Code not setting value=0

I don't understand this odd little thing in my code. I have simplified the function but this is basically what it's doing:

A=10;
fun=@(x) x+1/(A.*x);
z=fzero(fun,0)

This runs fine and gives me the result.

However if I do this...

A=10;
B=0;
fun=@(x) x+1/((A.*x)+(B.*x));
z=fzero(fun,0)

... then I get problems. I thought that this should give the same result! I don't get why defining B=0 doesn't seem to be taken into account when fzero solves fun.

The idea is that I can change B once I know that for B=0 I get the same result as for if there was no B term in the equation.

--- update ---

I have identified that the problem is with the x in the B.*x term. If I just do (A.*x)+B then it works fine. Because the fzero variable is x it seems to not recognise that whilst there is an additional x dependance it should not take it into account since B=0 makes that term 0.

[Matt J]

This runs fine and gives me the result.

No, even your first example, doesn't work. The function is undefined at x=0 and fails as it should

    z=fzero(fun,0)
    Error using fzero (line 309)
    Function value at starting guess must be finite and real.

[Matt J]

The function

 fun=@(x) x+1/((A.*x)+(B.*x));

can never have any roots when A+B>0. In the region x>0, both terms being summed in the function are strictly positive. The function is undefined at x=0. Inn the region x<0, all terms are strictly negative.

Answer 1

[Azzi Abdelmalek]

A=10;
B=1;
fun=@(x) x+1/((A.*x)+(B.*x));
z=fzero(fun,eps)

[James]

That doesn't seem to work. Still getting the problem that the B=0 result is not the same as when no B in the equation.

Answer 2

[Matt J]

I think you might be missing some parentheses

fun=@(x) (x+1)/((A.*x)+(B.*x));

Otherwise, the function can't have any roots (see my Comments above).