Graphs Help (1 Viewer)

[king.rafa]

Hey was just wondering what is the differene between y=x and y=x^2/x. Like if you mulitply a function by say "x/x" is it the same or does the new function have a discontinuity. Like if you change y=x to y=x^2/x, then does the domain and range remain but at the discontinuity x=1, you evaluate in this old function y=x?

Thanks

 

[vds700]

Hey was just wondering what is the differene between y=x and y=x^2/x. Like if you mulitply a function by say "x/x" is it the same or does the new function have a discontinuity. Like if you change y=x to y=x^2/x, then does the domain and range remain but at the discontinuity x=1, you evaluate in this old function y=x?

Thanks

no difference as x/x = 1, so you are basically multiplying the function by 1.

 

[king.rafa]

no difference as x/x = 1, so you are basically multiplying the function by 1.

but by multiplying in x/x arent u introducing a discontinuity?

 

[vds700]

but by multiplying in x/x arent u introducing a discontinuity?

No I dont think so because you are not really changing the function at all, just multiplying it by 1,

x^2/x = x because you can cancel one x from the top and bottom.

 

[lyounamu]

Hey was just wondering what is the differene between y=x and y=x^2/x. Like if you mulitply a function by say "x/x" is it the same or does the new function have a discontinuity. Like if you change y=x to y=x^2/x, then does the domain and range remain but at the discontinuity x=1, you evaluate in this old function y=x?

Thanks

That is basically same as y = x but the graph does not touch x = 0.

 

[vds700]

That is basically same as y = x but the graph does not touch x = 0.

Actually, x can be 0 as 0/0 = 0. So its defined. Have a look at the graph of x^2/x.

 

[tommykins]

Actually, x can be 0 as 0/0 = 0. So its defined. Have a look at the graph of x^2/x.

That graph assumes the x's cancel out, hence why you're able to place a point at (0,0). If we did not simplify x^2/x, there would be a hole at (0,0).

I'd say the graph should be touching (0,0) anyways.

 

[tunghungvuong]

:grinshak: . It is interesting that you have asked that question !
the function y=x is obviously a continuous function. The domain and range takes all values of x and y. However, the function y=x^2/x is discontinous at x=0. That is, lim f(x) = 0 as x approaches to zero. The point (0,0) is discontinous and should be shown by an openned circle rather than a closed-circle point. Is that answer your question ????:rofl:

 

[wendus]

Actually, x can be 0 as 0/0 = 0. So its defined.

0/0 is undefined! anything on zero is undefined. if you multiply a graph by x/x and it becomes x^2/x theoretically it wouldn't touch zero. i think.

you can cancel out the x's but that'd give you a different graph.

 

[vds700]

0/0 is undefined! anything on zero is undefined. if you multiply a graph by x/x and it becomes x^2/x theoretically it wouldn't touch zero. i think.

you can cancel out the x's but that'd give you a different graph.

oops my mistake, i kinda based that on the graph. I also thought that if 0/0 = 0, then as 0 x 0 = 0(I think ?), then 0/0 = 0. Just like if 6/3 = 2, then 2 x 3 = 6. That was my thinking but obviously its wrong- sorry.

 

[lyounamu]

That graph assumes the x's cancel out, hence why you're able to place a point at (0,0). If we did not simplify x^2/x, there would be a hole at (0,0).

I'd say the graph should be touching (0,0) anyways.

What I meant by "not touching" is that there might be a some kind of un-coloured dot on the origin. Anyway, is my answer right or what?

 

[tommykins]

What I meant by "not touching" is that there might be a some kind of un-coloured dot on the origin. Anyway, is my answer right or what?

If you take x²/x to JUST be x²/x, the graph would look like y = x, but at the origin there will be an open circle.

You're right.

 

[Tully B.]

Wow, it's cool to see what year 12 MX1 is like... I'm enjoying Y11, and this seems pretty interesting too.
Good luck with HSC