# Silly Question about systems of equations

1. Sep 29, 2007

I had a thought while doing some calculus last night and it now concerns me that for some reason I cannot answer it. If given the equations $$y=x$$ and $$y=x^2+1$$ it would appear that since the range of the latter "reaches" infinity "first" that these two curves never meet. Now that thought did not occur to me until I sketched them after I tried solving them as a system of equations.

Is there some way (not intuitively, but abstractly) to see that these two curves never intersect?

Thanks,
Casey

2. Sep 29, 2007

### Staff: Mentor

These two equations specifically, sure. x2+1=x obviously has no solutions in the reals.

Any system of equations, obviously not. There is no general solution for finding the zeros of a quintic, let alone any arbitrary function.

3. Sep 29, 2007

I am not interested in finding any zeros; I am interested in knowing if there is a way to determine that there are not any zeros.

Casey

4. Sep 29, 2007

### HallsofIvy

Staff Emeritus
Look at f(x)= x2+1- x. f'(x)= 2x- 1 which is positive for x> 1/2 so it is always increasing. When x= 1/2, that has a value of 1/4+ 1- 1/2= 3/4. As long as there is no zero between 0 and 1/2, the function cannot have a zero and so the equation
x2+ 1= x cannot have a solution.

Last edited by a moderator: Oct 2, 2007
5. Sep 29, 2007

### Staff: Mentor

There is no general method to determine whether some function $f(x)$ has any zeroes. Suppose you evaluate the function at two points, $x_1$ and $x_2$. If the function value is positive at both points (or negative at both points), there might well be no points or two points or four points (or ...) between $x_1$ and $x_2$ where the function is zero. (Note: In saying this I am counting double zeros such as x2=0 as two zeros). You don't know which is the case and there is no general way of knowing.

Suppose you get lucky and find that $f(x_1)$ and $f(x_2)$ have opposite signs. You still don't know that the function passes through zero between $x_1$ and $x_2$. Consider $f(x)=1/x$ for example. $f(-2)=-1/2$ and $f(1)=1$, but the function does not pass through zero between -2 and 1. In the special case tht $f(x)[/tex] is continuous and exists everywhere between [itex]x_1$ and $x_2$ then finding points with opposite signs does guarantee that the function is zero at at least one point between $x_1$ and $x_2$.

6. Sep 29, 2007