2008 Paper 2 Question 4

Sketch.

$f^{-1}(x)=\sqrt{x-1}+4.$
$D_{f^{-1}}=R_f=(1,\infty ).$

Sketch.

Line $y=x.$

$x=\frac{9+\sqrt{13}}{2}.$

$$\begin{array}{c}\text{Let }y={(x-4)}^2+1\\ {(x-4)}^2=y-1\\ \text{Since }x>4,\\ x-4=\sqrt{y-1}\\ x=\sqrt{y-1}+4\end{array}$$

$$f^{-1}(x)=\sqrt{x-1}+4■$$

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

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 intersection between the two curves is also the intersection between the graph of $y=f(x)$ and the line $y=x.$

Hence the solution to $f(x)=f^{-1}(x)$ can be found by solving$$\begin{array}{c}f(x)=x\\ {(x-4)}^2+1=x\\ x^2-8x+17=x\\ x^2-9x+17=0\end{array}$$

$$\begin{array}{rl}undefined& =\frac{-(-9)\pm \sqrt{{(-9)}^2-4\left(1\right)\left(17\right)}}{2\left(1\right)}\\ & =\frac{9\pm \sqrt{13}}{2}\\ & =\frac{9\pm \sqrt{13}}{2}\end{array}$$

Since $x>4,$ $$x=\frac{9+\sqrt{13}}{2}∎$$