In the expression (x-1) why is f(1/3) undefined?

Question

anonymous · Answer

Is that the entire question?

anonymous · Answer

No, (x-1) is what i got after i simplified the expression. and then it said to explain why f(1) = 0, f(0) = -1, and f(-1) = -2, yet f(1/3) is undefined.

anonymous · Answer

what was your expression?

anonymous · Answer

anyhoo here are 2 cases where you might have undefined

1/0
sqrt(x)  where x<0

ooh and a third one but i doubt it is the case here

0^0

anonymous · Answer

okkk, i think i get it

anonymous · Answer

so it could be

f(1/3)=1/((1/3)-(1/3))

jdoe0001 · Answer

hmm, what's the original expression?

anonymous · Answer

3x^2-4x+1  /  3x-1

anonymous · Answer

i think you got to where the denominator dissapears but it is still a hole there where  3x-1=0   

sovling for x you'd get 1/3         

This will come up again   when you work with limits. Which are the foundation of calculus.

anonymous · Answer

|dw:1377294951110:dw|

here is graph of x-1

anonymous · Answer

(3x^2-4x+1  )/(  3x-1)
Is the same however
However 

|dw:1377295036238:dw|

at 1/3 there will be a hole :)