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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`

Calculus, MET2 2011 VCAA 3

  1. Consider the function  `f: R -> R, f(x) = 4x^3 + 5x-9`.

     

    1. Find  `f^{prime}(x).`   (1 mark)

    2. Explain why  `f^{prime}(x) >= 5` for all `x`.   (1 mark)

  2. The cubic function `p` is defined by  `p: R -> R, p(x) = ax^3 + bx^2 + cx + k`, where `a`, `b`, `c` and `k` are real numbers.

     

    1. If `p` has `m` stationary points, what possible values can `m` have?   (1 mark)

    2. If `p` has an inverse function, what possible values can `m` have?   (1 mark)

  3. The cubic function `q` is defined by  `q:R -> R, q(x) = 3-2x^3`.

     

    1. Write down a expression for  `q^(-1)(x)`.   (2 marks)

    2. Determine the coordinates of the point(s) of intersection of the graphs of  `y = q(x)`  and  `y = q^(-1)(x)`.   (2 marks)

  4. The cubic function `g` is defined by  `g: R -> R, g(x) = x^3 + 2x^2 + cx + k`, where `c` and `k` are real numbers.

     

    1. If `g` has exactly one stationary point, find the value of `c`.   (3 marks)

    2. If this stationary point occurs at a point of intersection of  `y = g(x)`  and  `g^(−1)(x)`, find the value of `k`.   (3 marks)

Show Answers Only
    1. `f^{prime}(x) = 12x^2 + 5`
    2. `text(See Worked Solutions)`
    1. `m = 0, 1, 2`
    2. `m = 0, 1`
    1. `q^(-1)(x) = root(3)((3-x)/2), x ∈ R`
    2. `(1, 1)`
    1. `4/3`
    2. `-10/27`
Show Worked Solution

a.i.   `f^{prime}(x) = 12x^2 + 5`
  

a.ii.  `text(S)text(ince)\ \ x^2>=0\ \ text(for all)\ x,`

♦ Mean mark 47%.
` 12x^2` `>= 0`
`12x^2 + 5` `>=  5`
`f^{prime}(x)` `>=  5\ \ text(for all)\ x`

 

b.i.   `p(x) = text(is a cubic)`

♦♦♦ Mean mark part (b)(i) 9%, and part (b)(ii) 20%.
MARKER’S COMMENT: Good exam strategy should point students to investigate earlier parts for direction. Here, part (a) clearly sheds light on a solution!

`:. m = 0, 1, 2`

`text{(Note: part a.ii shows that a cubic may have no SP’s.)}`

 

b.ii.   `text(For)\ p^(−1)(x)\ text(to exist)`

`:. m = 0, 1`

 

c.i.   `text(Let)\ y = q(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= 3-2y^3`
`y^3` `= (3-x)/2`

`:. q^(-1)(x) = root(3)((3-x)/2), \ x ∈ R`
  

c.ii.  `text(Any function and its inverse intersect on)`

   `text(the line)\ \ y=x.`

`text(Solve:)\ \ 3-2x^3` `= xqquadtext(for)\ x,`
`x` `= 1`

 

`:.\ text{Intersection at (1, 1)}`
  

♦ Mean mark part (d)(i) 44%.
d.i.    `g^{prime}(x)` `= 0`
  `3x^2 + 4x + c` `= 0`
  `Delta` `= 0`
  `16-4(3c)` `= 0`
  `:. c` `= 4/3`

 

d.ii.   `text(Define)\ \ g(x) = x^3 + 2x^2 + 4/3x + k`

♦♦♦ Mean mark part (d)(ii) 14%.

  `text(Stationary point when)\ \ g^{prime}(x)=0`

`g^{prime}(x) = 3x^2+4x+4/3`

`text(Solve:)\ \ g^{prime}(x)=0\ \ text(for)\ x,`

`x = -2/3`

`text(Intersection of)\ g(x)\ text(and)\ g^(-1)(x)\ text(occurs on)\ \ y = x`

`text(Point of intersection is)\  (-2/3, -2/3)`

`text(Find)\ k:`

`g(-2/3)` `= -2/3\ text(for)\ k`
`:. k` ` = -10/27`