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Calculus, 2ADV C3 2022 HSC 22

Find the global maximum and minimum values of  `y=x^(3)-6x^(2)+8`, where  `-1 <= x <= 7`.   (4 marks)

Show Answers Only

`text{Global max = 57}`

`text{Global min = – 24}`

Show Worked Solution
`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}`

Calculus, 2ADV C3 2020 HSC 16

Sketch the graph of the curve  `y = −x^3 + 3x^2 - 1`, labelling the stationary points and point of inflection. Do NOT determine the `x`-intercepts of the curve.  (4 marks)

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Show Worked Solution
`y` `= −x^3 + 3x^2 – 1`
`(dy)/(dx)` `= −3x^2 + 6x`
`(d^2y)/(dx^2)` `= −6x + 6`

 
`text(SP’s when)\ (dy)/(dx) = 0`

`−3x^2 + 6x` `= 0`
`−3x(x – 2)` `= 0`

`x = 0\ \ text(or)\ \ 2`

 
`text(When)\ \ x = 0,`

`y = −1`

`(d^2 y)/(dx^2) = 6 > 0`

 
`:. text(MIN at)\ \ (0, −1)`
 

`text(When)\ \ x = 2,`

`y` `= −8 + 12 – 1 = 3`
`(d^2y)/(dx^2)` `= −6 xx 2 + 6 = −6 < 0`

 
`:. text(MAX at)\ \ (2, 3)`
 

`(d^2y)/(dx^2) = 0\ text(when)`

`−6x + 6` `= 0`
`x` `= 1`

 
`text(Checking change of concavity)`

`text(Concavity changes either side of)\ x = 1`

`:. text(POI at)\ (1, 1)`
 

Calculus, 2ADV C3 2019 HSC 14b

The derivative of a function  `y = f(x)`  is given by  `f^{′}(x) = 3x^2 + 2x-1`.

  1. Find the `x`-values of the two stationary points of  `y = f(x)`, and determine the nature of the stationary points.  (2 marks)

  2. The curve passes through the point  `(0, 4)`.

     

    Find an expression for  `f(x)`.  (2 marks)

  3. Hence sketch the curve, clearly indicating the stationary points.  (2 marks)

  4. For what values of `x` is the curve concave down?  (1 mark)

Show Answers Only
  1. `x = 1/3\ \ text{(min)}`
    `x = -1\ \ text{(max)}`
  2. `f(x) = x^3 + x^2-x + 4`
  3. `text(See Worked Solution)`
  4. `x < -1/3`
Show Worked Solution

a.    `f^{′}(x) = 3x^2 + 2x-1`

`f^{″}(x) = 6x + 2`

`text(S.P.’s when)\ \ f^{′}(x) = 0`

`3x^2 + 2x-1` `= 0`
`(3x-1)(x + 1)` `= 0`

 
`x = 1/3 or -1`

`text(When)\ x = 1/3,`

`f^{″}(x) = 4 > 0 =>\ text(MIN)`
 

`text(When)\ x = -1,`

`f^{″}(x)= -4 < 0 =>\ text(MAX)`

 

b.    `f(x)` `= int f^{′}(x)\ dx`
    `= int 3x^2 + 2x-1\ dx`
    `= x^3 + x^2-x + c`

 
`(0, 4)\ \ text(lies on)\ \ f(x)\ \ =>\ \ c = 4`

`:. f(x) = x^3 + x^2-x + 4`

 

c.    `text(When)\ \ x = -1,\ \ y = 5`
  `text(When)\ \ x = 1/3,\ \ y = 103/27`

 

 

d.   `text(Concave down when)\ f^{″}(x) < 0`

♦ Mean mark 36%.

`6x + 2` `< 0`
`6x` `< -2`
`x` `< -1/3`

Calculus, 2ADV C3 2018 HSC 13a

Consider the curve  `y = 6x^2 - x^3`.

