Find the global maximum and minimum values of `y=x^(3)-6x^(2)+8`, where `-1 <= x <= 7`. (4 marks)
--- 8 WORK AREA LINES (style=lined) ---
Calculus, 2ADV C3 2006 HSC 5a
A function `f(x)` is defined by `f(x) =2x^2(3-x)`.
- Find the coordinates of the turning points of `y =f(x)` and determine their nature. (3 marks)
--- 6 WORK AREA LINES (style=lined) ---
- Find the coordinates of the point of inflection. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
- Hence sketch the graph of `y =f(x)`, showing the turning points, the point of inflection and the points where the curve meets the `x`-axis. (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
- What is the minimum value of `f(x)` for `-1 ≤ x ≤4`? (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
Show Answers Only
a. `text(Min)\ (0, 0),\ text(Max)\ (2, 8)`
b. `text(P.I. at)\ (1, 4)`
c.
d. `-32`
Show Worked Solution
| a. | `f(x)` | `= 2x^2 (3-x)` |
| `= 6x^2-2x^3` | ||
| `f^{prime} (x)` | `= 12x-6x^2` | |
| `f^{primeprime}(x)` | `= 12-12x` |
`text(S.P.’s when)\ f^{prime}(x) = 0`
| `12x-6x^2` | `= 0` |
| `6x(2-x)` | `= 0` |
`x = 0 or 2`
`text(When)\ x = 0`
| `f(0)` | `= 0` |
| `f^{primeprime}(0)` | `= 12-0 = 12 > 0` |
`:.\ text(MIN at)\ (0, 0)`
`text(When)\ x = 2`
| `f(2)` | `= 2 xx 2^2 (3-2)` | `= 8` |
| `f^{primeprime}(2)` | `= 12-(12 xx 2)` | `= -12 < 0` |
`:.\ text(MAX at)\ (2, 8)`
b. `text(P.I. when)\ f^{primeprime}(x) = 0`
| `12-12x` | `= 0` |
| `12x` | `= 12` |
| `x` | `= 1` |
| `f^{primeprime}(0.5)` | `=6>0` |
| `f^{primeprime}(1.5)` | `=-6<0` |
`text(S)text(ince concavity changes)\ \ ->\ text(P.I. exists)`
`f(1)= 2 xx 1^2(3-1)=4\ \ :.\ text(P.I. at)\ (1, 4)`
c. `f(x)\ text(meets)\ x text(-axis when)\ f(x) = 0`
`2x^2 xx (3-x) = 0`
`x = 0 or 3`
d. `text(The graph clearly shows that in the given range)`
`-1<= x<=4,\ text(the minimum will occur when)\ x = 4`
| `:.\ text(Minimum` | `= 2 xx 4^2 (3-4)` |
| `= -32` |