Theorem: Let $A \subset \mathbb{R}$ be a nonempty set, let $c \in A$ and let $f: A \rightarrow \mathbb{R}$ be a function. Suppose that $f$ is continuous at $c$.
1. If $f(c)>0$, then there is some $M>0$ and some $\delta>0$ s.t. $x \in A$ and $|x-c|<\delta \implies f(x)> M$
2. If $f(c)<0$, then there is some $N<0$ and some $\delta>0$ s.t. $x \in A$ and $|x-c|<\delta \implies f(x)<N$
Proof
So, I've completed the following so far,
Since $f$ is a continuous at $c$; $\exists \delta >0$ and $\forall x \in A$ $$ |x-c|<\delta \implies |f(x)-f(c)|<M $$
Suppose we assume $\delta = \frac{\delta}{2}$ since $f(c)>0 \implies M>0$. Now $f$ is continuous at $x=c$ and $\exists \delta >0$ s.t. $|x-c|<\delta \implies |f(x)-f(c)|<M=\frac{f(c)}{2}$ So, $$ |f(x) - f(c)|<\frac{f(c)}{2} $$ $$ \implies 0<-\frac{f(c)}{2} <|f(x) - f(c)|<\frac{f(c)}{2}$$ $$ \implies 0<f(c)-\frac{f(c)}{2} <f(x) <f(c)+\frac{f(c)}{2} $$ $$ \implies 0<\frac{f(c)}{2} <f(x) <\frac{3f(c)}{2} $$ $$ \implies f(x)>M $$
Just from simplification.
I was wondering how I could prove the second part or does the following proof work for both cases.
To prove the second part from the first part, let $h(x)=-f(x)$, then there exists an $M$ such that for some $\delta >0, s.t. x \in A$ and $|x-c|< \delta$, we have
$$h(x) > M$$ $$-f(x) > M$$
Now let $N=-M$ and you are done.