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Calculus, 2ADV C3 2025 HSC 16

Consider the function  \(f(x)=\dfrac{x^2}{e^x}\).

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

  2. A partially completed graph of  \(f(x)=\dfrac{x^2}{e^x}\)  is shown.
  3. Use your answer from part (a) to complete the graph.   (1 mark)


     

Show Answers Only

a.   \(\text{MIN at}\ \ (0,0)\)

\(\text{MAX at}\ \ \left(2,\dfrac{4}{e^2}\right)\)

b.   
     

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{x^2}{e^x}\)
  \(f^{\prime}(x)\) \(=2 x \cdot e^x-e^x \cdot x^2\)
    \(=\dfrac{x e^x(2-x)}{e^{2 x}}\)
    \(=\dfrac{x(2-x)}{e^x}\)

 
\(\text{Find} \ x\ \text{when} \ \ f^{\prime}(x)=0:\)

\(x(2-x)=0\)

\(x=0 \ \text {or} \ 2\)

\(\text{When} \ \ x=0 \ \Rightarrow \ f(0)=0\)

\(\text {When}\ \  x=2 \ \Rightarrow \ f(2)=\dfrac{4}{e^2}\)

\(\text {Checking nature of SP’s:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline x & -1 & 0 & 1 & 2 & 3 \\
\hline f^{\prime}(x) & -3 e & 0 & \dfrac{1}{e} & 0 & -\dfrac{3}{e^3} \\
\hline
\end{array}

\(\therefore \ \text{MIN at}\ \ (0,0)\)

\(\quad \ \ \text{MAX at}\ \ \left(2,\dfrac{4}{e^2}\right)\)
 

b.
     

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 2023 HSC 30

Let  \(f(x)=e^{-x} \sin x\).

  1. Find the coordinates of the stationary points of \(f(x)\) for  \(0\leq x\leq 2\pi\). You do NOT need to check the nature of the stationary points.  (3 marks)

  2. Without using any further calculus, sketch the graph of  \(y=f(x)\), for  \(0\leq x\leq 2\pi\), showing stationary points and intercepts.  (2 marks)
     


Show Answers Only

a.    \( \Big(\dfrac{\pi}{4},\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}}\Big)\ \text{and}\  \Big(\dfrac{5\pi}{4},\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}}\Big)\)

b.    
         

Show Worked Solution

a.    \(f(x)=e^{-x} \sin x\)

\(f^{′}(x)=e^{-x} \cos x-e^{-x} \sin x = e^{-x}( \cos x-\sin x) \)

\(\text{SPs when}\ f^{′}(x)=0: \)

\(e^{-x}=0\ \ \rightarrow \ \text{no solution} \)

\(\cos x-\sin x\) \(=0\)  
\(1-\tan x\) \(=0\)  
\(\tan\) \(=1\)  
Mean mark (a) 54%.

\(x=\dfrac{\pi}{4}, \dfrac{5\pi}{4} \)
 

\(f(\dfrac{\pi}{4})=e^{-\frac{\pi}{4}}\sin \frac{\pi}{4}=\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}} \)

\(f(\dfrac{5\pi}{4})=e^{-\frac{5\pi}{4}}\sin \frac{5\pi}{4}=\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}} \)
 

\(\therefore\ \text{SPs at}\ \Big(\dfrac{\pi}{4},\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}}\Big)\ \text{and}\  \Big(\dfrac{5\pi}{4},\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}}\Big)\)

 
b.    \(\ x\text{-intercepts at}\ \ x=0, \pi,\ 2\pi \)
 

 

♦♦ Mean mark (b) 34%.

Calculus, 2ADV C3 2022 HSC 27

Let  `f(x)=xe^(-2x)`.

It is given that  `f^(′)(x)=e^(-2x)-2xe^(-2x)`.