  1. Find the stationary points and determine their nature.  (3 marks)

  2. Given that the point  (2,16)  lies on the curve, show that it is a point of inflection.  (2 marks)

  3. Sketch the curve, showing the stationary points, the point of inflection and the `x` and `y` intercepts.  (2 marks)

Show Answers Only
  1. `text(MIN at)\ (0, 0);\ text(MAX at)\ (4, 32)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(See Worked Solutions)`
Show Worked Solution

i.    `y = 6x^2 – x^3`

`(dy)/(dx) = 12x – 3x^2`

`(d^2y)/(dx^2) = 12 – 6x`
 

`text(S.P.s occur when)\ \ (dy)/(dx) = 0`

`12x – 3x^2 = 0`

`3x(4 – x) = 0`

`x = 0 or 4`
 

`text(When)\ \ x = 0,\ (d^2y)/(dx^2) > 0`

`:.\ text(MIN at)\ (0, 0)`
 

`text(When)\ \ x = 4,\ (d^2y)/(dx^2) < 0`

`:.\ text(MAX at)\ (4, 32)`

 

ii.  `text(P.I. occur when)\ \ (d^2y)/(dx^2) = 0,`

`12 – 6x` `= 0`
`x` `= 2`

 
`text(When)\ \ x = 2,\ y = 16`

 
`text(S)text(ince the concavity changes)`

`=>\ text(P.I. occurs at)\ \ (2, 16)`

 

iii.  

Calculus, 2ADV C3 2017 HSC 13b

Consider the curve  `y = 2x^3 + 3x^2 - 12x + 7`.

  1. Find the stationary points of the curve and determine their nature.  (4 marks)

  2. Sketch the curve, labelling the stationary points.  (2 marks)

  3. Hence, or otherwise, find the values of `x` for which `(dy)/(dx)` is positive.  (1 mark)

Show Answers Only
  1. `text(maximum at)\ (-2, 27)`

     

    `text(minimum at)\ (1, 0)`

  2.    
  3. `x < -2 and x > 1`
Show Worked Solution
i.   `y` `= 2x^3 + 3x^2 – 12x + 7`
  `(dy)/(dx)` `= 6x^2 + 6x – 12`
  `(d^2y)/(dx^2)` `= 12x + 6`

 

`text(S.P. when)\ (dy)/(dx)` `= 0`
`6x^2 + 6x – 12` `= 0`
`x^2 + x – 2` `= 0`
`(x + 2) (x – 1)` `= 0`

 
`x = -2 or 1`
 

`text(When)\ \ x = –2, (d^2y)/(dx^2) < 0`

`:.\ text(MAX at)\ (–2, 27)`
 

`text(When)\ \ x = 1, (d^2y)/(dx^2) > 0`

`:.\ text(MIN at)\ (1, 0)`

 

ii.  

 

iii.  `text(Solution 1)`

`text(From graph, gradient is positive for)`

`x < –2 and x > 1`

`:. (dy)/(dx) > 0\ \ text(for)\ \ x < –2 and x > 1`

 

`text(Solution 2)`

`(dy)/(dx) > 0`

`6x^2 + 6x – 12` `> 0`
`(x + 2) (x – 1)` `> 0`

 
 
`:. x < –2 and x > 1`

Calculus, 2ADV C3 2015 HSC 13c

Consider the curve  `y = x^3 − x^2 − x + 3`.

  1. Find the stationary points and determine their nature.   (4 marks)

  2. Given that the point  `P (1/3, 70/27)`  lies on the curve, prove that there is a point of inflection at  `P`.  (2 marks)

  3. Sketch the curve, labelling the stationary points, point of inflection and `y`-intercept.  (2 marks)

Show Answers Only
  1. `text(MAX at)\ (-1/3, 86/27); \ text(MIN at)\ (1, 2)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3.  
Show Worked Solution
i. `y` `= x^3 – x^2 – x + 3`
  `(dy)/(dx)` `= 3x^2 – 2x – 1`
  `(d^2y)/(dx^2)` `= 6x – 2`

`text(S.P.’s when)\ (dy)/(dx) = 0`

`3x^2 – 2x – 1` `= 0`
`(3x + 1) (x – 1)` `= 0`

`x = -1/3 or 1`

 

`text(When)\ \ x = -1/3`

`f(-1/3)` `= (-1/3)^3 – (-1/3)^2 – (-1/3) + 3`
  `= -1/27 – 1/9 + 1/3 + 3`
  `= 86/27`
`f″(-1/3)` `= (6 xx -1/3) – 2 = -4 < 0`

`:.\ text(MAX at)\ \ (-1/3, 86/27)`

 

`text(When)\ \ x = 1`

`f(1)` `= 1^3 – 1^2 – 1 + 3 =2`
`f″(1)` `= (6 xx 1) – 2 = 4 > 0`

`:.\ text(MIN at)\ \ (1, 2)`

 

ii.  `(d^2y)/(dx^2) = 0\ \ text(when)`

`6x-2` `=0`
`x` `=1/3`

 

`text(Checking change of concavity)`

`text(Concavity changes either side of)\ x = 1/3`

`:.\ (1/3, 70/27)\ \ text(is a P.I.)`

 

iii.  `text(When)\ \ x` `= 0`
`y` `= 3`

Calculus, 2ADV C3 2004 HSC 4b

Consider the function  `f(x) = x^3 − 3x^2`.

