Prove there exists a delta

Question

Let be a real-valued continuous function defined on a set S and suppose that for some c∈S we have f(c) < 0. Prove that there exists > 0 such that f(x) < 0 ∀∈(c−,c+).

I tried:

To prove - ( is continuous at c) and (c)<0 ⟹ ∃>0, ()<0 0<|−c|< Assume is continuous at c and (c)<0

Then, ∀>0,∃>0, s.t. |−c|<⟹|()−(c)|< s.t.|()|−|(c)|=|()|−(c)≤|()−(c)|< |()|<+(c)

This is where I stopped I was not sure as to where to go from here or if I am on the right path.

Answer 1

f is continuous at x=c. Therefore,

$\forall \epsilon>0, $ there exists a $\delta$ such that if $x \in (c-\delta,c+\delta)$, then $f(x) \in (f(c)-\epsilon,f(c)+\epsilon)$. In particular, you can choose $\epsilon$ to be $-\frac{1}{2} f(c)$, (epsilon is positive, and f(c) is negative), in which case you get $x \in (c-\delta,c+\delta)$, then $f(x) \in (\frac{3}{2}f(c),\frac{1}{2}f(c))$, so f(x)<0 for x.

Answer 2

Guide:

From continuity, we can find a $\delta >0$, such that $\forall x \in (c-\delta, c+\delta)$ such that $$|f(x)-f(c)|< -\frac{f(c)}2$$

Try to simplify the last inequality to get an upper bound on $f(x)$ in that neighborhood.