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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

i.    `text(maximum at)\ (-2, 27)`

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

ii.     

iii.  `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 2010 HSC 8d

Let  `f(x) = x^3-3x^2 + kx + 8`, where `k` is a constant.

Find the values of `k` for which `f(x)` is an increasing function.   (2 marks)

Show Answers Only

`k>3`

Show Worked Solution
`f(x)` `= x^3-3x^2 + kx + 8`
`f^{prime}(x)` `= 3x^2-6x + k`

  
`f(x)\ text(is increasing when)\ \ f^{prime}(x) > 0`

`-> 3x^2-6x + k > 0`

♦♦ Mean mark 28%.
MARKER’S COMMENT: The arithmetic required to solve `36-12k<0`  proved the undoing of many students.

  
`f^{prime}(x)\ text(is always positive)`

`-> f^{prime}(x)\ text(is a positive definite.)`

`text(i.e. when)\ \ a > 0\ text(and)\ Delta < 0`
 

`a=3>0`

`Delta = b^2-4ac`

`(-6)^2-(4 xx 3 xx k)` `<0`
`36-12k` `<0`
`12k` `>36`
`k` `>3`

 

`:.\ f(x)\ text(is increasing when)\ \ k > 3.`

Calculus, 2ADV C3 2012 HSC 14a

A function is given by  `f(x) = 3x^4 + 4x^3-12x^2`. 

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

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

  3. For what values of  `x`  is the function increasing?   (1 mark)

  4. For what values of  `k`  will  `f(x) = 3x^4 + 4x^3-12x^2 + k = 0`  have no solution?   (1 mark)

Show Answers Only

a.    `text{MAX at (0,0), MINs at (1, –5) and (–2, –32)}`

b.    
        2UA HSC 2012 14ai
 

c.    `f(x)\ text(is increasing for)\ -2 < x < 0\ text(and)\ x > 1`

d.    `text(No solution when)\ k > 32`

Show Worked Solution
a.     `f(x)` `= 3x^4 + 4x^3-12x^2`
  `f^{prime}(x)` `= 12x^3 + 12x^2-24x`
  `f^{primeprime}(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^{primeprime}(0)= -24 < 0`

`:.\ text{MAX at  (0,0)}`
  

`text(When)\ x=1`

`f(1)` `= 3+4-12 = -5`
`f^{primeprime}(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^{primeprime}(-2)` `= 36(-2)^2 + 24(-2)-24`
  `=144-48-24 = 72 > 0`

  
`:.\ text{MIN at  (–2, –32)}`
  

b.     2UA HSC 2012 14ai
♦ Mean mark (c) 42%
MARKER’S COMMENT: Be careful to use the correct inequality signs, and not carelessly include ≥ or ≤ by mistake.

  
c.    `f(x)\ text(is increasing for)`

`-2 < x < 0\ text(and)\ x > 1`
  

d.    `text(Find)\ k\ text(such that)`

♦♦♦ Mean mark (d) 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`