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

  2. Sketch the curve showing where it meets the axes.   (2 marks)

  3. Find the values of  `x`  for which the curve  `y = f(x)`  is concave up.   (2 marks)

Show Answers Only
  1. `text(MAX at)\ (0,0),\ \ text(MIN at)\ (2,-4)`
  2.  
    1. Geometry and Calculus, 2UA 2004 HSC 4b Answer
  3. `f(x)\ text(is concave up when)\ x>1`
Show Worked Solution
(i)    `f(x)` `= x^3 – 3x^2`
  `f'(x)` `= 3x^2 – 6x`
  `f″(x)` `= 6x – 6`

 

`text(S.P.’s  when)\ \ f'(x) = 0`

`3x^2 – 6x` `= 0`
`3x (x – 2)` `= 0`
`x` `= 0\ \ text(or)\ \ 2`

 

`text(When)\ x = 0`

`f(0)` `= 0`
`f″(0)` `= 0 – 6 = -6 < 0`
`:.\ text(MAX at)\ (0,0)`

 

`text(When)\ x = 2`

`f(2)` `= 2^3 – (3 xx 4) = -4`
`f″(2)` `= (6 xx 2) – 6 = 6 > 0`
`:.\ text(MIN at)\ (2, -4)`

 

(ii)   `f(x) = x^3 – 3x^2\ text(meets the)\ x text(-axis when)\ f(x) = 0`
`x^3 – 3x^2` `= 0`
`x^2 (x-3)` `= 0`
`x` `= 0\ \ text(or)\ \ 3`

 Geometry and Calculus, 2UA 2004 HSC 4b Answer

(iii)   `f(x)\ text(is concave up when)`
`f″(x)` `>0`
`6x – 6` `>0`
`6x` `>6`
`x` `>1`

 

`:. f(x)\ text(is concave up when)\ \ x>1`

Calculus, 2ADV C3 2005 HSC 4b

A function  `f(x)`  is defined by  `f(x) = (x + 3)(x^2- 9)`.

  1. Find all solutions of  `f(x) = 0`  (2 marks)

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

  3. Hence sketch the graph of  `y = f(x)`, showing the turning points and the points where the curve meets the `x`-axis.  (2 marks)

  4. For what values of `x` is the graph of  `y = f(x)`  concave down?  (1 mark)

Show Answers Only
  1. `−3 or 3`
  2. `text{53.2 cm  (to 1 d.p.)}`
  3. `text(See worked solutions)`
  4. `x < −1`
Show Worked Solutions
i.    `f(x)` `= (x + 3)(x^2 − 9)`
    `= (x + 3)(x +3)(x − 3)`
  `:. f(x)` `= 0\ text(when)\ \ x=–3\ text(or)\ 3`

 

ii.   `f (x)` `= (x +3)(x^2 − 9)`
    `= x^3 − 9x + 3x^2 − 27`
    `= x^3 + 3x^2 − 9x − 27`
  `f′(x)` `= 3x^2 + 6x − 9`
  `f″(x)` `= 6x + 6`

 

`text(S.P.’s  when)\ \ f′(x) = 0`

`3x^2 + 6x − 9` `= 0`
`3(x^2 + 2x − 3)` `= 0`
`3(x − 1)(x + 3)` `= 0`

 

`text(At)\ x =1`

`f(1)` `= (4)(−8)=−32`
 `f″(1)` `= 6 + 6=12>0`
`:.\ text(MIN at)\ (1, −32)` 

 

`text(At)\ x = −3`

`f(-3)` `= 0`
`f″(−3)` `= (6 xx −3) + 6 = −12 <0`
`:.\ text(MAX at)\ (−3, 0)`

 

iii.   Geometry and Calculus, 2UA 2005 HSC 4b Answer

 

iv.  `f(x)\ \ text(is concave down when)`

`f″(x)` `< 0`
`6x + 6` `< 0`
`6x` `< −6`
`x` `< −1`

Calculus, 2ADV C3 2006 HSC 5a

A function  `f(x)`  is defined by  `f(x) =2x^2(3-x)`.

