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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 SM-Bank 23

The function  `f: [0,oo) → R`  with rule  `f(x) = 1/(1 + x^2)`  is drawn below.

Inverse Functions, EXT1 2004 HSC 5b

  1. Copy or trace this diagram into your writing booklet.
  2. On the same set of axes, sketch  `y=f^(-1)(x)`  where  `f^(-1)` is the inverse function of  `f(x)`.   (1 mark)

  3. Find the domain of the inverse  `f^(-1)`.   (1 mark)

  4. Find an expression for  `y = f^(-1)(x)`  in terms of  `x`.   (2 marks)

  5. The graphs of  `y = f(x)`  and  `y = f^(-1)(x)`  meet at exactly one point  `P`.Let  `α`  be the `x`-coordinate of  `P`. Explain why  `α`  is a root of the equation
  6.      `x^3 + x-1 = 0`.   (1 mark)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `0 < x ≤ 1`
  3. `y = sqrt((1-x)/x), y > 0`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.  

Inverse Functions, EXT1 2004 HSC 5b Answer

b.   `text(Range of)\ \ f(x):\ (0,1]`

`:.\ text(Domain of)\ \ f^(-1)(x):  (0,1]`

 

c.  `f(x) = 1/(1 + x^2)`

`text(Inverse: swap)\  x harr y`

`x` `= 1/(1 + y^2)`
`x(1 + y^2)` `= 1`
`1 + y^2` `= 1/x`
`y^2` `= 1/x-1`
  `= (1-x)/x`
`y` `= ± sqrt((1-x)/x)`

 

`:.y = sqrt((1-x)/x), \ \ y >= 0`

 

d.   `P\ \ text(occurs when)\ \ f(x)\ \ text(cuts)\ \ y = x`

`text(i.e. where)`

`1/(1 + x^2)` `= x`
`1` `= x(1 + x^2)`
`1` `= x + x^3`
`x^3 + x-1` `= 0`

 

`=>\ text(S)text(ince)\ α\ text(is the)\ x\ text(-coordinate of)\ P,`

`text(it is a root of)\ \ \ x^3 + x-1 = 0`

Graphs, MET1 SM-Bank 20

The rule for  `f` is  `f(x) = x-1/2 x^2`  for  `x <= 1`.  This function has an inverse,  `f^(-1) (x)`.

  1. Sketch the graphs of  `y = f(x)`  and  `y = f^(-1) (x)`  on the same set of axes. (Use the same scale on both axes.)   (2 marks)

  2. Find the rule for the inverse function  `f^(-1) (x)`.    (2 marks)

  3. Evaluate  `f^(-1) (3/8)`.    (1 mark)

Show Answers Only
  1.  
    Inverse Functions, EXT1 2008 HSC 5a Answer
  2. `y = 1-sqrt(1-2x)`
  3. `1/2`
Show Worked Solution
a. 

Inverse Functions, EXT1 2008 HSC 5a Answer
b.   `y = x-1/2 x^2,\ \ \ x <= 1`

 

`text(For the inverse function, swap)\ \ x↔y,`

`x` `= y-1/2 y^2,\ \ \ y <= 1`
`2x` `= 2y-y^2`
`y^2-2y + 2x` `= 0`

 

`text(Using quadratic formula,)`

`y` `= (2 +- sqrt( (-2)^2-4 * 1 * 2x) )/2`
  `= (2 +- sqrt(4-8x))/2`
  `= (2 +- 2 sqrt(1-2x))/2`
  `= 1 +- sqrt (1-2x)`

 

`:. y = 1-sqrt(1-2x), \ \ (y <= 1)`

 

c.    `f^(-1) (3/8)` `= 1-sqrt(1-2(3/8))`
    `= 1-sqrt(1-6/8)`
    `= 1-sqrt(1/4)`
    `= 1-1/2`
    `= 1/2`