Determine \(D\), the largest possible domain of \(f\).

Show that the graph of \(f\) has three asymptotes and state their equations.

Show that \(f'(x) =  - \frac{{3{{\text{e}}^x}}}{{{{(2{{\text{e}}^x} - 1)}^2}}}\).

Use your answers from parts (b) and (c) to justify that \(f\) has an inverse and state its domain.

Find an expression for \({f^{ - 1}}(x)\).

Consider the region \(R\) enclosed by the graph of \(y = f(x)\) and the axes.

Find the volume of the solid obtained when \(R\) is rotated through \(2\pi \) about the \(y\)-axis.

attempting to solve either \(2{{\text{e}}^x} - 1 = 0\) or \(2{{\text{e}}^x} - 1 \ne 0\) for \(x\)     (M1)

\(D = \mathbb{R}\backslash \left\{ { - \ln 2} \right\}\) (or equivalent eg \(x \ne  - \ln 2\))     A1

Note: Accept \(D = \mathbb{R}\backslash \left\{ { - 0.693} \right\}\) or equivalent eg \(x \ne  - 0.693\).

[2 marks]

considering \(\mathop {\lim }\limits_{x \to  - \ln 2} f(x)\)     (M1)

\(x =  - \ln 2{\text{ }}(x =  - 0.693)\)    A1

considering one of \(\mathop {\lim }\limits_{x \to  - \infty } f(x)\) or \(\mathop {\lim }\limits_{x \to  + \infty } f(x)\)     M1

\(\mathop {\lim }\limits_{x \to  - \infty } f(x) =  - 2 \Rightarrow y =  - 2\)    A1

\(\mathop {\lim }\limits_{x \to  + \infty } f(x) =  - \frac{1}{2} \Rightarrow y =  - \frac{1}{2}\)    A1

Note: Award A0A0 for \(y =  - 2\) and \(y =  - \frac{1}{2}\) stated without any justification.

[5 marks]

\(f'(x) = \frac{{ - {{\text{e}}^x}(2{{\text{e}}^x} - 1) - 2{{\text{e}}^x}(2 - {{\text{e}}^x})}}{{{{(2{{\text{e}}^x} - 1)}^2}}}\)    M1A1A1

\( =  - \frac{{3{{\text{e}}^x}}}{{{{(2{{\text{e}}^x} - 1)}^2}}}\)    AG

[3 marks]

\(f'(x) < 0{\text{ (for all }}x \in D) \Rightarrow f\) is (strictly) decreasing     R1

Note: Award R1 for a statement such as \(f'(x) \ne 0\) and so the graph of \(f\) has no turning points.

one branch is above the upper horizontal asymptote and the other branch is below the lower horizontal asymptote     R1

\(f\) has an inverse     AG

\( - \infty  < x <  - 2 \cup  - \frac{1}{2} < x < \infty \)    A2

Note: Award A2 if the domain of the inverse is seen in either part (d) or in part (e).

[4 marks]

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

Note: Award M1 for interchanging \(x\) and \(y\) (can be done at a later stage).

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

\({{\text{e}}^y}(2x + 1) = x + 2\)    A1

\({f^{ - 1}}(x) = \ln \left( {\frac{{x + 2}}{{2x + 1}}} \right){\text{ }}\left( {{f^{ - 1}}(x) = \ln (x + 2) - \ln (2x + 1)} \right)\)    A1

[4 marks]

use of \(V = \pi \int_a^b {{x^2}{\text{d}}y} \)     (M1)

\( = \pi \int_0^1 {{{\left( {\ln \left( {\frac{{y + 2}}{{2y + 1}}} \right)} \right)}^2}{\text{d}}y} \)    (A1)(A1)

Note: Award (A1) for the correct integrand and (A1) for the limits.

\( = 0.331\)    A1

[4 marks]