f is continuous at x = a 
Let $f$ be a continuous function defined on $\mathbb{R}$



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*In case of $f(0)=-1$ Prove that there exists values  $x>0$  with $f(x)<0$

*In case of $f(1)=1$ Prove that there exists values  $0<x<1$  with $f(x)>0$


Indeed,
 I'm supposed to use the definition of continuity here
$f$ is continuous at x = a iff
$$\forall \varepsilon > 0 \quad \exists \eta > 0 \quad \forall x \in I \quad \Big[|x - a| <\eta \Rightarrow|f(x) - f(a)|<\varepsilon\Big].$$


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*To prove question $1$  it's suffice to take $a=0,\ \varepsilon= 1 \implies \exists \eta > 0 \  |x| <\eta \Rightarrow|f(x) - f(0)|<1 $
$|f(x)-f(0)|<1 \implies  f(x) <f(0) + |f(x) - f(0)| < 0$ 


Then $f(x)<0$ and $\exists \eta > 0, \text{ such that } |x| <\eta$ here i'm stuck how can i say that $\exists \eta > 0, \text{ such that } |x| <\eta  \implies x>0$ i can't see 


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*To prove question 2: it's suffice to take $a=1,\ \varepsilon= 1 \implies \exists \eta > 0 \  |x| <\eta \Rightarrow|f(x) - f(1)|<1 $
$|f(x)-f(1)|<1 \implies  f(x)> f(1) - |f(x) - f(1)| > 0$ 


Then $f(x)>0$ and $\exists \eta > 0, \text{ such that } |x-1| <\eta$ here i'm stuck how can i say that $\exists \eta > 0, \text{ such that } |x-1| <\eta  \implies 0<x<1$ i can't see 
any help would be greatly appreciated
 A: question 1): For any $x$ with $\left|x\right|<\eta$ it is
true that $f\left(x\right)<0$. So it is true for e.g. $x=\frac{1}{2}\eta>0$.
question 2): For any $x$ with $\left|x-1\right|<\eta$ it is
true that $f\left(x\right)>0$. So it is true for e.g. $x=\max\left(1-\frac{1}{2}\eta,\frac{1}{2}\right)\in\left(0,1\right)$.
A: Suppose
$f(0)=-1$ and $f(x)\ge0$ for all $x>0$. 
Then for every $\delta>0$, 
we have $f(\delta)-f(0)\ge1$, 
so f cannot be continous at x=0 (Choose $\epsilon=\frac{1}{2}$) 
Analogue, you can show that
 f(1)=1 implies that f(x)>0 for some x with $0<x<1$ by using $f(1)-f(1-\delta)\ge1$ for all $\delta$ with $0<\delta<1$, if $f(x)\le0$ for all x with $0<x<1$
A: For $1$, we have concluded $\forall x\;(\;|x|<\eta\implies |f(x)+1|<1\;)$ for some $\eta>0$.
Thus, for $x_0=\frac12 \eta>0$, since $|x_0|<\eta$, we conclude, $|f(x_0)+1|<1$ which implies $f(x_0)<0$.
Similarly for $2$, we know  $\forall x\;(\;|x-1|<\eta\implies |f(x)-1|<1\;)$ for some $\eta>0$. Can you find an $x$ that is within $(0,1)$ but at a distance less than $\eta$ from $1$?
A: To prove question 1, all you have to do is look at $f(x)$ on $I = \mathbb{R}_{\ge 0}$, (the set of all non-negative real numbers), and then you can see there exists an $x > 0$ such that $f(x)<1$ using your method. This is because if $f$ is continuous on $\mathbb R$ it is continuous on any subset of $\mathbb R$.
On question 2, just examine $f$ on the closed interval $I = [0,1]$, again $f$ is continuous on this interval and the desired result would follow immediately from what you have already proven.
