Determine if valid initial guess for closed numerical method

If I'm given a function f(x), how can I determine whether say [0 1] or [1,2 ] are valid initial guesses for a closed numerical method, of solving f(x)=0.

Answer 1

[Torsten]

If you want to use "fzero", you can first evaluate f in the endpoints of the interval [a b] you provide. If sign(f(a)*f(b)) < 0, you provide a valid initial guess.

Or did you want to ask something else - because your question was quite unclear ?

Best wishes

Torsten.

[Rachel Dawn]

Well, I'm trying to write code for bisection method, to find the roots of a function. It's for an assignment, and they are asking of us to first check if the guesses are valid and then to "determine the result such that f(x)< abs(E) where E=0.1." I'm guessing I should do the "if" statement and below it a bunch of code intended to find the root and then after that "else disp('Invalid guess')". But I was looking at some notes from class, and they wrote it out like this:

xa=10
xb=21
iteration=1
while iteration<= 30 && abs(xb-xa)>=10^-4
xc= (xa+xb)/2
if f(xc)*f(xa)<0
xb=xc
else
xa=xc
end
iteration=iteration+1
end

This confuses me because they have if f(xc)*(xa)<0 (which makes sure it's a valid guess), but then else (if >0), they say that xa=xc ? Shouldn't it be that the guess wasn't valid?

[Rachel Dawn]

Oh wait! I just realized that it says f(xc), not f(xb), so I would do if f(xa)*f(xb)>0 code to find root and else *disp('invalid guess')

But I'm still not sure about the f(x)<abs(E) part. What does this imply? & should it be placed at the end of the code?

[Rachel Dawn]

Also, sorry for all the questions, but in the following code:

while(iteration <=100 && abs(xleft-xright)>10^-3

it's part of the bisection method, but what does that abs part imply? why is it necessary?

[Torsten]

Try

xa = 10;
xb = 21;
fxa = f(xa);
fxb = f(xb);
if fxa*fxb > 0
  disp('Invalid initial guess.')
end
iteration = 1;
E = 0.1;
error_f = 1.0;
while (iteration < 100 && error_f >= E)
  xc = (xa+xb)/2;
  fxc = f(xc);
  if fxc*fxa<0
    xb = xc;
  else
    xa = xc;
    fxa = fxc;
  end
  error_f = abs(fxc);
  iteration = iteration+1;
end
xc

You do not need to include the abs-part from above in the while-statement because the number of iterations already determines the minimum length of the interval in which you search for the root.

Best wishes

Torsten.