Let $f(x)$ be defined on an interval $I$ (not assumed compact), and assume that its secants have bounded slope, i.e., for any two distinct points on the graph of $f(x)$ over $I,$ the slope $\lambda$ of the line joining them is bounded: $|\lambda| < K,$ where $K$ is some fixed number not depending on the two points selected.
$(a)$ Prove $f(x)$ is uniformly continuous on $I.$
$(b)$ Does $\sqrt{x}$ on $[0,1]$ satisye the above hypothesis on $f(x)?$ Is it uniformly continuous on the interval?
2. For the first one we have that for all $x,y \in I$ with $x \neq y$ $\frac{ \vert f(x)-f(y) \vert }{ \vert x-y \vert } \leq K$ so $\vert f(x) - f(y) \vert \leq K \vert x-y \vert$ and so $f$ is Lipschitz (taking $\delta = \epsilon /K$ it follows that $f$ is uniformly continous).
For the second one take $x=0$ then $\frac{ \sqrt{y} }{y} = \frac{1}{ \sqrt{y} }$ and it follows that the slope of a line through $0$ and $f(y)=\sqrt{y}$ can be made arbitrarily large taking $y$ close to $0$ so $\sqrt{x}$ doesn't satisfy the property in $[0,1]$, although it is unif. cont. in $[0,1]$ since it's continous on a compact interval.