A restricted parabola and its inverse

The graph is the increasing right branch of the parabola with vertex (4,1), but the vertex is excluded because x>4. Draw an open endpoint at (4,1) and show the axes and origin. The graph lies to the right of x=4 and above y=1.

Let $y=f(x)$, so $f^{-1}(y)=x$. From $y=(x-4)^2+1$, obtain $(x-4)^2=y-1$ and $x=4\pm\sqrt{y-1}$. The negative branch fails x>4. Therefore $f^{-1}(x)=4+\sqrt{x-1}$, with $D_{f^{-1}}=R_f=(1,\infty)$. The inverse has range (4,∞).

Reflect the original curve in y=x. The inverse is the increasing square-root curve with open endpoint (1,4). Keep both curves and the reflection line on one set of axes.

The reflection line is y=x. The source sketch has one intersection, lying on this line. Solve $(x-4)^2+1=x$, giving $x^2-9x+17=0$, so $x=(9\pm\sqrt{13})/2$. Reject the smaller value because x>4; the exact solution is $(9+\sqrt{13})/2$. As an independent justification, a strictly increasing function cannot have an off-diagonal two-cycle: f(x)>x would imply f(f(x))>f(x), and f(x)<x gives the reverse contradiction. Thus equality with its inverse here forces a fixed point.