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: To show it is decreasing you need to show that if $x < y$ then ${1 \over x} > {1 \over y}$. 
But writing the expression ${1 \over x} > {1 \over y}$ with a common denominator yields ${y \over xy} > {x \over xy}$, which follows from the fact that $x < y$ just by dividing through by $xy$.
Part b) is done by first showing $f(x)$ is one-to-one, and then do what you are describing.
A: 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 $x$.
A: b) $f$ is a strictly decreasing continuous function on $(0,\infty)$, so it is surely invertible. The image is $(0,\infty)$ as given. Yes, you can swith $x$ and $y$ to find $f^{-1}$. To find the domain and image of $f^{-1}$, you can as well swith the domain and range of $f$. 
A: An elementary way of showing it
a)
Function is strictly decreasing if for every $f(x_1)<f(x_2) \implies x_1>x_2$$$f(x_1)< f(x_2)\\ \frac{1}{x_1} < \frac{1}{x_2}\\x_1>x_2$$
Last line because both $x_1,x_2>0$
b) $f$ is invertible if it's a bijection,prove it's 1 on 1 $$f(x_1)=f(x_2)\implies x_1=x_2\\\ \frac{1}{x_1}=\frac{1}{x_2}\\x_2=x_1$$
So $f$ is 1-1,now show that $f$ is onto
$$f(x)=y\\\ \frac{1}{x}=y\\ \frac{1}{y}=x$$
Which shows that for every $y$ there is a corresponding $x$ so function is onto now to find inverse
$$f^{-1}(f(x))=x\\f^{-1}(\frac{1}{x})=x\\ \frac{1}{x}=t\\x=\frac{1}{t}\\f^{-1}(t)=\frac{1}{t}\\f^{-1}(x)=\frac{1}{x}$$
