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The function \(y=f(x)\) has an inverse function \(y=f^{-1}(x)\).
The tangent to \(y=f(x)\) at the point \((2,3)\), \(\ell\), has a gradient of 1.
Show that the tangent to \(y=f^{-1}(x)\) at the point \((3,2)\) is parallel to \(\ell\). (3 marks)
--- 9 WORK AREA LINES (style=lined) ---
\(\text{Since tangent to}\ f(x) \ \text {touches at}\ (2,3):\)
\(f(2)=3 \ \ \Rightarrow\ \ f^{-1}(3)=2\)
\(\text{For inverse functions,}\ \ \left(f^{-1}\right)^{\prime}(x)=\dfrac{1}{f^{\prime}\left(f^{-1}(x)\right)}\)
\(\text{Since tangent to} \ \ f(x) \ \ \text{at} \ \ x=2 \ \ \text{has gradient 1}\)
\(m_{\ell}=1 \ \Rightarrow \ f^{\prime}(2)=1\)
\(\text{Find gradient of}\ f^{-1}(x) \ \text{at} \ \ x=3:\)
\((f^{-1})^{\prime}(3)=\dfrac{1}{f^{\prime}\left(f^{-1}(3)\right)}=\dfrac{1}{f^{\prime}(2)}=1\)
\(\therefore \text{Gradient of tangent to} \ f^{-1}(x) \ \text {at } x=3\ \ \ \text {is parallel to} \ \ell\).
\(\text{Since tangent to}\ f(x) \ \text {touches at}\ (2,3):\)
\(f(2)=3 \ \ \Rightarrow\ \ f^{-1}(3)=2\)
\(\text{For inverse functions,}\ \ \left(f^{-1}\right)^{\prime}(x)=\dfrac{1}{f^{\prime}\left(f^{-1}(x)\right)}\)
\(\text{Since tangent to} \ \ f(x) \ \ \text{at} \ \ x=2 \ \ \text{has gradient 1}\)
\(m_{\ell}=1 \ \Rightarrow \ f^{\prime}(2)=1\)
\(\text{Find gradient of}\ f^{-1}(x) \ \text{at} \ \ x=3:\)
\((f^{-1})^{\prime}(3)=\dfrac{1}{f^{\prime}\left(f^{-1}(3)\right)}=\dfrac{1}{f^{\prime}(2)}=1\)
\(\therefore \text{Gradient of tangent to} \ f^{-1}(x) \ \text {at } x=3\ \ \ \text {is parallel to} \ \ell\).
The polynomial \(h(x)=x^3+2x+1\) passes through the point \((1,4)\).
Find the gradient of the tangent to \(f(x)=x h^{-1}(x)\) at the point where \(x=4\). (3 marks)
--- 12 WORK AREA LINES (style=lined) ---
\(\dfrac{9}{5}\)
\(h(1)=4 \ \ \Rightarrow\ \ h^{-1}(4)=1\)
\(f(x)=x h^{-1}(x)\)
\(\text{Using the product rule:}\)
\(f^{\prime}(x)=h^{-1}(x)+x\cdot \dfrac{d}{dx}\left(h^{-1}(x)\right)\)
\(\text{Find}\ \dfrac{d}{dx}\left(h^{-1}(x)\right):\)
\(\text{Let}\ \ y=h^{-1}(x)\ \ \Rightarrow\ \ x=h(y)\)
\(\dfrac{dx}{dy}=h^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{dy}{dx}=\dfrac{1}{h^{\prime}(y)}\)
\(\dfrac{d}{dx}\left(h^{-1}(x)\right)=\dfrac{1}{h^{\prime}\left(h^{-1}(x)\right)}\)
\(f^{\prime}(x)=h^{-1}(x)+\dfrac{x}{h^{\prime}\left(h^{-1}(x)\right)}\)
\(f^{\prime}(4)=h^{-1}(4)+\dfrac{4}{h^{\prime}\left(h^{-1}(4)\right)}=1+\dfrac{4}{h^{\prime}(1)}\)
\(h(x)=x^3+2x+1\ \ \Rightarrow\ \ h^{\prime}(x)=3x^2+2\)
\(f^{\prime}(4)=1+\dfrac{4}{3(1)^2+2}=\dfrac{9}{5}\)
\(\therefore\ \text{Gradient of tangent}=\dfrac{9}{5}\)
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. \(f(x)=2 x+\ln x\) \(f^{′}(x)=2+\dfrac{1}{x} \) \(\text{In domain}\ x \gt 0\ \ \Rightarrow f^{-1}(x) \gt 0 \ \ (f(x)\ \text{is monotonically increasing}) \) \(\text{Since}\ f(x)\ \text{is one-to-one,}\ f^{-1}(x)\ \text{is a function.} \) \(f(g(x))=x\) \(\text{Differentiate both sides:}\) \(\Rightarrow g(2)=1 \ \text{(by inverse definition)}\)
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Mean mark (i) 54%.
ii. \(g(x)=f^{-1}(x) \)
\(g^{′}(x)\ f^{′}(g(x))\)
\(=1\)
\(g^{′}(x)\)
\(=\dfrac{1}{f^{′}(g(x))}\)
\(g^{′}(2)\)
\(=\dfrac{1}{f^{′}(g(2))}\)
\(f(1)=2 \times 1 + \ln1 = 2 \)
\(\therefore g^{′}(2)\)
\(= \dfrac{1}{f^{′}(1)} \)
\(=\dfrac{1}{2+\frac{1}{1}}\)
\(=\dfrac{1}{3} \)
♦♦ Mean mark (ii) 32%.
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)
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\(\text{Gradient of tangent}=\dfrac{10}{7}\)
\(g(1)=3 \ \ \Rightarrow\ \ g^{-1}(3)=1\)
\(f(x)=x g^{-1}(x)\)
\(\text {Using product rule:}\)
\(f^{\prime}(x)=g^{-1}(x)+x \cdot \dfrac{d}{d x}\left(g^{-1}(x)\right)\)
\(\text{Find}\ \ \dfrac{d}{d x}\left(g^{-1}(x)\right):\)
\(\text{Let} \ \ y=g^{-1}(x) \ \ \Rightarrow\ \ x=g(y)\)
\(\dfrac{d x}{d y}=g^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{d y}{d x}=\dfrac{1}{g^{\prime}(y)}\)
\(\dfrac{d}{d x}\left(g^{-1}(x)\right)=\dfrac{1}{g^{\prime}\left(g^{-1}(x)\right)}\)
\(f^{\prime}(x)=g^{-1}(x)+\dfrac{x}{g^{\prime}\left(g^{-1}(x)\right)}\)
\(f^{\prime}(3)=g^{-1}(3)+\dfrac{3}{g^{\prime}\left(g^{-1}(3)\right)}=1+\dfrac{3}{g^{\prime}(1)}\)
\(g(x)=x^3+4 x-2\ \ \Rightarrow\ \ g^{\prime}(x)=3 x^2+4\)
\(f^{\prime}(3)=1+\dfrac{3}{3(1)^2+4} = \dfrac{10}{7}\)
\(\therefore \ \text{Gradient of tangent}=\dfrac{10}{7}\)