Proving f(x)=0 by Least Upper Bound on [a,b]

[andilus]

if f is continuous on [a,b] with f(a)<0<f(b), show that there is a largest x in [a,b] with f(x)=0

i think it can be done by least upper bounds, but i dun know wat is the exact prove.

 

[l'Hôpital]

Look up "Intermediate Value Theorem" or "Bolzano's Theorem."

 

[LCKurtz]

As another poster suggested, the intermediate value theorem guarantees there is an x in [a,b] where f(x) = 0. And your idea of using the lub is a good one. So let

$z =$ lub $\{ x \in [a,b] | f(x) = 0\}$

So what you need to show to finish the problem is:

1. z is in [a,b]
2. f(z) = 0
3. No value x > z in [a,b] satisfies f(x) = 0.

 

[LCKurtz]

I now see that this is a duplicate of an identical thread in the Calculus & Beyond section. Please don't do that. It wastes our time answering questions that have already been answered elsewhere.