The graph of `f(x) = cos^2(x) + cos(x) + 1` over the domain `0 <= x <= 2pi` is shown below.
- i. Find `f'(x)`. (1 mark)
- ii. Hence, find the coordinates of the turning points of the graph in the interval `(0, 2pi)`. (2 marks)
- Sketch the graph of `y = 1/(f(x))` on the set of axes above. Clearly label the turning points and endpoints of this graph with their coordinates. (3 marks)
Show Answers Only
- i. `−sin(x)(2cos(x) + 1)`
- ii. `(pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`
Show Worked Solution
| a.i. | `f(x)` | `= cos^2(x) + cos(x) + 1` |
| `f'(x)` | `= −2sin(x)cos(x) – sin(x)` | |
| `= −sin(x)(2cos(x) + 1)` |
a.ii. `text(SP when)\ \ sin(x) = 0\ \ text(or)\ \ 2cos(x) + 1 = 0`
`sin(x) = 0 \ => \ x = pi\ \ (x = 0\ \ text{not in domain})`
| `2cos(x) + 1` | `= 0` |
| `cos(x)` | `= −1/2` |
| `x` | `= (2pi)/3, (4pi)/3` |
`text(When)\ \ cos(x) = −1/2 \ => \ f(x) = 1/4 – 1/2 + 1 = 3/4`
`:.\ text(Turning Points:)\ (pi, 1), ((2pi)/3,3/4), ((4pi)/3, 3/4)`
| b. | |