
uniform continous
I have the function f: AR such that f is uniformly continuous. abs(f(x)) greater than or = to k greater than 0 for all x in A. How would I show that (1/f) is also uniformly continuous on A?
I also need to find an example where the conclusion fails for a function which only satisfies abs(g(x)) greater than 0 for all x in A. Would the function
g(x)=(x1)/(x+1) where 1/g is (x+1)/(x1) be an example since it would not be continuous at 1?

Here is a brief outline of what must be done.
$\displaystyle \left {f(x)} \right \ge k > 0\quad \Rightarrow \quad \frac{1}{{\left {f(x)} \right}} \le \frac{1}{k}.$
$\displaystyle \left {\frac{1}{{f(x)}}  \frac{1}{{f(y)}}} \right = \frac{{\left {f(y)  f(x)} \right}}{{\left {f(x)} \right\left {f(y)} \right}} \le \frac{{\left {f(y)  f(x)} \right}}{{k^2 }}.$
Now you have control over $\displaystyle {\left {f(y)  f(x)} \right}.$

Thank you. I'm just having trouble finding a suitable delta.

What epsilon and delta would work here?
Does delta/2 hold to make the function continuous?

By uniform continuity of f
$\displaystyle \left[ {\varepsilon > 0} \right]\left( {\exists \delta :\left {x  y} \right < \delta \Rightarrow \left {f(x)  f(y)} \right < \frac{{\varepsilon k^2 }}{2}} \right).$