2011 Paper 2 Question 3

$f^{-1}(x)=\frac{{\mathrm{e}}^{x-3}-1}{2}.$
$D_{f^{-1}}=R_f=(-\infty ,\infty ).$

Sketch.

$x=-0.4847,5.482.$

$$\begin{array}{c}\text{Let }y=\ln (2x+1)+3\\ \ln (2x+1)=y-3\\ 2x+1={\mathrm{e}}^{y-3}\\ 2x={\mathrm{e}}^{y-3}-1\\ x=\frac{{\mathrm{e}}^{y-3}-1}{2}\end{array}$$$$f^{-1}(x)=\frac{{\mathrm{e}}^{x-3}-1}{2}■$$

$$\begin{array}{rl}{D}_{f^{-1}}& =R_f\\ & =(-\infty ,\infty )■\end{array}$$

The graph of $y=f^{-1}(x)$ is a reflection of the graph of $y=f(x)$ about the line $y=x.$

We observe that the intersections between the two curves are also the intersections between the graph of $y=f(x)$ and the line $y=x.$

Hence the solutions to $f(x)=f^{-1}(x)$ can be found by solving$$\begin{array}{c}f(x)=x\\ \ln (2x+1)+3=x\\ \ln (2x+1)=x-3∎\end{array}$$

Using a GC, the $x\text{-coordinates}$ of the points of intersection are

$$x=-0.4847∎,x=5.482∎$$