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

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

  3. 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)

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

Show Answers Only
  1. `text(Min)\ (0, 0),\ text(Max)\ (2, 8)`
  2. `text(P.I. at)\ (1, 4)`
  3.  
  4. `-32`
Show Worked Solution
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`

2UA HSC 2006 5a

 

(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 2009 HSC 10

`text(Let)\ \ f(x) = x - (x^2)/2 + (x^3)/3`

  1. Show that the graph of  `y = f(x)`  has no turning points.   (2 marks)

  2. Find the point of inflection of  `y = f(x)`.     (1 mark)

  3. i. Show that `1 - x + x^2 - 1/(1 + x) = (x^3)/(1 + x)`  for  `x != -1`.   (1 mark)

     

    ii. Let  `g(x) = ln (1 + x)`.

     

        Use the result in part c.i. to show that  `f prime (x) >= g prime (x)`  for all  `x >= 0`.   (2 marks)

  1. Sketch the graphs of  `y = f(x)`  and  `y = g(x)`  for  `x >= 0`.    (2 marks)

  2. Show that  `d/(dx) [(1 + x) ln (1 + x) - (1 + x)] = ln (1 + x)`.   (2 marks)

  3. Find the area enclosed by the graphs of  `y = f(x)`  and  `y = g(x)`, and the straight line  `x = 1`.   (2 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `(1/2, 5/12)`
  3. i. `text{Proof  (See Worked Solutions)}`

     

    ii. `text{Proof  (See Worked Solutions)}` 

  4.  
        Geometry and Calculus, 2UA 2009 HSC 10 Answer
  5. `text{Proof  (See Worked Solutions)}`
  6. `1 5/12\ – 2ln2\ \ text(u²)`
Show Worked Solution
a.    `f(x) = x\ – (x^2)/2 + (x^3)/3`
♦♦ Mean mark 28% for all of Q10 (note that data for each question part is not available).
 

`text(Turning points when)\ f prime (x) = 0`

`f prime (x) = 1\ – x + x^2`

`x^2\ – x + 1 = 0`

`text(S)text(ince)\ \ Delta` `= b^2\ – 4ac`
  `= (–1)^2\ – 4 xx 1 xx 1`
  `= -3 < 0 => text(No solution)`

 
`:.\ f(x)\ text(has no turning points)`

 

b.    `text(P.I. when)\ f″(x) = 0`
`f″(x)` `= -1 + 2x = 0`
`2x` `= 1`
`x` `= 1/2`

`text(Check for change in concavity)`

`f″(1/4)` `= -1/2 < 0`
`f″(3/4)` `= 1/2 > 0`

`=>\ text(Change in concavity)`

`:.\ text(P.I. at)\ \ x = 1/2`

 

`f(1/2)` `= 1/2\ – ((1/2)^2)/2 + ((1/2)^3)/3`
  `= 1/2\ – 1/8 + 1/24`
  `= 5/12`

`:.\ text(Point of Inflection at)\ (1/2, 5/12)`

 

c.i.    `text(Show)\ 1\ – x + x^2\ – 1/(1 + x) = (x^3)/(1 + x),\ \ \ x != -1` 
`text(LHS)` `= (1+x)/(1+x)\ – (x(1+x))/(1+x) + (x^2(1+x))/((1+x))\ – 1/(1+x)`
  `= (1 + x\ – x\ – x^2 + x^2 + x^3\ – 1)/(1+x)`
  `= (x^3)/(1+x)\ \ \ text(… as required)`

 

c.ii.   `text(Let)\ g(x) = ln(1+x)`
  `g prime (x) = 1/(1 + x)`
`f prime (x)\ – g prime (x)` `= 1\ – x + x^2\ – 1/(1+x)`
  `= (x^3)/(1 + x)\ \ text{(using part (i))}`

`text(S)text(ince)\ (x^3)/(1 + x) >= 0\ text(for)\ x >= 0`

`f prime (x)\ – g prime (x) >= 0`
`f prime (x) >= g prime (x)\ text(for)\ x >= 0`
MARKER’S COMMENT: When 2 graphs are drawn on the same set of axes, you must label them. 
 

d. 

