smc-5205-50-Find intersection

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

Copy or trace this diagram into your writing booklet.
On the same set of axes, sketch `y=f^(-1)(x)` where `f^(-1)` is the inverse function of `f(x)`. (1 mark)
Find the domain of the inverse `f^(-1)`. (1 mark)
Find an expression for `y = f^(-1)(x)` in terms of `x`. (2 marks)
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
`x^3 + x-1 = 0`. (1 mark)

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`text(See Worked Solutions)`
`0 < x ≤ 1`
`y = sqrt((1-x)/x), y > 0`
`text(See Worked Solutions)`

Show Worked Solution

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`

Consider the function `f: R -> R, f(x) = 4x^3 + 5x-9`.
1. Find `f^{′}(x)` (1 mark)
2. Explain why `f^{′}(x) >= 5` for all `x`. (1 mark)
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)
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)
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)

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1. `f^{′}(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^{′}(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^{′}(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^{′}(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^{′}(x)=0`

`g^{′}(x) = 3x^2+4x+4/3`

`text(Solve:)\ \ g^{′}(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`

Algebra, MET1 SM-Bank 23

Calculus, MET2 2011 VCAA 3