Let `f:[0,2] -> R`, where `f(x) = 1/sqrt2 sqrtx`.
- Find the domain and the rule for `f^(-1)`, the inverse function of `f`. (2 marks)
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The graph of `y = f(x)`, where `x ∈ [0, 2]`, is shown on the axes below.
- 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)
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- 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)
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Show Answers Only
- `text(Domain) = [0, 1]`
`f^(-1)(x) = 2x^2`
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- `(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)` |