Geometry and Calculus, 2UA 2009 HSC 10 Answer
e.    `text(Show)\ d/(dx) [(1 + x) ln (1 + x)\ – (1 + x)] = ln (1 + x)`
  `text(Using)\ d/(dx) uv=uv′+vu′`
`text(LHS)` `= (1+x) xx 1/(1 + x) + ln(1+x)xx1 +  – 1`
  `= 1+ ln(1+x)\ – 1`
  `= ln(1+x)`
  `=\ text(RHS    … as required)`

 

f.    `text(Area)` `= int_0^1 f(x)\ – g(x)\ dx`
    `= int_0^1 (x\ – (x^2)/2 + (x^3)/3\ – ln(x+1))\ dx`
    `= [x^2/2\ – x^3/6 + (x^4)/12\ – (1 + x) ln (1+x) + (1+x)]_0^1`
    `text{(using part (e) above)}`
    `= [(1/2 – 1/6 + 1/12 – (2)ln2 + 2) – (ln1 + 1)]`
    `= 5/12\ – 2ln2 + 2\ – 1`
    `= 1 5/12\ – 2 ln 2\ \ text(u²)`

Calculus, 2ADV C3 2010 HSC 6a

Let  `f(x) = (x + 2)(x^2 + 4)`.

  1. Show that the graph  `y=f(x)`  has no stationary points.   (2 marks)

  2. Find the values of  `x`  for which the graph  `y=f(x)`  is concave down, and the values for which it is concave up.    (2 marks)

  3. Sketch the graph  `y=f(x)`,  indicating the values of the  `x`  and  `y` intercepts.   (2 marks)

Show Answers Only
  1. `text(Proof)  text{(See Worked Solutions)}`
  2.  `f(x)\ text(is concave down when)\ x < -2/3`

     

    `f(x)\ text(is concave up when)\ x > -2/3`

  3.  
Show Worked Solution

i.  `text(Need to show no  S.P.’s)`

`f(x)` `= (x+2)(x^2 + 4)`
  `=x^3 + 2x^2 + 4x + 8`
`f prime (x)` `= 3x^2 + 4x + 4`

 

`text(S.P.s  occur when)\ \ f prime (x) =0,`

`3x^2 + 4x + 4 =0`

`Delta` `= b^2\ – 4ac`
  `=4^2\ – (4 xx 3 xx 4)`
  `=16\ – 48`
  `= -32 < 0`

 

`text(S)text(ince)\ \ Delta < 0,\ \ text(No Solution)`

`:.\ text(No  S.P.’s  for)\ \ f(x)`

 

ii.  `f(x)\ text(is concave down when)\ f″(x) < 0`

MARKER’S COMMENT: The significance of the sign of the second derivative was not well understood by most students.

`f″(x) = 6x + 4`

`=> 6x + 4` `< 0`
`6x` `< -4`
`x` `< -2/3`

`:.\ f(x)\ text(is concave down when)\ x < -2/3`

`f(x)\ text(is concave up when)\ f″(x) > 0`

`f″(x) = 6x + 4`

`=> 6x + 4` `> 0`
`6x` `> -4`
`x` `> -2/3`

`:. f(x)\ text(is concave up when)\ x > -2/3`

 

♦♦ Mean mark 33%.
MARKER’S COMMENT: Students are reminded to bring a ruler to the exam and use it to draw the axes for graphing and to help with an appropriate scale.

iii.  `y text(-intercept) =2 xx4=8`

`x text(-intercept)=–2` 

 

Calculus, 2ADV C3 2011 HSC 7a

Let  `f(x) = x^3-3x + 2`. 

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

  2. Hence, sketch the graph  `y = f(x)`  showing all stationary points and the  `y`-intercept.   (2 marks)

Show Answers Only
  1. `text(MIN at)\ \ (1,0);\ text(MAX at)\ \ (–1,4)`
  2.  
    Geometry and Calculus, 2UA 2011 HSC 7a
Show Worked Solution
i.    `f(x)` `= x^3-3x + 2`
  `f^{′}(x)` `= 3x^2-3`
  `f^{″}(x)` `=6x`

 

`text(Stationary points when)\ f prime (x) = 0`

`3x^2-3` `=0`
`3 (x^2-1)` `= 0`
`:. x^2` `=1`
`x` `=+- 1`

 
`text(When)\ x = 1`

`f(1)` `= 1-3 + 2 = 0`
`f^{″}(1)` `= 6 > 0` 
`:.\ text(MIN S.P. at)\ \ (1,0)`

 

`text(When)\ \ x= -1`

`f(–1)` `= -1 + 3 + 2 = 4`
`f^{″}(–1)` `= –6 < 0`
`:.\ text(MAX S.P. at)\ \ (–1,4)`
MARKER’S COMMENT: Graphs should be large (around ½ page), axes drawn with a ruler, with intercepts and turning points clearly shown. The scale can be different on each axe for clarity, as shown in the Worked Solution.

 

ii.    `y = x^3-3x + 2`
  `y text(-intercept) = 2`

Geometry and Calculus, 2UA 2011 HSC 7a