Show that $f$ is uniformly continuous given that $f'$ is bounded. Let $D\subset \mathbb{R}$ be an interval and $f:D\rightarrow \mathbb{R}$ a continuously differentiable function such that its derivative $f':D\rightarrow \mathbb{R}$ is bounded. Show that $f$ is uniformly continuous.
I will admit before I begin; I am stabbing in the dark.
If we restrict the domain of $f$ to some interval $[-X,X] \subset D$, then $f:[-X,X]\rightarrow \mathbb{R}$ is uniformly continuous. 
Now since we know that $f'$ is bounded we can say that for some arbitrary $y \in D$, $|f'(x)| = \lim_{x \rightarrow y}\frac{|f(x) - f(y)|}{|x-y|} \leq K   ( \forall x \in D:x \neq y) $ for some $K >0 $
This is not going to sounds mathematical, but I need to somehow get rid of the limit so I can proceed with something like $\frac{|f(x) - f(y)|}{|x-y|} \leq K$
 A: Let $D$ be the interval on which $f$ is defined, and $f'$ the derivative, assumed bounded on $D$ (left/right derivatives at endpoints). So there is $K>0$ for which $|f'(t)|<K$ for all $t \in D.$ Now suppose $x<y$ are two points in $D.$ By the mean value theorem, $(f(y)-f(x))/(y-x)=f'(c)$ for some $c$ with $x<c<y.$ This gives what you want since here $|f'(c)|<K.$
Note here the $c$ at which $f'$ gets evaluated is not an endpoint of $D$ since it is strictly between two elements of $D$ [Maybe that doesn't matter, but its the boundedness of the derivative at the collection of interior points of $D$ which is needed in the above argument.]
A: Another more abstract argument is the following one: a differentiable function with bounded derivative is Lipschitz continuous with Lipschitz constant given by the derivative's bound. And a Lipschitz continuous function is easily seen to be uniformly continuous.
A: Let $f:[-X,X]\rightarrow \mathbb{R}$ has the conditions that you want. Let $\epsilon>0$, then we want to find $\delta>0$ such that:
$$\forall x,y\in [-X,X](|x-y|<\delta \rightarrow |f(x)-f(y)|<\epsilon)$$
By Mean value theorem for every $x,y\in [-X,X],x<y$, there exists $c\in (-X,X)$ such that
$$\frac{f(y)-f(x)}{y-x}=f'(c)$$ 
Because for every $x\in (-X,X)$, we have $|f'(x)|<K$, therefore for every $x,y\in [-X,X],x<y$
$$\frac{|f(y)-f(x)|}{|y-x|}<K$$
so we choose $\delta = \frac{\epsilon}{K}$
