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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)

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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.    \(\dfrac{1}{3}\)

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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.
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`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 + 4x - 2`  passes through the point (1, 3).

Find the gradient of the tangent to  `f(x) = xg^(-1)(x)`  at the point where  `x = 3`.  (2 marks)

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`10/7`

Show Worked Solution

`g(x) = x^3 + 4x – 2`

♦♦♦ Mean mark 10%.

`g′(x) = 3x^2 + 4`

`f(x) = x g^(-1)(x)`

`f′(x) = x · d/(dx) g^(-1)(x) + g^(-1)(x)`

COMMENT: The reciprocal relationship of gradients between `g(x)` and `g^(-1)(x)` is critical here.

 

`g(x)\ text(passes through)\ (1, 3)`

`=> g^(-1)(x)\ text(passes through)\ (3, 1)`

`g′(1) = 3 + 4 = 7`

`=> d/(dx) g^(-1)(3) = 1/(d/(dy) g(y)) = 1/(g′(1)) = 1/7`

`:. f′(x)|_(x = 3)` `= 3 · 1/7 + 1`
  `= 10/7`