SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, EXT1 C2 2025 HSC 10 MC

For the function \(f(x)\), it is known that  \(f(3)=1, f^{\prime}(3)=2\)  and  \(f^{\prime \prime}(3)=4\).

Let  \(g(x)=f^{-1}(x)\).

What is the value of \(g^{\prime \prime}(1)\) ?

  1. \(\dfrac{1}{4}\)
  2. \(-\dfrac{1}{4}\)
  3. \(-\dfrac{1}{2}\)
  4. \(-1\)
Show Answers Only

\(C\)

Show Worked Solution

\(f(3)=1, f^{\prime}(3)=2, f^{\prime \prime}(3)=4\)

\(\text{Given} \ \ g(x)=f^{-1}(x):\)

\(f(g(x))=x \ \ \text{(Definition of an inverse fn)}\)

♦♦♦ Mean mark 22%.

\(\text{Differentiate both sides:}\)

\(g^{\prime}(x) \cdot f^{\prime}(g(x))=1 \ \ \Rightarrow \ \ g^{\prime}(x)=\dfrac{1}{f^{\prime}(g(x))}\)

\(g^{\prime \prime}(x)=\dfrac{d}{d x}\left(\dfrac{1}{f^{\prime}(g(x))}\right)=-\dfrac{f^{\prime \prime}(g(x)) \cdot g^{\prime}(x)}{\left[f^{\prime}(g(x))\right]^2}\)
 

\(\text{When}\ \ x=1:\)

\(g^{\prime}(1)\) \(=\dfrac{1}{f^{\prime}(g(1))}=\dfrac{1}{f^{\prime}(3)}=\dfrac{1}{2}\)
\(g^{\prime \prime}(1)\) \(=-\dfrac{f^{\prime \prime}(g(1)) \cdot g^{\prime}(1)}{\left[f^{\prime}(g(1))\right]^2}=-\dfrac{f^{\prime \prime}(3) \cdot \dfrac{1}{2}}{\left[f^{\prime}(3)\right]^2}=-\dfrac{4 \times \dfrac{1}{2}}{2^2}=-\dfrac{1}{2}\)

 
\(\Rightarrow C\)

Calculus, EXT1 C2 2023 HSC 14a

Let  \(f(x)=2 x+\ln x\), for \(x>0\).

  1. Explain why the inverse of \(f(x)\) is a function.  (1 mark)

  2. Let  \(g(x)=f^{-1}(x)\). By considering the value of \(f(1)\), or otherwise, evaluate \(g^{\prime}(2)\).  (2 mark)

Show Answers Only

i.     \(f(x)=2 x+\ln x\)

\(f^{′}(x)=2+\dfrac{1}{x} \)

\(\text{In domain}\ x \gt 0\ \ \Rightarrow f^{-1}(x) \gt 0 \ \ (f(x)\ \text{is monotonically increasing}) \)

\(\text{Since}\ f(x)\ \text{is one-to-one,}\ f^{-1}(x)\ \text{is a function.} \)
 

ii.    \(\dfrac{1}{3}\)

Show Worked Solution

i.     \(f(x)=2 x+\ln x\)

\(f^{′}(x)=2+\dfrac{1}{x} \)

\(\text{In domain}\ x \gt 0\ \ \Rightarrow f^{-1}(x) \gt 0 \ \ (f(x)\ \text{is monotonically increasing}) \)

\(\text{Since}\ f(x)\ \text{is one-to-one,}\ f^{-1}(x)\ \text{is a function.} \)

Mean mark (i) 54%.

 
ii.    \(g(x)=f^{-1}(x) \)

\(f(g(x))=x\)

\(\text{Differentiate both sides:}\)

\(g^{′}(x)\ f^{′}(g(x))\) \(=1\)  
\(g^{′}(x)\) \(=\dfrac{1}{f^{′}(g(x))}\)  
\(g^{′}(2)\) \(=\dfrac{1}{f^{′}(g(2))}\)  

 
\(f(1)=2 \times 1 + \ln1 = 2 \)

\(\Rightarrow g(2)=1 \ \text{(by inverse definition)}\)

\(\therefore g^{′}(2)\) \(= \dfrac{1}{f^{′}(1)} \)  
  \(=\dfrac{1}{2+\frac{1}{1}}\)  
  \(=\dfrac{1}{3} \)  
♦♦ Mean mark (ii) 32%.

Calculus, EXT1 C2 2022 HSC 9 MC

A given function  `f(x)`  has an inverse  `f^{-1}(x)`.

The derivatives of  `f(x)`  and  `f^{-1}(x)`  exist for all real numbers `x`.

The graphs  `y=f(x)`  and  `y=f^{-1}(x)`  have at least one point of intersection.

Which statement is true for all points of intersection of these graphs?

  1. All points of intersection lie on the line  `y=x`.
  2. None of the points of intersection lie on the line  `y=x`.
  3. At no point of intersection are the tangents to the graphs parallel.
  4. At no point of intersection are the tangents to the graphs perpendicular.
Show Answers Only

`D`

Show Worked Solution

`text{By Elimination:}`

`text{Consider}\ \ f(x)=x\ \ =>\ \ f^(-1)(x)=x:`

`text{All POI lie on}\ \ y=x\ \ text{and all tangents are parallel}`

`text{→ Eliminate B and C}`
 

`text{Consider}\ \ f(x)=-x\ \ =>\ \ f^(-1)(x)=-x:`

`text{All POI lie on}\ \ y=-x`

`text{→ Eliminate A}`

`=>D`


♦♦♦ Mean mark 13%.

Calculus, EXT1 C2 2021 HSC 14e

The polynomial  \(g(x)=x^3+4 x-2\)  passes through the point \((1,3)\).

Find the gradient of the tangent to  \(f(x)=x g^{-1}(x)\)  at the point where  \(x=3\).   (3 marks)

Show Answers Only

\(\text{Gradient of tangent}=\dfrac{10}{7}\)

Show Worked Solution

\(g(1)=3 \ \ \Rightarrow\ \ g^{-1}(3)=1\)

\(f(x)=x g^{-1}(x)\)

\(\text {Using product rule:}\)

\(f^{\prime}(x)=g^{-1}(x)+x \cdot \dfrac{d}{d x}\left(g^{-1}(x)\right)\)

♦♦♦ Mean mark 10%.

\(\text{Find}\ \ \dfrac{d}{d x}\left(g^{-1}(x)\right):\)

\(\text{Let} \ \ y=g^{-1}(x) \ \ \Rightarrow\ \ x=g(y)\)

\(\dfrac{d x}{d y}=g^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{d y}{d x}=\dfrac{1}{g^{\prime}(y)}\)

\(\dfrac{d}{d x}\left(g^{-1}(x)\right)=\dfrac{1}{g^{\prime}\left(g^{-1}(x)\right)}\)
 

\(f^{\prime}(x)=g^{-1}(x)+\dfrac{x}{g^{\prime}\left(g^{-1}(x)\right)}\)

\(f^{\prime}(3)=g^{-1}(3)+\dfrac{3}{g^{\prime}\left(g^{-1}(3)\right)}=1+\dfrac{3}{g^{\prime}(1)}\)
 

\(g(x)=x^3+4 x-2\ \ \Rightarrow\ \ g^{\prime}(x)=3 x^2+4\)

\(f^{\prime}(3)=1+\dfrac{3}{3(1)^2+4} = \dfrac{10}{7}\)

\(\therefore \ \text{Gradient of tangent}=\dfrac{10}{7}\)