smc-969-60-Range defined

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}`

Show Worked Solution

`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)`.

Find the coordinates of the turning points of `y =f(x)` and determine their nature. (3 marks)

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Find the coordinates of the point of inflection. (1 mark)

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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)

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What is the minimum value of `f(x)` for `–1 ≤ x ≤4`? (1 mark)

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Show Worked Solution

`text(S.P.’s when)\ f^{′}(x) = 0`

`12x-6x^2`	`= 0`
`6x(2-x)`	`= 0`

`x = 0 or 2`

`text(When)\ x = 0`

`text(When)\ x = 2`

ii. `text(P.I. when)\ f^{″}(x) = 0`

`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`

Calculus, 2ADV C3 2022 HSC 22