Show that a continuous function f either has a root hence $f(c)=0$ or $|f(x)|> e$ for $e>0$. From what I understand, I am being asked to show that a function $f$ on an interval $[a,b]$ either has a root $c$ such that $f(c)=0$ or it does not have a root hence $|f(x)|\ge e$.
Should I apply the Intermediate Value Theorem for the first part?

 A: This is a consequence of the continuity of $f$. Since $[a,b]$ is compact, the function $|f|:[a,b]\longrightarrow \mathbb{R}, ~x\mapsto |f(x)|$ attains its minimum on $[a,b]$. Note that $|f(x)|\geq 0$ always holds. 
If the minimum of $|f|$ is zero, you have $f(x)=0$ for some $x\in [a,b]$, hence $x$ is a root. If not, the second part holds by the definition of a minimum.
A: Let $f:[a,b]\to\mathbb{R}$ be continuous. We want to prove that either
$(a)$ there exists $c\in[a,b]$ such that $f(c)=0$
or
$(b)$ there exists $\varepsilon>0$ such that $|f(x)|\geq\varepsilon$ for all $x\in [a,b]$.
It is enough to prove that if $(b)$ is false then $(a)$ is true (or vice versa).
Assume $(b)$ is false. Then, for all $n\in\mathbb{N}$, there is $x_n\in[a,b]$ such that $|f(x_n)|<\frac{1}{n}$. It follows from the Squeeze Theorem that $\lim f(x_n)=0$. Since $[a,b]$ is compact, there exists $c\in[a,b]$ and a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $\lim x_{n_k}=c$. As $f$ is continuous, we conclude that
$$f(c)=f(\lim x_{n_k})=\lim f(x_{n_k})=0$$
and thus $(a)$ holds.
A: It is continuous so for each $x\in\mathbb{R}$, f is defined and has a value like $y\in\mathbb{R}$. Using $x$ and $y=f(x)$, f can be represented in a cartesian system. It is clear that $\exists{x_1,x_2}: f(x_1)<0<f(x_2) \implies \exists{x_3}: f(x_3)=0$ because at some point f has to move from $y_{>0}$ part of the plane to the $y_{<0}$ part and since it is continuous, there has to be a defined point along this transition that actually crosses $x$ axis and that would be our $x_3$.
For the other part, $\nexists{x}: f(x)=0 \implies \forall{x}: f(x)<0 \:\lor f(x)>0$ this is obvious, that if some function doesnt contain 0 as the result, it either has to result in negative or positive number even if it is not continuous. then it's obvious that $\forall{x}: f(x)\ne0 \implies |f(x)|>0 \implies \exists\varepsilon:  |f(x)|\geqslant\varepsilon>0$, even if they are not continuous.
A: Two basic properties of a continuous $f:[a,b]\to R$, with $a<b$ :... (1).$ M=\max \{f(x) :x\in [a,b]\}$ and $m=\min \{f(x):x\in [a,b]\}$ exist....(2).The Intermediate Value Property (IVP) : $$\{f(x):x\in [a,b]\}= [m,M].$$ Therefore, since $m\leq M$, we have $$ 0 \not \in [m,M] \iff (M<0\lor m>0).$$ Now if $M<0$ then $$x\in [a,b]\implies f(x)\leq M<0\implies |f(x)|\geq -M>0.$$ While if $m>0$ then $$x\in [a,b]\implies f(x)\geq m>0.$$ On the other hand, by the IVP we have $$0\in [m,M]\iff 0\in \{f(x) :x\in [a,b]\}\iff \exists x\in [a,b]\;(f(x)=0).$$ 
