# Thread: First Year Mathematics A (Differentiation & Linear Algebra)

1. ## Re: MATH1131 help thread Originally Posted by leehuan Not sure if troll on spelling
Drongoski rarely does sarcasm, but it is clear here.  Reply With Quote

2. ## Re: MATH1131 help thread

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?  Reply With Quote

3. ## Re: MATH1131 help thread Originally Posted by Flop21 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?
test some points and values of x and y, it's a perfectly cromulent inequality  Reply With Quote

4. ## Re: MATH1131 help thread Originally Posted by Flop21 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?
It's the interior and edges of the triangle with vertices: (0,0), (2,0), (2,4).  Reply With Quote

5. ## Re: MATH1131 help thread Originally Posted by Flop21 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?
Additional input.  Reply With Quote

6. ## Re: MATH1131 help thread

Can someone do the working out for this inequality

x >= 6/(x-1)

seems simple but I'm not getting the right answer thanks  Reply With Quote

7. ## Re: MATH1131 help thread Originally Posted by Flop21 Can someone do the working out for this inequality

x >= 6/(x-1)

seems simple but I'm not getting the right answer thanks  Reply With Quote

8. ## Re: MATH1131 help thread Originally Posted by InteGrand Thanks so much. So we cannot use the 'multiply by (x-1)^2' method?  Reply With Quote

9. ## Re: MATH1131 help thread  Reply With Quote

10. ## Re: MATH1131 help thread Originally Posted by InteGrand You can try that too. I just didn't want to deal with a cubic since in general it may not be easy to factorise.
That's why you don't expand the cube out and just factor out (x-1) unexpanded  Reply With Quote

11. ## Re: MATH1131 help thread

sweet!  Reply With Quote

12. ## Re: MATH1131 help thread

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 ^^^?  Reply With Quote

13. ## Re: MATH1131 help thread Originally Posted by Flop21 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 ^^^?  Reply With Quote

14. ## Re: MATH1131 help thread

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  Reply With Quote

15. ## Re: MATH1131 help thread Originally Posted by leehuan 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?)?  Reply With Quote

16. ## Re: MATH1131 help thread Originally Posted by Flop21 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.  Reply With Quote

17. ## Re: MATH1131 help thread

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.  Reply With Quote

18. ## Re: MATH1131 help thread Originally Posted by InteGrand 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.
Ohhh haha right, I understand.

Thanks all  Reply With Quote

19. ## Re: MATH1131 help thread Originally Posted by leehuan Additional input.

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  Reply With Quote

20. ## Re: MATH1131 help thread Originally Posted by Flop21 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.  Reply With Quote

21. ## Re: MATH1131 help thread

How do I show that f is continuous at 0... f(x) = |x|  Reply With Quote

22. ## Re: MATH1131 help thread Originally Posted by Flop21 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.  Reply With Quote

23. ## Re: MATH1131 help thread Originally Posted by InteGrand 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.
Is that using the formula proof of a limit? [fml I was hoping that was just one small part of the topic and have skipped over it].  Reply With Quote

24. ## Re: MATH1131 help thread Originally Posted by Flop21 Is that using the formula proof of a limit? [fml I was hoping that was just one small part of the topic and have skipped over it].
I did it (proved the required limit) via the epsilon-delta definition of a limit.  Reply With Quote

25. ## Re: MATH1131 help thread Originally Posted by InteGrand 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.
So knowing the definition of a continuous function, can you just say since the its lim = 0 from both the left and right hand side, it is continuous since as x->a f(x) = f(a).

Would writing just that be acceptable as an answer instead of using the eps-delta method?  Reply With Quote

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