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Calculus, 2ADV C3 2024 HSC 19

Sketch the curve  \(y=x^4-2 x^3+2\)  by first finding all stationary points, checking their nature, and finding the points of inflection.   (5 marks)

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Show Worked Solution
\(y\) \( = x^4-2 x^3+2\)  
\(y^{\prime}\) \( = 4 x^3-6 x^2 =2 x^2(2 x-3)\)  
\(y^{\prime\prime}\) \( = 12 x^2-12 x =12 x(x-1)\)  

 
\(\text{SP’s when}\ \ y^{′}=0:\)

\(2 x^2(2 x-3) =0 \ \Rightarrow \ \ x =0\ \ \text{or}\ \ \dfrac{3}{2}\)

\(\text{At}\ \ x=0, \ y^{\prime \prime} =0\)

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -1 & \ \ \ \ 0\ \ \ \  & 1 \\
\hline
\rule{0pt}{2.5ex} y^{′} \rule[-1ex]{0pt}{0pt} & -10 & 0 & -5 \\
\hline
\end{array}

\(\Rightarrow\ \text{Horizontal POI at}\ (0,2)\)
 

\(\text{At}\ \ x=\dfrac{3}{2}:\)

\(\ y^{\prime \prime} =12 \times \dfrac{3}{2}\left(\dfrac{3}{2}-1\right)=9 \gt 0, \ \ y=\Bigg(\dfrac{3}{2}\Bigg)^{4}-2\Bigg(\dfrac{3}{2}\Bigg)^{3}+2=\dfrac{5}{16}\)

\(\Rightarrow \text{MIN at}\ \left(\dfrac{3}{2}, \dfrac{5}{16}\right)\)
 

\(\text{POI when}\ \ y^{″}=0:\)

\(12 x(x-1)=0 \ \Rightarrow\ \  x=1\ \ \text{or}\ \ x=0\ \text{(see above)}\)

\(\text{Test concavity change at}\ \ x=1:\)

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \dfrac{1}{2} & \ \ \ \ 1\ \ \ \  &  \dfrac{3}{2} \\
\hline
\rule{0pt}{2.5ex} y^{″} \rule[-1ex]{0pt}{0pt} & \ \ -3\ \  & 0 & \ \ \ 9\ \ \  \\
\hline
\end{array}

\(\Rightarrow\ \text{POI at}\ (1,1)\)
 

Calculus, 2ADV C3 2016 HSC 13a

Consider the function  `y = 4x^3 - x^4.`

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

  2. Sketch the graph of the function, clearly showing the stationary points and the `x` and `y` intercepts.  (2 marks)

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i.   `text{P.I. at (0, 0) and Max at (3, 27)}`

ii.
 ext2-hsc-2016-13bi

Show Worked Solution
i.    `y` `= 4x^3 – x^4`
  `y prime` `= 12x^2 – 4x^3`
  `y″` `= 24x – 12x^2`

 

`text(S.P.’s when)\ \ y prime = 0,`

`12x^2 – 4x^3` `= 0`
 `4x^2 (3 – x)` `= 0`
`:. x = 0 or 3`

 

`text(When)\ \ x = 0,\ \ y″ (0) = 0`

`:.\ text(P.I. at)\ \ (0, 0)`

 

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

`y″ (3) = 24(3) – 12 (9) = -36 < 0`

`:.\ text(MAX)\ \ text(at)\ \ (3, 27)`

 

ii.  ext2-hsc-2016-13bi

Calculus, 2ADV C3 2007 HSC 6b

Let  `f (x) =x^4 - 4x^3`.

  1. Find the coordinates of the points where the curve crosses the axes.  (2 marks)

  2. Find the coordinates of the stationary points and determine their nature.  (4 marks)

  3. Find the coordinates of the points of inflection.  (1 mark)

  4. Sketch the graph of  `y = f (x)`, indicating clearly the intercepts, stationary points and points of inflection.  (3 marks)

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  1. `(0, 0)\ , \ (4, 0)`
  2. `text(Minimum S.P. at)\ (3,\ text(-27))`
  3. `(0, 0) and (2, 16)`
  4.  
Show Worked Solution

i.  `f (x) = x^4 – 4x^3`

`text(Cuts)\ x text(-axis when)\ f(x) = 0`

`x^4 – 4x` `= 0`
`x^3 (x – 4)` `= 0`

`x = 0 or 4`

`:. text(Cuts the)\ x text(-axis at)\ (0, 0)\ ,\ (4, 0)`

 

`text(Cuts the)\ y text(-axis when)\ x = 0`

`:. text(Cuts the)\ y text(-axis at)\ (0, 0)`

 

ii.   `f(x) = x^4 – 4x^3`

`f prime (x) = 4x^3 – 12x^2`

`f″ (x) = 12x^2 – 24x`

 

