(*Plot the function*)
k1[x_]:= 8/9 - ((4/3-x)^2)/2;
k2[x_]:= 1 - 1/(6 x^2);
Show[Plot[k1[x],{x,0,1}],Plot[k2[x],{x,1,10}],
	PlotLabel->"y = K(x)",AxesLabel->{"x","y"},AxesOrigin->{-0.2,-0.02},PlotRange->{{0,10},{0,1}}]
(*Plot the inverse function the slow way*)
Solve[k1[x]==y,x]
l1[y_]:= x/.Solve[k1[x]==y,x][[1]];
l1[y]
Solve[k2[x]==y,x]
l2[y_]:= x/.Solve[k2[x]==y,x][[2]];
l2[y]
Show[Plot[l1[y],{y,0,5/6}],Plot[l2[y],{y,5/6,599/600}],
	PlotLabel->"x = L(y)",AxesLabel->{"y","x"},AxesOrigin->{-0.02,-0.2},PlotRange->{{0,1},{0,10}}]
(*Plot the inverse function the fast way*)
Show[ParametricPlot[{k1[x],x},{x,0,1}],ParametricPlot[{k2[x],x},{x,1,10}],
	PlotLabel->"x = L(y)",AxesLabel->{"y","x"},AxesOrigin->{-0.02,-0.2},PlotRange->{{0,1},{0,10}}]