Find the global maximum and minimum values of `y=x^(3)-6x^(2)+8`, where `-1 <= x <= 7`. (4 marks)
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Find the global maximum and minimum values of `y=x^(3)-6x^(2)+8`, where `-1 <= x <= 7`. (4 marks)
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`text{Global max = 57}`
`text{Global min = – 24}`
`y` | `=x^3-6x^2+8` | |
`dy/dx` | `=3x^2-12x` | |
`(d^2y)/(dx^2)` | `=6x-12` |
`text{SP’s when}\ \ dy/dx=0:`
`3x^2-12x` | `=0` | |
`3x(x-4)` | `=0` |
`x=0\ \ text{or}\ \ 4`
`text{When}\ \ x=0,\ \ y=8,\ \ (d^2y)/(dx^2)<0`
`=>\ text{Local Max at}\ \ (0,8)`
`text{When}\ \ x=4,\ \ y=4^3-6(4^2)+8=-24,\ \ (d^2y)/(dx^2)>0`
`=>\ text{Local Min at}\ \ (4,-24)`
`text{Check ends of domain:}`
`text{When}\ \ x=-1,\ \ y=-1-6+8=1`
`text{When}\ \ x=7,\ \ y=7^3-6(7^2)+8=57`
`:.\ text{Global max = 57}`
`:.\ text{Global min = – 24}`
A function `f(x)` is defined by `f(x) =2x^2(3 - x)`.
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i. `f(x)` | `= 2x^2 (3 – x)` |
`= 6x^2 – 2x^3` | |
`f prime (x)` | `= 12x – 6x^2` |
`f″(x)` | `= 12 – 12x` |
`text(S.P.’s when)\ f′(x) = 0`
`12x – 6x^2` | `= 0` |
`6x(2 – x)` | `= 0` |
`x = 0 or 2`
`text(When)\ x = 0`
`f(0)` | `= 0` |
`f″(0)` | `= 12 – 0 = 12 > 0` |
`:.\ text(MIN at)\ (0, 0)` |
`text(When)\ x = 2`
`f(2)` | `= 2 xx 2^2 (3 – 2)` | `= 8` |
`f″(2)` | `= 12 – (12 xx 2)` | `= -12 < 0` |
`:.\ text(MAX at)\ (2, 8)` |
ii. `text(P.I. when)\ f″(x) = 0`
`12 – 12x` | `= 0` |
`12x` | `= 12` |
`x` | `= 1` |
`f″(0.5)` | `=6>0` |
`f″(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)`
iii. `f(x)\ text(meets)\ x text(-axis when)\ f(x) = 0`
`2x^2 xx (3 – x) = 0`
`x = 0 or 3`
(iv) `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` |