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Let $f$ be a continuous function defined on $\mathbb{R}$

  1. In case of $f(0)=-1$ Prove that there exists values $x>0$ with $f(x)<0$
  2. 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 $$\forall \varepsilon > 0 \quad \exists \eta > 0 \quad \forall x \in I \quad \Big[|x - a| <\eta \Rightarrow|f(x) - f(a)|<\varepsilon\Big].$$

  • 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

  • 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

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    $\begingroup$ About 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. $x=\frac{1}{2}\eta>0$. $\endgroup$ – drhab May 26 '14 at 14:09
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    $\begingroup$ About 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. $x=\max\left(1-\frac{1}{2}\eta,\frac{1}{2}\right)\in\left(0,1\right)$. $\endgroup$ – drhab May 26 '14 at 14:15
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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. $x=\max\left(1-\frac{1}{2}\eta,\frac{1}{2}\right)\in\left(0,1\right)$.

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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 $0<x<1$

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  • $\begingroup$ Thanks but My goal to see $\exists \eta > 0, \text{ such that } |x| <\eta \implies x>0$ $\endgroup$ – Adam May 26 '14 at 13:59
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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 $1$?

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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.

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