## A Tricky Domain Problem and Wolfram Alpha

What is the domain of the function $f(x)=\frac{1}{\frac{x-3}{x-4}}$?  The answer is $\{x|x \neq 3 \: and \: x \neq 4\}$ because if $x = 3$ or $x= 4$ a divide by zero situation occurs.  One is sorely tempted to simplify $f(x)$ to $f(x)=\frac{x-4}{x-3}$ which would be incorrect as it stands since now $x\neq 4$ needs to be explicitly stated.  So $f(x)=\frac{x-4}{x-3},x\neq 4$.

Wolfram Alpha handles this function curiously.  If the original functions is input as $f(x)=\frac{1}{\frac{x-3}{x-4}}$,  Alpha simplifies it to  $f(x)=\frac{x-4}{x-3}$ with nary a mention that $x\neq 4$.  This is clearly wrong.  Alpha also shows the graph of the function as if $f(4)$ exists instead of displaying a hole.

If the expression $\frac{1}{\frac{x-3}{x-4}}$ is entered, Alpha still simplifies it to $\frac{x-4}{x-3}$ and shows a function plot without a hole at $(4,0)$. However in a box entitled “Properties as a real function” the domain $\{x\epsilon\Re|x \neq 3 \: and \: x \neq 4\}$ is correct.  I have alluded to problems with Wolfram Alpha in this post and this post.