Find a linear function that goes through function $g(x)=(3-2\sqrt2){x}$ in point T($x_0$,1) and does that perpendicullary to the function..
$$\phi=\pi/2$$ $$\pi/2=\frac{(3-2\sqrt2-k)}{1+(3-2\sqrt2)k}$$ $$\frac{\pi}{2}(1+(3-2\sqrt2)k)-3+2\sqrt{2}=\frac{-k}{1}$$ Then we get: $$k=\frac{-3+2\sqrt{2}+2\pi}{\sqrt{2}\pi-1}$$ The point in which the x-axis intercept should be is the zero of the function $g$, so it is $x=0$ $$h(x)=(\frac{-3+2\sqrt{2}+2\pi}{\sqrt{2}\pi-1})x$$
I find it very strange that such a function would be the solution.
If $g(x)$ is linear (which it is) and has slope $m$, then the line perpendicular to it has slope $-\frac{1}{m}$. So the line perpendicular $h(x)$ has slope $\frac{-1}{3-2\sqrt{2}}$. The point $\big(\frac{1}{3-2\sqrt2},1\big)$ is on $g(x)$, and $g(x)$ and $h(x)$ share that point, so the line perpendicular, by point-slope, is $$h(x) = \frac{-1}{3-2\sqrt{2}}\Big(x-\frac{1}{3-2\sqrt2}\Big)+1$$ If I interpreted the question wrong please let me know, I was confused over some of the notation.