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 \(g^{\prime \prime}(1)\) ?
- \(\dfrac{1}{4}\)
- \(-\dfrac{1}{4}\)
- \(-\dfrac{1}{2}\)
- \(-1\)
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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 \(g^{\prime \prime}(1)\) ?
\(C\)
\(f(3)=1, f^{\prime}(3)=2, f^{\prime \prime}(3)=4\)
\(\text{Given} \ \ g(x)=f^{-1}(x):\)
\(f(g(x))=x \ \ \text{(Definition of an inverse fn)}\)
\(\text{Differentiate both sides:}\)
\(g^{\prime}(x) \cdot f^{\prime}(g(x))=1 \ \ \Rightarrow \ \ g^{\prime}(x)=\dfrac{1}{f^{\prime}(g(x))}\)
\(g^{\prime \prime}(x)=\dfrac{d}{d x}\left(\dfrac{1}{f^{\prime}(g(x))}\right)=-\dfrac{f^{\prime \prime}(g(x)) \cdot g^{\prime}(x)}{\left[f^{\prime}(g(x))\right]^2}\)
\(\text{When}\ \ x=1:\)
| \(g^{\prime}(1)\) | \(=\dfrac{1}{f^{\prime}(g(1))}=\dfrac{1}{f^{\prime}(3)}=\dfrac{1}{2}\) |
| \(g^{\prime \prime}(1)\) | \(=-\dfrac{f^{\prime \prime}(g(1)) \cdot g^{\prime}(1)}{\left[f^{\prime}(g(1))\right]^2}=-\dfrac{f^{\prime \prime}(3) \cdot \dfrac{1}{2}}{\left[f^{\prime}(3)\right]^2}=-\dfrac{4 \times \dfrac{1}{2}}{2^2}=-\dfrac{1}{2}\) |
\(\Rightarrow C\)
Let \(f(x)=2 x+\ln x\), for \(x>0\). --- 4 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) --- 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. \(\dfrac{1}{3}\) i. \(f(x)=2 x+\ln x\) \(f^{′}(x)=2+\dfrac{1}{x} \) \(\text{In domain}\ x \gt 0\ \ \Rightarrow f^{-1}(x) \gt 0 \ \ (f(x)\ \text{is monotonically increasing}) \) \(\text{Since}\ f(x)\ \text{is one-to-one,}\ f^{-1}(x)\ \text{is a function.} \) \(f(g(x))=x\) \(\text{Differentiate both sides:}\) \(\Rightarrow g(2)=1 \ \text{(by inverse definition)}\)
ii. \(g(x)=f^{-1}(x) \)
\(g^{′}(x)\ f^{′}(g(x))\)
\(=1\)
\(g^{′}(x)\)
\(=\dfrac{1}{f^{′}(g(x))}\)
\(g^{′}(2)\)
\(=\dfrac{1}{f^{′}(g(2))}\)
\(f(1)=2 \times 1 + \ln1 = 2 \)
\(\therefore g^{′}(2)\)
\(= \dfrac{1}{f^{′}(1)} \)
\(=\dfrac{1}{2+\frac{1}{1}}\)
\(=\dfrac{1}{3} \)
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?
`D`
`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`
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)
--- 5 WORK AREA LINES (style=lined) ---
`10/7`
`g(x) = x^3 + 4x – 2`
`g′(x) = 3x^2 + 4`
`f(x) = x g^(-1)(x)`
`f′(x) = x · d/(dx) g^(-1)(x) + g^(-1)(x)`
`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` |