Continuity and Differentiability question regarding non-degenerate intervals Let $ I\subset \mathbb{R}$ be a non-degenerate open interval, let $c \in I$ and let $f: I \rightarrow \mathbb{R}$ be a function. Suppose that $f$ is differentiable at $c$.


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*Prove that if $f'(c)>0$, then there is some $\delta>0$ such that $(c-\delta,c+\delta) \subset I$, that $c \in (c-\delta,c)$ implies $f(x)<f(c)$, and that $x \in [c,c+\delta) \implies f(x)>f(c)$.


Proof I have so far
Clearly, if $f'(c)>0$ and $f$ is differentiable at $c$ then $f$ is also continuous at $c$ so, $\forall \epsilon >0:\exists \delta >0| (c-\delta, c+\delta) \subset I$. This seams correct as it meets the definition of continuity.
The second and third parts confuse me as I'm unsure how 
$c \in (c-\delta,c)$ implies $f(x)<f(c)$. Is it because our interval is restricted $\delta$ close to zero and $f(x)$ must be $0 \leq f(x) \leq f(c-\delta)\leq f(c)$. I'm unsure how to prove this.
 A: Some $\delta>0$ is such that 
\begin{align*}
\left|\dfrac{f(x)-f(c)}{x-c}-f'(c)\right|<\dfrac{f'(c)}{2},~~~~0<|x-c|<\delta,
\end{align*}
for $x\in(c,c+\delta)$, then 
\begin{align*}
\dfrac{f(x)-f(c)}{x-c}-f'(c)>-\dfrac{f'(c)}{2},
\end{align*}
so $f(x)-f(c)>\dfrac{1}{2}f'(c)(x-c)>0$, so $f(x)>f(c)$.
A: You have to get the understanding behind symbols. Once you have the understanding in your mind, translating that into symbols is trivial job. You are given that the derivative of $f$ at $c$ ie $f'(c) $ is positive. Next you know that the derivative $f'(c)$ is the limit of difference quotient $(f(x) - f(c)) /(x-c) $ as $x\to c$. Thus combining the two facts above we can see that the limit of difference quotient is positive.
Next we need to understand the idea of limit. If the limit of something is positive the thing also has to be positive in some neighborhood of the point where limit is being considered. Why? Because by very definition of limit, the thing can be ensured to be as close to the limit as we want and if you get too close to a positive number, you will reach positive values.
Thus we have a neighborhood of type $(c-\delta, c+\delta) \subset I$ such that the difference quotient $(f(x) - f(c)) /(x-c)$ is positive whenever $x$ lies in this (deleted/punctured) neighborhood $(c-\delta, c+\delta)-\{c\} $. This is exactly what you are asked to prove (you might need sometime to figure out that your second third parts can be combined to say that the difference quotient is positive).

Your approach does not seem to be heading anywhere. You are trying to infer continuity from differentiability (which is mathematically correct) but we don't need the continuity here, rather what is needed is the behavior of difference quotient. Further if $I$ is open interval and $c\in I$ then by definition of open intervals there is some $\delta>0$ such that $(c-\delta, c+\delta) \subset I$. This is nothing but the defining property of open intervals and has nothing to do with continuity / differentiability. Our question is not about finding just some $\delta>0$ for which $(c-\delta, c+\delta) \subset I$ but the same $\delta$ should ensure that the second and third parts also hold. 
A: It is an analytic proof but rather in a more "geometric" way of understanding.
Remember that differentiability at point $c$ means geometrically, that the graph can be approximated by a small piece of line with slope $f^{\prime}(c)$. More precisely, we have
$$f(x) = f(c) + f^{\prime}(c)(x-c) + o(x-c) \mbox{ where } \frac{o(x-c)}{x-c}\stackrel{x\rightarrow c}{\rightarrow}0$$
In order to use that last property of the $o(x-c)$, you may write
$$f(x) = f(c) + f^{\prime}(c)(x-c) + \frac{o(x-c)}{x-c}(x-c) = f(c) + \left( f^{\prime}(c) + \frac{o(x-c)}{x-c} \right)(x-c)$$
Now, looking at $\left( f^{\prime}(c) + \frac{o(x-c)}{x-c} \right)$ you see that for any $\varepsilon < f^{\prime}(c)$ there is a $\delta_{\varepsilon}$ with $|\frac{o(x-c)}{x-c}| \leq \varepsilon$ for $x \in (c-\delta_{\varepsilon}, c+\delta_{\varepsilon})$ which gives you
$$f^{\prime}(c) + \frac{o(x-c)}{x-c} > 0 \mbox{ for } x \in (c-\delta_{\varepsilon}, c+\delta_{\varepsilon})$$
From this your claim follows immediately.
A: Perhaps proof by contradiction is easier for you. Suppose there is no $\delta >0$ such that $f(x)>f(c)$ for all $x \in [c,c+\delta)$. Then taking $\delta =1/n$ we see that there is some point $x_n \in [c,c+1/n)$ with $f(x_n) \leq f(c)$ Now $f'(c) =lim \frac {f(x_n) -f(c)} {x_n-c} \leq 0$ since the numerator is $\leq 0$ for all n. This is a contradiction to the hypothesis. A similar argument can be used to show that $f(x) <f(c)$ for all $x \in (c-\delta, c]$ if $\delta$ is sufficiently small.
