `= int_0^(1/2) 1/sqrt2 sqrtx-2x^2 dx + int_(1/2)^1 2x^2-1/sqrt2 sqrtx\ dx`
  `= [sqrt2/3 x^(3/2)-2/3 x^3]_0^(1/2) + [2/3 x^3-sqrt2/3 x^(3/2)]_(1/2)^1`
  `= [sqrt2/3 (1/sqrt2)^3-2/3(1/2)^3] + [(2/3-sqrt2/3)-(2/24-sqrt2/3 · 1/(2sqrt2))]`
  `= (1/6-1/12) + 2/3-sqrt2/3-(1/12-1/6)`
  `= 1/12 + 2/3-sqrt2/3 + 1/12`
  `= (1 + 8-4sqrt2 + 1)/12`
  `= (5-2sqrt2)/6\ \ text(u²)`

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`
Show Answers Only

`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`
Show Answers Only

`=> 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`
Show Answers Only

`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`