# ambiguity?

• June 4th 2009, 04:15 AM
the kopite
ambiguity?
in the function 1/ (1-x), the obvious value to remove from the domain of f is 1. this will be shown on the graph by the function being translated one unit to the positive direction of the x-axis. is this contradictory to what we are all taught that 'in the case of f(x+c) and c>0 then the graph will be of f(x) translated c units to the NEGATIVE direction of the x-axis'. ?????
would this be due to the negative nature of x in f(x)?
• June 4th 2009, 05:32 AM
HallsofIvy
Quote:

Originally Posted by the kopite
in the function 1/ (1-x), the obvious value to remove from the domain of f is 1. this will be shown on the graph by the function being translated one unit to the positive direction of the x-axis.

Surely it is simpler to jus note that if x= 1, the denominator is 0 so the fraction is not defined.

Quote:

is this contradictory to what we are all taught that 'in the case of f(x+c) and c>0 then the graph will be of f(x) translated c units to the NEGATIVE direction of the x-axis'. ?????
would this be due to the negative nature of x in f(x)?
$\frac{1}{1- x}= \frac{1}{-(x-1)}= \frac{-1}{x-1}= \frac{-1}{x-(-1)}$. It does NOT contradict the statement you give because the statement includes the condition "c> 0". Here c is -1 and is not greater than 0.
• June 4th 2009, 09:40 PM
the kopite
so x must be positive for the statement to apply? is that it?