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Let $f: \mathbb R \to \mathbb R$ function, that is differentiable on $x=0$ and $f'(0) \neq 0$. Show by the definition of derivative that there exists $\delta > 0$ such that

$f(x) \neq f(0)$ for all $x \in ]-\delta, \delta[, x \neq 0$.

Okay, I tried to apply epsilon-delta to this, but I'm quite unsure even what to do, can you help me out?

2 Answers 2

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Suppose to the contrary, that to each $ n \in \mathbb N$ we find some $x_n \in (-\frac{1}{n}, \frac{1}{n})$ such that $x_n \ne 0$ and $f(x_n)=f(0)$. Then we get

$$f'(0)= \lim_{n \to \infty}\frac{f(x_n)-f(0)}{x_n-0}=0,$$

a contradiction !

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HINT: let the $\epsilon{-}\delta$ definition of differentiability for some function $f$ at the point $y$:

$$\forall\epsilon>0,\exists\delta>0:0<|x-y|<\delta\implies\left|\frac{f(x)-f(y)}{x-y}-f'(y)\right|<\epsilon$$

Then for the point $y=0$ we have that for any $\epsilon>0$ exists some $\delta>0$ such that

$$0<|x|<\delta\implies \left|\frac{f(x)-f(0)}{x}-f'(0)\right|<\epsilon$$

Now:

  1. Choose some $\epsilon\le|f'(0)|$ (we can do this because the definition of differentiability hold for any $\epsilon>0$ and $f'(0)\neq0$).

  2. Observe what happen if it would exists some $0<|x_0|<\delta$ such that $f(x_0)-f(0)=0$.

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    I can't finish it.. – 2017-01-12
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    @repulsive23 what happen if, for some $\epsilon<|f'(0)|$ (as I said above) exists some $0<|x_0|< \delta$ such that $f(x_0)=f(0)$? – 2017-01-12
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    It isnt differentiable anymore? I have no idea – 2017-01-12
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    $|f'(x_0)| < $ epsilon? But it says in the question that it doesnt happen? – 2017-01-12
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    @repulsive23 Im showing a proof by contradiction. If such $x_0$ would exists then you have the inequality $|f'(0)|<|f'(0)|$, what cannot be possible, hence cannot exists such $x_0$. – 2017-01-12
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    Thank you so much for this help!!! I think I got it now – 2017-01-12