`text(S.P.’s when)\ f prime (x) = 0`

`4x^3 – 12x^2` `= 0`
`4x^2 (x – 3)` `= 0`

`x = 0 or 3`

`text(When)\ x = 0`

`f(0) = 0`

`f″(0) = 0`

`text(S)text(ince concavity changes, a P.I.)`

`text(occurs at)\ (0, 0)`

 

`text(When)\ x = 3`

`f (3)` `= 3^4 – 4 xx 3^3`
  `= -27`
`f″ (3)` `= 12 xx 3^2 – 24 xx 3`
  `= 36 > 0`

 

`:. text(Minimum S.P. at)\ (3,\ text(–27))`

 

iii.  `text(P.I. when)\ f″(x) = 0`

`12x^2 – 24x` `= 0`
`12x(x – 2)` `= 0`

`x = 0 or 2`

`text(P.I. at)\ (0, 0)\ \ \ text{(from(ii))}`

 

`text(When)\ x = 2`

`text(S)text(ince concavity changes, a P.I.)`

`text(occurs when)\ x = 2`

`f (2)` `= 2^4 – 4 xx 2^3`
  `= 16`

 

`:. text(P.I.’s at)\ (0, 0) and (2, 16)`

 

iv.

 2UA HSC 2007 6b

Calculus, 2ADV C3 2008 HSC 8a

Let  `f(x) = x^4 − 8x^2`.

  1. Find the coordinates of the points where the graph of  `y = f(x)`  crosses the axes.  (2 marks)

  2. Show that  `f(x)`  is an even function.   (1 mark)

  3. Find the coordinates of the stationary points of  `f(x)`  and determine their nature.   (4 marks)

  4. Sketch the graph of  `y = f(x)`.   (1 mark)

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  1. `(-2 sqrt 2,0), (2 sqrt 2, 0)`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(S.P.s at)\ (0,0), (2,-16), (-2,-16)`
  4.  
  5. 2UA HSC 2008 8ai
Show Worked Solution
i.    `f(x) = x^4 – 8x^2`

`y text(-intercept when)\ x = 0`

`:.\ text(Cuts)\ y\ text(axis at)\ (0,0)`

`x\ text(-intercept when)\ f(x) = 0`

`x^4 – 8x^2` `= 0`
`x^2 (x^2 – 8)` `= 0`

`x^2 – 8 = 0\ \ \ \ \ text(or)\ \ \ \ \ x = 0`

`x^2` `= 8`
`x` `= +- sqrt 8 = +- 2 sqrt 2`

 

`:.\ text(Cuts)\ x text(-axis at)\ (-2 sqrt 2,0),\ \ text(and)\ \ (2 sqrt 2, 0)`

`text{(Note that it only touches}\ xtext{-axis at (0,0))}`

 

ii.    `f(x)` `= x^4 – 8x^2`
  `f(–x)` `= (–x)^4 – 8(–x)^2`
    `= x^4 – 8x^2`
    `= f(x)`

 

`:.\ f(x)\ text(is an even function.)`

 

iii.   `f(x)` `= x^4 – 8x^2`
  `f'(x)` `= 4x^3 – 16x`
  `f″(x)` `=12x^2-16`

 

`text(S.P.  when)\ \ f'(x) = 0`

`4x^3 – 16x` `= 0`
`4x (x^2 – 4)` `= 0`
`x^2 – 4` `=0\ \ \ \ \ x = 0`
`x^2` `= 4`
`x` `= +-2`
`text(At)\ x=2\ \ `  `f(x)`  `=(2)^4 – 8(2)^2 = -16`
  `f″(x)` `=12(2^2)-16>0`

`:.\ text{MIN at (2, –16)}`

`:.\ text{MIN at (–2, –16)},\ \ \ (f(x)\ text(is even))`

 

`text(At)\ x=0\ \ `  `f(x)`  `=(0)^4 – 8(0)^2 = 0`
  `f″(x)` `=12(0^2)-16<0`

`:.\ text{MAX at (0,0)}`

 

iv. 

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)

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  1. `text{MAX at (0,0), MINs at (1, –5) and (–2, –32)}`
  2.  
        
        2UA HSC 2012 14ai
     
  3. `f(x)\ text(is increasing for)\ -2 < x < 0\ text(and)\ x > 1`
  4. `text(No solution when)\ k > 32`
Show Worked Solution
i. `f(x)` `= 3x^4 + 4x^3-12x^2`
  `f^{′}(x)` `= 12x^3 + 12x^2-24x`
  `f^{″}(x)` `= 36x^2 + 24x-24`

 

`text(Stationary points when)\ f^{′}(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^{″}(0)` `= -24 < 0`
`:.\ text{MAX at  (0,0)}`

 

`text(When)\ x=1`

`f(1)` `= 3+4-12 = -5`
`f^{″}(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^{″}(-2)` `= 36(-2)^2 + 24(-2)-24`
  `=144-48-24 = 72 > 0`
`:.\ text{MIN at  (–2, –32)}`

 

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

 

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

 

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

♦♦♦ Mean mark 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`