f(x) = (x^2 - 1)/(x+1)
= (x+1)(x-1)/(x+1)
= x-1 , x ≠ -1
so the graph will actually be a straight line
y = x-1 , with a hole at (-1, -2)
the graph will be continuous for all values of x, except x = -1
f(x)= ⎨x^2-1 / x+1, for x not = -1
⎨8, for x=-1
I'm trying to graph and determine the values of x for which the function is continuous.
Does the graph looks like a u shape opening up or a line.
3 answers
So the function is defined or not defined.
The question wasn't about a function being defined or not.
You wanted to know if it was continuous.
As I said, it is defined and continuous for all values of x , except x = -1
for all values of x, x ≠-1 , you have the function
f(x) = x-1, which is a straight line
the point (-1,-2) is a hole in the straight line and it is being replaced with (-1,8) , a single point lying above the line.
Strange function.
You wanted to know if it was continuous.
As I said, it is defined and continuous for all values of x , except x = -1
for all values of x, x ≠-1 , you have the function
f(x) = x-1, which is a straight line
the point (-1,-2) is a hole in the straight line and it is being replaced with (-1,8) , a single point lying above the line.
Strange function.