How do you do this?
"Sketch the set of points (x,y) which satisfy the following relation: 0 <= y <= 2x and 0 <= x <= 2"
Don't get it because there's both x and y in the one thing?
How do you do this?
"Sketch the set of points (x,y) which satisfy the following relation: 0 <= y <= 2x and 0 <= x <= 2"
Don't get it because there's both x and y in the one thing?
Can someone do the working out for this inequality
x >= 6/(x-1)
seems simple but I'm not getting the right answer thanks
sweet!
Can someone help me with ranges? I understand domain fine, but always have trouble with the range.
For example, 1/[sqrt(3-x)].
What is the range of that? The worked answers I found online say y>0 but why is that, isn't it a hyperbola that has part of the graph above and below y=0??
When I put it into wolfram it told me the bottom of the graph was "imaginary" what does that mean.
I suppose the only way to do these questions is to graph the question - so I just gotta remember 1/x is hyperbola and all that jazz, but again the issue is why is it y>0 ^^^?
You probably haven't done complex numbers yet Flop. It's after vectorial geometry for your course.
Just remember that with real numbers you can never square root a negative
Yeah I understand that rule, thanks.
So I'm just not understanding why the sqrt on the bottom is making y have to be positive. I get that it involves the x but how does it involve the y?
A way of doing this, do we rearrange the function so that x is the subject, and from that we can easily see that y cannot be negative (and is this a good method for solving other functions like this?)?
Because of the square root, the denominator is by definition positive (it can't be 0 since we can't have 0 on a denominator; and remember, the symbol √(t) refers to the positive positive square root of t, where t is a positive real number).
Since y is equal to 1/(something positive), y itself is positive. In other words, 1/t > 0 whenever t > 0.
sqrt(1) = 1
sqrt(2) = sqrt(2)
sqrt(3) = sqrt(3)
sqrt(4) = 2
...
I just picked some positive integers.
Effectively, you cannot use the square root to force out a negative y value.
Last edited by leehuan; 2 Apr 2016 at 3:58 PM.
Sorry can someone further explain this.
The part I don't understand is actually getting the points to draw, I understand enough to figure out what quadrant etc though.
So you say sketching y<=2x, but how do you do that?
E.g. I have another similar question: sketch the set of points in the (x,y) plane satisfying 0<x<3y and 0<y<2
To sketch y ≤ 2x, first sketch y = 2x. This is the boundary of the region. Then the places where y is less than (or equal to) 2x will be the places below (or on) this line. If the restriction is 0 ≤ x ≤ 2, then we only take the part where x is between 0 and 2.
To sketch the set of points in the (x,y) plane satisfying 0<x<3y and 0<y<2, first we note 0 < y < 2. Therefore, the region must be in the horizontal band strictly between the lines y = 0 and y = 2, so you can draw these lines (in dashed form), as our region will be between these.
Then sketch 0 < x < 3y, taking only the part in the horizontal band. To sketch 0 < x < 3y, first sketch x = 3y (this is the line y = x/3). Its intersection with the dashed line y = 2 is at the point (6, 2). Then, since we want only the places where x > 0 and x < 3y, take the part to the left of this line, but to the right of the y-axis (as x > 0).
So the final region is the interior of the triangle formed by the points (0,0), (0,2), (6,2), and excluding the boundaries.
How do I show that f is continuous at 0... f(x) = |x|
Note f(0) = 0, so we need to show that f(x) -> f(0) = 0 as x -> 0. So we need to show that given eps > 0, there exists a delta > 0 such that if 0 < |x-0| < delta (i.e. 0 < |x| < delta), then |f(x) - 0| < eps.
To do this, let delta = eps, for any given eps > 0. Then suppose 0 < |x-0| < delta.
Then we have |x| < delta. Since |x| = ||x|| = ||x| - 0|, we have ||x| - 0| < delta, i.e. |f(x) - 0| < delta. But delta = eps, so |f(x) - 0| < eps, as we wanted.
Hence f is continuous at 0.
Last edited by InteGrand; 3 Apr 2016 at 9:10 PM.
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