# 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 $y$?

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$f'(x) = - \frac 1 {x^2}$ now you must show that this is negative for all $x > 0$. So $-\frac 1 {x^2} <0 \iff \frac 1 {x^2} > 0 \iff \ldots$ –  flawr Aug 5 '14 at 13:01
You can use the rule with $f'(x)$ because $f$ has a continuous derivative ($C'$-function) on $(0,\infty)$. –  Steven Van Geluwe Aug 5 '14 at 13:06
Start with $x^2\gt0$, then take the inverse, then multiply both sides with $-1$ and you get $f^\prime(x)\lt0$. –  Hakim Aug 5 '14 at 13:07

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 $$f^{-1}(f(x))=x\\f^{-1}(\frac{1}{x})=x\\ \frac{1}{x}=t\\x=\frac{1}{t}\\f^{-1}(t)=\frac{1}{t}\\f^{-1}(x)=\frac{1}{x}$$ - 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)$$f'(x)=-\frac{1}{x^2}<0 \ \ \ \forall x \in (0, \infty)$$ 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$x$. - 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\$.