how did we obtain the value of f(xr)

pls show how in both iteration 1 and 2

Q=n(B+2H)2/3S(BH)5/3 where, Q=flow=5m3/s,B=width=20m,H=depthofliquidS=slope=0.0002,n=roughnesscoefficient=0.03 1) Suggest two initial guesses for the root (H). Justify your choice the depth of the wootar depends on Ke depth ot be ol and sinc it is not sentioned how deep he clobnel to on chagese ang numba trom oto Assuning xL=0 and xu=1 leks check it ow initiol gaesses cre corred letH=x5=0.03(20+2x)2/3(2104)(20x)5/30=0.03(20+2x)2/3(2104)(20x)5/35f(x2)=5f(xu)=3.847 2) Would you use the Bisection or False position method to solve this problem? False position becouse it mostly converger faster Van Bisection mebod 3) Perform three iterations using the False position method, Determine for the last iteration. 0=0.03(20+2x)2/3(2104)(20x)5/35 given xL=0xu=1 ileration 1 f(xL)=5f(xr)=1.489f(xu)=3.847xr1(5)(3.847)(3.847)(01)=0.565f(xL)f(xr)>0 " there is no root between XL and Xr" keep xu=1 updat xL=xr=0.565 jteration. 2. f(xL)=1.409xr=1(1.489)(3.847)(3.847)(0.5651)=0.686f(xu)=3.847f(xr)=0.182)f(xr)>0f(sthereisno.rootbetweenxLandxrkeepxu=1updatxL=xr=0.686 iteration 3 f(x2)=0.182f(x4)=3.847xr=1(0.882)(3.847)(3.847)(0.6861)0.70) a=xrxrxrold100a=0.7010.7010.686100a=2%