Answer: tangent \(y=9x-19\); inverse tangent gradient \(\dfrac19\); intersection \(\left(\dfrac{19}{8},\dfrac{19}{8}\right)\).

(a) Differentiate

\(p(x)=x^3-4x^2+6x-1.\)

This gives

\(p'(x)=3x^2-8x+6.\)

At \(x=3\),

\(p'(3)=3(3^2)-8(3)+6=27-24+6=9.\)

The tangent at \((3,8)\) has gradient \(9\), so

\(y-8=9(x-3).\)

Therefore

\(y=9x-19.\)

(b)(i) The tangent to \(y=p^{-1}(x)\) at \((8,3)\) has gradient equal to the reciprocal of the tangent gradient to \(y=p(x)\) at \((3,8)\).

Hence its gradient is

\(\frac19.\)

(b)(ii) The tangent to \(y=p^{-1}(x)\) at \((8,3)\) is

\(y-3=\frac19(x-8).\)

So

\(y=\frac{x+19}{9}.\)

Now solve the two tangent equations:

\(9x-19=\frac{x+19}{9}.\)

Multiplying by \(9\),

\(81x-171=x+19.\)

Thus

\(80x=190,\qquad x=\frac{19}{8}.\)

Substitute into \(y=9x-19\):

\(y=9\left(\frac{19}{8}\right)-19=\frac{19}{8}.\)

Therefore the point of intersection is

\(\left(\frac{19}{8},\frac{19}{8}\right).\)