How do I show that f is strictly decreasing on (0, infinity)? I have been asked to define $f: (0, \infty) \to (0, \infty)$ by $f(x) = \frac 1 x$
a) How do I show that f is strictly decreasing on $(0, \infty)$? I realize that I have to show that $f'(x)<0$, but I'm not entirely sure how to go about this. Would anyone be able to help or point me in the right direction?
b) How do I show that $f$ is invertible, and find $f^{-1}$? Do I switch $x$ and $y$ and solve for $y$?

a)$$f'(x)=-\frac{1}{x^2}<0 \ \ \ \forall x \in (0, \infty)$$
Since the first derivative is negative at the whole interval $(0,\infty)$ the function $f$ is strictly decreasing on this interval.
b) You have to show that the function $f$ is injective.
Then to find $f^{-1}$, you have set $f(x)=y$ then you have to switch $x$ and $y$ and solve for $x$.
