Find the largest possible domain \(D\) for \(f\) to be a function.

Sketch the graph of \(y = f(x)\) showing clearly the equations of asymptotes and the coordinates of any intercepts with the axes.

Explain why \(f\) is an even function.

Explain why the inverse function \({f^{ - 1}}\) does not exist.

Find the inverse function \({g^{ - 1}}\) and state its domain.

Find \(g'(x)\).

Hence, show that there are no solutions to \(g'(x) = 0\);

Hence, show that there are no solutions to \(({g^{ - 1}})'(x) = 0\).

\({x^2} - 1 > 0\)     (M1)

\(x < - 1\) or \(x > 1\)     A1

[2 marks]

shape     A1

\(x = 1\) and \(x = - 1\)     A1

\(x\)-intercepts     A1

[3 marks]

EITHER

\(f\) is symmetrical about the \(y\)-axis     R1

OR

\(f( - x) = f(x)\)     R1

[1 mark]

EITHER

\(f\) is not one-to-one function     R1

OR

horizontal line cuts twice     R1

Note:     Accept any equivalent correct statement.

[1 mark]

\(x = - 1 + \ln \left( {\sqrt {{y^2} - 1} } \right)\)     M1

\({{\text{e}}^{2x + 2}} = {y^2} - 1\)     M1

\({g^{ - 1}}(x) = \sqrt {{{\text{e}}^{2x + 2}} + 1} ,{\text{ }}x \in \mathbb{R}\)     A1A1

[4 marks]

\(g'(x) = \frac{1}{{\sqrt {{x^2} - 1} }} \times \frac{{2x}}{{2\sqrt {{x^2} - 1} }}\)     M1A1

\(g'(x) = \frac{x}{{{x^2} - 1}}\)     A1

[3 marks]

\(g'(x) = \frac{x}{{{x^2} - 1}} = 0 \Rightarrow x = 0\)     M1

which is not in the domain of \(g\) (hence no solutions to \(g'(x) = 0\))     R1

[2 marks]

\(({g^{ - 1}})'(x) = \frac{{{{\text{e}}^{2x + 2}}}}{{\sqrt {{{\text{e}}^{2x + 2}} + 1} }}\)     M1

as \({{\text{e}}^{2x + 2}} > 0 \Rightarrow ({g^{ - 1}})'(x) > 0\) so no solutions to \(({g^{ - 1}})'(x) = 0\)     R1

Note:     Accept: equation \({{\text{e}}^{2x + 2}} = 0\) has no solutions.

[2 marks]