A function is given by `f(x) = 3x^4 + 4x^3-12x^2`.
- Find the coordinates of the stationary points of `f(x)` and determine their nature. (3 marks)
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- Hence, sketch the graph `y = f(x)` showing the stationary points. (2 marks)
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- For what values of `x` is the function increasing? (1 mark)
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- For what values of `k` will `f(x) = 3x^4 + 4x^3-12x^2 + k = 0` have no solution? (1 mark)
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Show Answers Only
- `text(MAX at)\ (0,0)`
- `text(MIN at)\ text{(1,–5)}`
- `text(MIN at)\ text{(–2,–32)}`
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- `f(x)\ text(is increasing for)\ -2 < x < 0\ text(and)\ x > 1`
- `text(No solution when)\ k > 32`
Show Worked Solution
| i. | `f(x)` | `= 3x^4 + 4x^3 -12x^2` |
| `f^{′}(x)` | `= 12x^3 + 12x^2-24x` | |
| `f^{″}(x)` | `= 36x^2 + 24x-24` |
`text(Stationary points when)\ f prime (x) = 0`
| `=> 12x^3 + 12x^2-24x` | `=0` |
| `12x(x^2 + x-2)` | `=0` |
| `12x (x+2) (x -1)` | `=0` |
`:.\ text(Stationary points at)\ x=0,\ 1\ text(or)\ –2`
| `text(When)\ x=0,\ \ \ \ f(0)=0` | |
| `f^{″}(0)` | `= -24 < 0` |
| `:.\ text{MAX at (0,0)}` | |
`text(When)\ x=1`
| `f(1)` | `= 3+4\-12 = -5` |
| `f^{″}(1)` | `= 36 + 24\-24 = 36 > 0` |
| `:.\ text{MIN at (1,–5)}` | |
`text(When)\ x=–2`
| `f(–2)` | `=3(–2)^4 + 4(–2)^3-12(–2)^2` |
| `= 48 -32\-48` | |
| `= -32` | |
| `f^{″}(–2)` | `= 36(–2)^2 + 24(–2) -24` |
| `=144-48-24 = 72 > 0` | |
| `:.\ text{MIN at (–2,–32)}` | |
| ii. |  |
♦ Mean mark (HSC) 42%
MARKER’S COMMENT: Be careful to use the correct inequality signs, and not carelessly include `>=` or `<=` by mistake.
MARKER’S COMMENT: Be careful to use the correct inequality signs, and not carelessly include `>=` or `<=` by mistake.
| iii. | `f(x)\ text(is increasing for)` |
| `-2 < x < 0\ text(and)\ x > 1` |
iv. `text(Find)\ k\ text(such that)`
♦♦♦ Mean mark (HSC) 12%.
`3x^4 + 4x^3-12x^2 + k = 0\ \ text(has no solution)`
`k\ text(is the vertical shift of)\ \ y = 3x^4 + 4x^3-12x^2`
`=>\ text(No solution if it does not cross the)\ x text(-axis.)`
`:.\ text(No solution when)\ \ k > 32`
