# 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 $y$?

• $f'(x) = - \frac 1 {x^2}$ now you must show that this is negative for all $x > 0$. So $-\frac 1 {x^2} <0 \iff \frac 1 {x^2} > 0 \iff \ldots$ Commented Aug 5, 2014 at 13:01
• You can use the rule with $f'(x)$ because $f$ has a continuous derivative ($C'$-function) on $(0,\infty)$. Commented Aug 5, 2014 at 13:06
• Start with $x^2\gt0$, then take the inverse, then multiply both sides with $-1$ and you get $f^\prime(x)\lt0$. Commented Aug 5, 2014 at 13:07

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 $f$.

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 $$f^{-1}(f(x))=x\\f^{-1}(\frac{1}{x})=x\\ \frac{1}{x}=t\\x=\frac{1}{t}\\f^{-1}(t)=\frac{1}{t}\\f^{-1}(x)=\frac{1}{x}$$