I have a certain problem: When you have a function $f(x)$ and it can have some undefined values, the reciprocal function does some things I don't understand. For example:
i) $$f(x)=\tan(x) $$
or another one:
ii)
$$ f(x)=\frac {5}{x-3} $$
Clearly you can see that for ii) $x=3$ will be undefined and for i) $\pi/2, 3\pi/2,\ldots$
When I now work with the reciprocal functions my problem appears :
i) $$(f(x))^{-1}=\frac{1}{\tan x}=\cot x$$
ii) $$(f(x))^{-1}=\frac{1}{\frac {5}{x-3}}$$
Coming up to my questions:
i) The function of $\cot x$ is $0$ at those $x$-values where $\tan x$ is undefined. Why is $1/\text{undefined}=0$ and not undefined too?
ii) The same here; $f(x)$ is undefined at $x=3$, but depending on which graphic calculator you use it defines $$(f(3))^{-1}=0 $$ or as a discontinuity.
*What happens at those "magical" points when $f(x)=\text{undefined}$ and $(f(x))^{-1}= 1/\text{undefined?}$
What I got so far:
"If y = f (x) = 0 for some value of x, then 1/f (x) is undefined.
There is a jump or discontinuity in its graph for this value of x.
This means that, as f (x) gets close to 0, 1/f (x) will become very
large in value. Equally, if there is a jump or discontinuity in the
graph of y = f (x) for some value of x, then y = 1/f (x) = 0 for that
value of x."
That's a definition I have from "Jenny Olive: Math a
student's survival guide"
and of course : Why is cot(x)=0 instead of undefined
and my own thesis:
Since you can convert $$ y= \frac {1}{\frac {5}{x-3}}=1\cdot\frac{x-3}{5}$$ the value for x=3 is defined
Same for $\cot(x) = \cos(x) / \sin(x)$ , where $x= \pi/2$ is defined.
Is my thesis working or am I hurting mathematics at this point?
Best Answer
The reciprocal of $5/(x-3)$ is $(x-3)/5$ except at $x=3$ where it is undefined.
So you look near $x=3$, and you find it approaches $0$ both when $x<3$ and when $x>3$.
That is called a 'removable discontinuity'.
You might as well define $g(x)$ to be $(x-3)/5$ except at $x=3$, and $0$ at $x=3$. But that is just $g(x)=(x-3)/5$ for all $x$.