Theorem: if f is continuous in [a, b] and f(a) < 0, f(b) > 0, then there exists c in [a, b] such that f(c) = 0.
The most popular proof on the website is proof by contradiction. Thus, we have two cases, f(c) < 0 or f(c) > 0.
first, I suppose f(c) > 0. How can I use continuity to prove that there is a contradiction? The following is my attempt.
$\forall \epsilon > 0, \exists \delta > 0$ s.t $\forall x: |x - c| < \delta$ we have $|f(x) - f(c)| < \epsilon$ Also, I let S = {x $\in$ [a, b] : f(x) < 0}. Since S is bounded, I got a least upper bound c. Then, how can I proceed?
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$\begingroup$In the next section we a prove a proposition by contradiction that can be used by the OP.
Proposition 1: Let $g:[a,b] \to \Bbb R$ be a function satisfying $g(x) \lt 0$ for $x \lt b$.
If $g$ is continuous at $x = b$ then $g(b) \lt 0$ or $g(b) = 0$.
Proof
To arrive at a contradiction, assume that $g(b) \gt 0$.
Let
$\quad \varepsilon = g(b)$
Since $g$ is continuous at $b$ we can find a $\delta \gt 0$ such that
$\tag 1 \big (\forall x \in [a,b]\big ) \; |x - b| < \delta \text{ implies } |g(x) - g(b)| < \varepsilon$
It follows that we can find a $x_0 \in [a,b)$ such that
$\tag 2 |g(x_0) - g(b)| < \varepsilon = g(b)$
But then $g(x_0) \gt 0$, a contradiction. $\quad \blacksquare$
The OP can now prove the following theorem.
Theorem 2: Let $f:[a,b] \to \Bbb R$ be a continuous function satisfying $f(a) \lt 0$ and $f(b) \gt 0$. Then there exists one and only one number $c \in (a,b)$ such that $f(x) \lt 0$ for $x \in [a,c)$ and $f(c) = 0$.
$\endgroup$ $\begingroup$Another possible proof is to successively halve the interval. In this way we construct two sequences: nondecreasing $(x_n)$ and nonincreasing $(y_n)$ s.t. $x_n\le y_n$ for any $n$ and $f(x_n)$ is of a constant sign, $f(y_n)$ is of constant opposite sign. We prove that $x_n,y_n$ tend to the common limit, which is a zero of $f$.
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