f(x)=(x)/(x^2 - x) find the x values at which f is not continuous. which of the discontinuities are removable. ...what exactly do they mean by that?

Question

f(x)=(x)/(x^2  - x) find the x values at which f is not continuous. which of the discontinuities are removable. ...what exactly do they mean by that?

amistre64 · Answer

|dw:1314933040112:dw|

anonymous · Answer

Do you have a graphing calculator? If so you can graph it and figure out where their is a break in the graph and select that value for where it isn't continuous

anonymous · Answer

There are functions that can be removable. |dw:1314932996154:dw| Like this, this can be removable by simply defining the function when x=0

amistre64 · Answer

a removable discont is one that is produced by a bad zero in the denominator that can cancel with the same number up top

amistre64 · Answer

(x^2+5x)/x  would have a "hole" at x=0 since it factors out in the end

anonymous · Answer

|dw:1314933069298:dw|
This is not removable because there is no way you can "fix" it by defining it more.

amistre64 · Answer

$$\frac{x}{(x^2  - x)}$$
$$\frac{x}{x(x -1)}$$

you see the bottom factors out a like term with the top; it make a hole at x=0

amistre64 · Answer

but at x=1, there is nothing to control the flow, and it goes haywire into infinity

anonymous · Answer

so the nonremovable discontinuity would be the vertical asymptope in this case? and what exactly do you mean by fixing it to make it removable?

anonymous · Answer

Oops, when I said, "There are functions that can be removable", I meant discontinuities.

amistre64 · Answer

a vertical asymptote is like running next to a moving train; you get close to it, but can never get across it

amistre64 · Answer

a "hole" is created by a spot in the graph where it just looks to be missing a part, a piece that might have fallen out of the box and your trying to finish the jigsaw puzzle

amistre64 · Answer

they are both created when the denominator goes zero; but in the case of a hole; you have a control in the top part that cancels it out evenly

anonymous · Answer

lets say there are two open circles at x=2, would those two y's be removable? just like the abovve drawing bydenebel

amistre64 · Answer

is this statement true?
$$\frac{x}{x(x-1)}=\frac{1}{(x-1)}$$

anonymous · Answer

and yes that statement is true

amistre64 · Answer

if there are open circles that just look to be missing a piece, then yes; they are removable