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Functions, MET2 2024 VCAA 6 MC

Consider the function  \(f(x)=\dfrac{2 x+1}{3-x}\)  with domain  \(x \in R \backslash\{3\}\)

The inverse of \(f\) is

  1. \(f^{-1}(x)=\dfrac{3 x-1}{x+2}\)  with domain  \(x \in R \backslash\{3\}\)
  2. \(f^{-1}(x)=3-\dfrac{7}{x+2}\)  with domain  \(x \in R \backslash\{-2\}\)
  3. \(f^{-1}(x)=3+\dfrac{5}{x+2}\)  with domain  \(x \in R \backslash\{-2\}\)
  4. \(f^{-1}(x)=\dfrac{1-3 x}{x+2}\)  with domain  \(x \in R \backslash\{-2\}\)
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\(B\)

Show Worked Solution

\(\text{Let}\ y=f(x)\)

\(\text{For inverse: swap }x\leftrightarrow y\)

\(x\) \(=\dfrac{2y+1}{3-y}\)
\(3x-xy\) \(=2y+1\)
\(3x-1\) \(=2y+xy\)
\(y(2+x)\) \(=3x-1\)
\(y\) \(=\dfrac{3x-1}{2+x}\)
  \(=\dfrac{6+3x-7}{2+x}\)
  \(=3-\dfrac{7}{2+x}\)

 
\(\Rightarrow B\)

♦ Mean mark 50%.

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. \( \begin {cases}f(x)=\log_e(x)+1        &\ \ x>0 \\ g(x)=e^{x-1}        &\ \ x \in R \\ \end{cases}\)
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\(C\)

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%.

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)

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  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)

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  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`