Let `y=f(x)` be a function defined for `0 <= x <= 6`, with `f(0)=0`.
The diagram shows the graph of the derivative of `f`, `y = f^{′}(x)`.
The shaded region `A_1` has area 4 square units. The shaded region `A_2` has area 4 square units.
- For which values of `x` is `f(x)` increasing? (1 mark)
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- What is the maximum value of `f(x)`? (1 mark)
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- Find the value of `f(6)`. (1 mark)
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- Draw a graph of `y =f(x)` for `0 <= x <= 6`. (2 marks)
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Show Answers Only
- `f(x)\ text(is increasing when)\ 0 <= x < 2`
- `text(MAX value of)\ f(x) = 4`
- `-6`
-
Show Worked Solution
| i. | `f(x)\ text(is increasing when)\ \ f^{′}(x) > 0` |
| `text(From the graph)` | |
| `f(x)\ text(is increasing when)\ 0 <= x < 2` |
| ii. | `f^{′}(x) = 0\ \ text(when)\ \ x=2` |
| `:.\ text(MAX at)\ \ x = 2` | |
| `int_0^2 f^{′}(x)\ dx = 4\ \ \ (text(given since)\ A_1 = 4 text{)}` | |
| `text(We also know)` |
| `int_0^2\ f^{′}(x)\ dx` | `= [f(x)]_0^2` |
| `= f(2)-f(0)` | |
| `= f(2)\ \ \ \ text{(since}\ f(0) = 0 text{)}` |
`=> f(2) = 4`
♦♦♦ Parts (ii) and (iii) proved particularly difficult for students with mean marks of 12% and 11% respectively.
`:.\ text(MAX value of)\ \ f(x) = 4`
| iii. | `int_0^4 f^{′}(x)` | `= A_1-A_2` |
| `=0` |
`text(We also know)`
| `int_0^4 f^{′}(x)\ dx` | `= int_2^4 f^{′}(x)\ dx + int_0^2 f^{′}(x)\ dx` |
| `=[f(x)]_2^4 + 4` | |
| `= f(4)-f(2) + 4\ \ \ (text(note)\ f(2)=4)` | |
| `=f(4)` |
`=> f(4) = 0`
`text(Gradient = – 3 from)\ \ x = 4\ \ text(to)\ \ x = 6`
| `:.\ f(6)` | `= -3 (6\ – 4)` |
| `= -6` |
♦♦ Mean mark 28%
EXAM TIP: Clearly identify THE EXTREMES when given a defined domain. In this case, the origin is obvious graphically, and the other extreme at `x=6`, is CLEARLY LABELLED!
EXAM TIP: Clearly identify THE EXTREMES when given a defined domain. In this case, the origin is obvious graphically, and the other extreme at `x=6`, is CLEARLY LABELLED!
| iv. |  |