  1. Show that  `f^(″)(x)=4(x-1)e^(-2x)`.  (2 marks)

  2. Find any stationary points of  `f(x)`  and determine their nature.  (2 marks)

  3. Sketch the curve  `y=x e^{-2 x}`, showing any stationary points, points of inflection and intercepts with the axes.  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Max S.P. at}\ \ (1/2, 1/(2e))`
  3.  
Show Worked Solution

a.   `f^(′)(x)=e^(-2x)-2xe^(-2x)`

`f^(″)(x)` `=-2e^(-2x)-[2x(-2)e^(-2x)+2e^(-2x)]`  
  `=-2e^(-2x)+4xe^(-2x)-2e^(-2x)`  
  `=4xe^(-2x)-4e^(-2x)`  
  `=4(x-1)e^(-2x)\ \ text{… as required}`  

 

b.   `text{S.P.’s occur when}\ \ f^(′)(x)=0`

`e^(-2x)-2xe^(-2x)` `=0`  
`e^(-2x)(1-2x)` `=0`  

 
`e^(-2x)=0\ \ =>\ \ text{No solution}`

`1-2x=0\ \ =>\ \ x=1/2`

`f^(″)(1/2)` `=4(1/2-1)e^(-2xx1/2)`  
  `=-2e^(-1)<0\ \ =>\ text{MAX}`  

 
`f(1/2)=1/2e^(-1)=1/(2e)`
 

c.   `text{POI occurs when}\ \ f^(″)(x)=0`

`f^(″)(x)=4(x-1)e^(-2x)=0\ \ =>\ \ x=1`

`text{Test for change in concavity:}`

`f^(″)(0)=4(0-1)e^0=-4`

`f^(″)(2)=4(2-1)e^(-4)>0\ \ =>\ text{Concavity changes}`

`:.\ text{POI exists at}\ \ (1,1/e^2)`

`text{As}\ \ x->oo\ \ =>\ \ y->0^+`
 


Mean mark part (c) 54%.

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 MET1 4

Given the function  `f(x) = log_e (x-3) + 2`,

  1. State the domain and range of `f(x)`.  (1 mark)

  2. i.  Find the equation of the tangent to the graph of  `f(x)` at  `(4, 2)`.  (2 marks)

     

    ii. On the axes below, sketch the graph of the function  `f(x)`, labelling any asymptote with its equation.

     

        Also draw the tangent to the graph of  `f(x)`  at  `(4, 2)`.  (4 marks)
     

Show Answers Only
  1. `x >3`

     

    `y in R`

  2. i.  `y = x-2`

     

    ii. `text(See Worked Solutions)`

Show Worked Solution
a.   `text(Domain)` `: \ x > 3`
  `text(Range)` `: \ y in R`

 

b.i.   `g(x)` `= log_e (x-3) + 2`
  `g^{\prime}(x)` `= 1/(x-3)`
  `g^{\prime}(4)` `= 1`

 
`text(Equation of tangent),\ m = 1\ \ text(through)\ (4, 2):`

`y-2` `= 1(x-4)`
`y` `= x-2`

 

b.ii.  

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 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)

Show Answers Only

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

b.
 ext2-hsc-2016-13bi

Show Worked Solution
a.      `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)`

 

b.     ext2-hsc-2016-13bi

Calculus, 2ADV C3 2004 HSC 9c

Consider the function  `f(x) = (log_e x)/x`, for  `x > 0`.

  1. Show that the graph of  `y = f(x)`  has a stationary point at  `x = e`.  (2 marks)

  2. By considering the gradient on either side of  `x = e`, or otherwise, show that the stationary point is a maximum.  (1 mark)

  3. Use the fact that the maximum value of  `f(x)`  occurs at  `x = e`  to deduce that  `e^x ≥ x^e`  for all  `x > 0`.  (2 marks)

Show Answers Only

a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a.    `f(x) = (log_e x)/x, text(for)\ x > 0`

`text(Using the quotient rule,)`

`f′(u/v)` `=(u′v-uv′)/v^2`
`f′(x)` `= (x · d/(dx) (log_e x) − log_e x · d/(dx) (x))/(x^2)`
  `= (x(1/x) − log_e x · 1)/(x^2)`
  `= (1 − log_e x)/(x^2)`

 

`text(SP’s occur when)\ \ f′(x)=0`

`(1 − log_e x)/(x^2)` `= 0`
`1 − log_e x` `= 0`
`log_e x` `= 1`
`:. x` `= e`

 

`:. text(There is a stationary point at)\ \ x = e.`

 

b.    

