Plotting a Function with Zeros in the Denominator

Question

Suppose you want to plot

 

f(x) = (x² - 1) / (x - 1)

 

The denominator is zero at x = 1.  However, if you factor (x² - 1), f(x) simplifies to (x - 1)(x + 1) / (x - 1) = x + 1.  Is it appropriate to just let f(x) = x+ 1 and conclude it is defined everywhere?  Why or why not?  If not, what is wrong with canceling the factor of (x - 1) in the numerator and denominator?

 

 

Answer 1

I don't think anything is wrong with what you're thinking.  You're saying that you've been asked to plot the f(x)= (x^2-1)/(x-1).  It doesn't ask you to find the domain and I do believe that even if you were taking the domain into consideration, it wouldn't be zero.  

 

As you've said the (x-1) in the numerator and the (x-1) in the denominator "cancel" each other.  This leaves you with (x+1)/(1), which is just the graph of (x+1), so I'd say yes, you can conclude that it is defined everywhere!  I think before you graph the equation, you should simplify the equation first.  

 

Let me know your thoughts.

Answer 2

Because the (x - 1) cancel each other out, you have no vertical asymptote at x=1.  However, you have a discontinuity at x=1.  So you draw the line   f(x) = x + 1   with an open circle at the point (1, 2).  This is what we call a hole at the value.