Here's the question I've been asked to prove:
Suppose that $f$ is differentiable at $c$ and that $f^{\prime } (c)
\ne 0$. Show that there exists a $\delta > 0$ such that
$0<|x-c|<\delta \Rightarrow f(x)\ne f(c)$.
Here's my attempt:
Assume the contrary that for every $\delta > 0$, there is a $x_{\delta} \in \text{dom} f$ such that $0<|x_{\delta}-c|<\delta$ with $f(x_{\delta})=f(c)$.
Now, we pick $\varepsilon$ such that $0< \varepsilon < |f^{\prime} (c)|$. Since $f$ is differentiable at $c$, there is an $\delta > 0$ such that if $0<|x-c|<\delta$ then we have $\left| \frac{f(x)-f(c)}{x-c} – f^{\prime} (c) \right| <\epsilon$. For this $\delta$, there is $x_{\delta} \in \text{dom} f$ such that $0<|x_{\delta}-c|<\delta$ with $f(x_{\delta})=f(c)$. Thus, we must have $\left| \frac{f(x_{\delta})-f(c)}{x-c} – f^{\prime} (c) \right| = \left| f^{\prime} (c) \right|<\epsilon$ which contradicts our choice of $\varepsilon$.
Is this proof okay? Is there any possible direct proof? This question is asked in my analysis textbook before "The Mean Value Theorem" section.
Best Answer
Your proof is fine. With only a small modification you can make it a direct argument:
For $0 < \epsilon < |f'(c)|$ there is (as you said) a $\delta > 0$ such that for $0 < |x - c | < \delta$ $$ \left| \frac{f(x)-f(c)}{x-c} - f'(c) \right| < \epsilon \, . $$ This implies (using the triangle inequality) $$ \left| \frac{f(x)-f(c)}{x-c} \right| > |f'(c)| - \epsilon > 0 $$ and therefore $f(x) \ne c$ for $0 < |x - c | < \delta$.