Geometry and Calculus, 2UA 2004 HSC 9c Answer

`text(When)\ \ x < e, log_e x<1\ \ \ text{(from graph)}`

`1 − log_e x` `> 0`
`(1 − log_e x)/(x^2)` `> 0`
`:. f′(x)` `> 0`

 

`text(When)\ x > e, log_e x>1\ \ \ text{(from graph)}`

`1 − log_e x` `< 0`
`(1 − log_e x)/(x^2)` `< 0`
`:. f′(x)` `< 0`

 

`:.\ text(The stationary point at)\ \ x = e\ \ text(is a maximum.)`

 

c.     `f(e)` `= (log_e x)/e`
    `= 1/e`

`:. f(x)\ \ text(has a maximum value at)\ \ (e,1/e)`

MARKER’S COMMENT: Very challenging for most students. Successful students recognised the link to part (ii).

 

`:. (log_e x)/x` `≤ 1/e`
`log_e x` `≤ x/e`
`e log_e x` `≤ x`
`log_e x^e` `≤ x`
`e^(log_e x^e)` `≤ e^x`
`x^e` `≤ e^x\ \ \ text{(using}\ e^(log x) = x)`
`:. e^x` `≥ x^e \ \ \ text{(for}\ \ x > 0text{)}`

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)

Show Answers Only

a.    `(0, 0)\ , \ (4, 0)`

b.    `text(Minimum S.P. at)\ (3,\ text(-27))`

c.    `(0, 0) and (2, 16)`

d. 
   

Show Worked Solution

a.   `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)`

 

b.    `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))`

 

c.    `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)`

 

d.    

 2UA HSC 2007 6b

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

a.    `text(MAX at)\ (-1/3, 86/27); \ text(MIN at)\ (1, 2)`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.
   

Show Worked Solution
a.     `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)`

 

b.    `(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.)`

 

c.    `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

a.    `text(MAX at)\ (0,0),\ \ text(MIN at)\ (2,-4)`

b. 

    Geometry and Calculus, 2UA 2004 HSC 4b Answer

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

Show Worked Solution
a.     `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)`

 

b.     `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

c.     `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

a.    `−3 or 3`

b.    `text{53.2 cm  (to 1 d.p.)}`

c.    `text(See worked solutions)`

d.    `x < −1`

Show Worked Solutions
a.     `f(x)` `= (x + 3)(x^2 − 9)`
    `= (x + 3)(x +3)(x − 3)`
  `:. f(x)` `= 0\ text(when)\ \ x=–3\ text(or)\ 3`

 

b.     `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)`

 

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

 

d.    `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

a.    `text(Min)\ (0, 0),\ text(Max)\ (2, 8)`

b.    `text(P.I. at)\ (1, 4)`

c.
   

d.    `-32`

Show Worked Solution
a.    `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)`

 

b.    `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)`

 

c.    `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

 

d.   `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 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)

Show Answers Only

a.    `(-2 sqrt 2,0), (2 sqrt 2, 0)`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(S.P.s at)\ (0,0), (2,-16), (-2,-16)`

d.    

  1. 2UA HSC 2008 8ai
Show Worked Solution
a.     `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))}`

 

b.     `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.)`

 

c.     `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)}`

 

d.    

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

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

b.    `(1/2, 5/12)`

c.i.  `text{Proof  (See Worked Solutions)}`

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

d.   

 Geometry and Calculus, 2UA 2009 HSC 10 Answer

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

f.    `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^{prime prime}(x) = 0`
`f^{prime prime}(x)` `=-1 + 2x = 0`
`2x` `= 1`
`x` `= 1/2`

`text(Check for change in concavity)`

`f^{prime prime}(1/4)` `=-1/2 < 0`
`f^{prime prime}(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^{prime}+vu^{prime}`
`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

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

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

 

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

 

Show Worked Solution

a.    `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)`

 

b.    `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.

c.    `y text(-intercept) =2 xx4=8`

`x text(-intercept)=–2` 

 

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^{′}(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)}`

 

b.     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.

 

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

 

d.    `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`

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

a.     `text(MIN at)\ \ (1,0);\ text(MAX at)\ \ (–1,4)`

b.    

    Geometry and Calculus, 2UA 2011 HSC 7a

Show Worked Solution
a.     `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.

 

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

Geometry and Calculus, 2UA 2011 HSC 7a