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)
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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)
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\(\dfrac{9}{5}\)
\(h(1)=4 \ \ \Rightarrow\ \ h^{-1}(4)=1\)
\(f(x)=x h^{-1}(x)\)
\(\text{Using the product rule:}\)
\(f^{\prime}(x)=h^{-1}(x)+x\cdot \dfrac{d}{dx}\left(h^{-1}(x)\right)\)
\(\text{Find}\ \dfrac{d}{dx}\left(h^{-1}(x)\right):\)
\(\text{Let}\ \ y=h^{-1}(x)\ \ \Rightarrow\ \ x=h(y)\)
\(\dfrac{dx}{dy}=h^{\prime}(y)\ \ \Rightarrow\ \ \dfrac{dy}{dx}=\dfrac{1}{h^{\prime}(y)}\)
\(\dfrac{d}{dx}\left(h^{-1}(x)\right)=\dfrac{1}{h^{\prime}\left(h^{-1}(x)\right)}\)
\(f^{\prime}(x)=h^{-1}(x)+\dfrac{x}{h^{\prime}\left(h^{-1}(x)\right)}\)
\(f^{\prime}(4)=h^{-1}(4)+\dfrac{4}{h^{\prime}\left(h^{-1}(4)\right)}=1+\dfrac{4}{h^{\prime}(1)}\)
\(h(x)=x^3+2x+1\ \ \Rightarrow\ \ h^{\prime}(x)=3x^2+2\)
\(f^{\prime}(4)=1+\dfrac{4}{3(1)^2+2}=\dfrac{9}{5}\)
\(\therefore\ \text{Gradient of tangent}=\dfrac{9}{5}\)