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]