Coordinate Geometry (Tangent, Derivative of Inverse Function)

 (CIE 0606/2021/m/22/Q7)
A curve has equation $y=p(x)$, where $p(x)=x^{3}-4 x^{2}+6 x-1$

(a) Find the equation of the tangent to the curve at the point $(3,8)$. Give your answer in the form $y=m x+c . \quad[5]$

(b) (i) Given that $\mathrm{p}^{-1}$ exists, write down the gradient of the tangent to the curve $y=\mathrm{p}^{-1}(x)$ at the point $(8,3) .$

(ii) Find the coordinates of the point of intersection of these two tangents.


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(a) $p(x)=x^{3}-4 x^{2}+6 x-1$

$\dfrac{dp}{d x}=3 x^{2}-8 x+6$

At $(3,8)$, Ite gradient of tangent $=3(3)^{2}-8(3)+6=9$ 

Tangent Fquation: $y-8=9(x-3)$ $y=9 x-19\cdots(1)$

(b) (i) At $(8,3)$, the gradient of tangent to $y=p^{-1}(x)$ is $\dfrac{1}{9}$ 

Tangent Equation: $\quad y-3=\dfrac{1}{9}(x-8) \cdots$ (2)

(ii) $(1)-(2): 3=9 x-19-\dfrac{1}{9}(x-8)$


Thus $\quad y=9\left(\dfrac{19}{8}\right)-19=\dfrac{19}{8}$

$\therefore$ Intersection point $=\left(\dfrac{19}{8}, \dfrac{19}{8}\right)$
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