For what value of x is the function g(x)=x/1-x not defined?

The function is definately defined at x=0 but not x=1.

But its just one part of the coordinate (x,y).

If the value of y or f(x) is considered, you'll see that it is never possible to attain f(x)=0. In other terms (x,y)= (0,0) is not a defined point in the graph of the function because the graph doesnt pass through that point.

Now I hope you understood what I meant!

Conclusion- The above function is not defined at all points in the space having the abscissa or x=1 in the coordinate and also at ordinate or y=0 in the coordinate.nation:

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: