To put it bluntly, those of your colleagues who don’t accept ‘$x=1$’ as a solution don’t speak English. That sort of hyperpedantry accomplishes nothing beyond making students think that mathematics is all about invoking the right (incomprehensible) magic formula(s). I’ll go so far as to say that I think it excessive pedantry to object to $x=\frac12\left(1\pm\sqrt5\right)$ as an answer to a question asking for the solution(s) to $x^2-x-1=0$.
It would be another matter if the question were written The solution set is __: that question clearly calls for a set. In that case, however, the script should accept $\{1\}$, $\{x\in\Bbb R:x=1\}$, $\{x:x=1\}$, $[1,1]$, and any other reasonably straightforward variant, but not $x=1$, $1$, or the like.
For the Newton-Raphson case, if the number were rounded to $8$ decimal places between iterations, then if you get the same number twice in a row, every successive number has to be the same after that. Since Newton-Raphson is quadratically convergent you normally get about twice the significant figures on every iteration after a pretty good approximation has been achieved. Of course you can make up examples where no convergence is reached or even force the algorithm to cycle between a limit set of values.
If you are tracking the values by hand you can see when you have over half the significant figures you want and perform the last iteration with extra precision to create a better probability of getting the last digit right. But you are right in thinking that it's a big problem to try to get the last digit right every time in floating point computations. In general you probably need about twice the significant figures of your final output to get the last digit right every time.
Let's look at an example with $\cosh x$ where we will run out all the $8$-digit numbers between $0$ and $1$ and then see how many might require $15$ digits to get the $8^{\text{th}}$ digit right. Out program searches for outputs with digits $9:15$ having values between $4999999$ and $5000001$. Since this is a range of $2$ out of $1\times10^7$, we might expect about $20$ hits for a program testing $1\times10^8$ inputs, and we are not far off in that estimate.
program round
use ISO_FORTRAN_ENV, only:wp=>REAL128,qp=>REAL128,wi=>INT64
implicit none
real(wp) x,y,z
real(qp) qx, qy
integer(wi) i
do i = 0,10_wi**8
x = 1.0e-8_wp*i
y = cosh(x)
z = y*1.0e8_wp+0.5_wp
if(abs(z-nint(z))<1.0e-7) then
qx = 1.0e-8_qp*i
qy = cosh(qx)
write(*,'(f10.8,1x,f22.20)') x,qy
end if
end do
end program round
Output:
0.00010000 1.00000000500000000417
0.00030000 1.00000004500000033750
0.05844226 1.00170823500000040322
0.07380594 1.00272489500000039410
0.44746231 1.10179282500000099007
0.45315675 1.10444463500000093431
0.47303029 1.11398059500000076845
0.47962980 1.11724437499999901204
0.49332468 1.12417258499999983893
0.49725888 1.12620181499999997563
0.51736259 1.13684395499999912340
0.59708506 1.18361444500000091792
0.60322956 1.18752751500000094494
0.64184055 1.21314873500000023866
0.72946242 1.27806675499999917355
0.75252442 1.29676328499999998383
0.90813720 1.44148689499999975793
0.93055850 1.46512925500000051185
The output for $0.00010000$ was expected from the Taylor series for $\cosh x$, but check out how close we got with $0.75252442$: we would have had to calculate the $17^{\text{th}}$ digit to round the $8^{\text{th}}$ digit correctly.
Best Answer
The graph shows were are, more or less, the roots.
Consider the first one and make $x=-2+t$ to obtain $$7+64 t-112 t^2+72 t^3-20 t^4+2 t^5=0$$ So, now, the powers of $t$ become negigibl and an appoximation is given by $$7+64t=0 \implies t=-\frac 7 {64}$$ Restart the same procedure with now $x=-2-\frac 7 {64}+t$ and neglecting again, the new approximation is $$-\frac{771453991}{536870912}+\frac{764957925 t}{8388608}=0 \implies t=\frac{771453991}{48957307200}$$ SO, at this point $$x=-2-\frac 7 {64}+\frac{771453991}{48957307200}=-\frac{25624466471}{12239326800}=-2.09362$$ while the exact solution is $-2.09323$
This is more than simplistic but this is how Newton invented Newton method