Let f(x) = (x^3-1)/(x-1) if x does not equal 1.?

Question

Is it possible to define f(1) in such a way that f is continous throughout the x axis? please show work

Answer 1

x^3 - 1 = (x - 1)(x^2 + x + 1)
So, using x^2 + x + 1 instead of f(x) gives a function continuous at x = 1, and its value is 3.
f(1) = 3.

Answer 2

Do the division, dividing the numerator by the denominator and this shows

F(x) = x^2 +x +1. Evaluate this at x= 1, F(1) = 3

Answer 3

no, that function is undifined. because it would end up working out to be y=(1^3-1)/(1-1) so, y=(1-1)/(1-1) so, y=0/0. anything divided by zero is undifined. the graph of that would be a vertical line

Answer 4

Sure, it can be defined...
f(x) = {0 for x=1, (x^3-1)/(x-1) for x<>1}

Not a very gratifying answer, but an answer nonetheless.