# 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.

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?

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

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).

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?

$The part pertaining to y\ge0, \, x\ge 0 tells me to use the first quadrant.\\ Then more specifically, the domain of the function is 0\le x \le 2. \\ Lastly, it's just a matter of sketching y \le 2x over these boundaries$

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

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
$\noindent We have x \geq \frac{6}{x-1} (note x\neq 1, since then we have a zero denominator). We'll do it in cases.$

$\underline{Case 1. x >1 }$

$\noindent Here, x-1>0, so when we multiply through the inequation by x-1, we don't change the sign. On doing so, we have x(x-1) \geq 6 \Longleftrightarrow x^2 - x-6 \geq 0 \Longleftrightarrow (x-3)(x+2) \geq 0 (and x>1 in this case). Sketching the parabola, the solution will be x \geq 3.$

$\underline{Case 2. x<1 }$

$\noindent Here, x-1<0, so when we multiply through the inequation by x-1, we \emph{do} change the sign. On doing so, we have x(1-x) \leq 6 \Longleftrightarrow x-x^2 \leq 6 \Longleftrightarrow x^2-x+6 \leq 0\Longleftrightarrow (x+2)(x-3) \leq 0 (and x<1 in this case). Sketching the parabola, the solution will require -2 \leq x \leq 3 and x<1. So -2 \leq x < 1 in this case.$

$\noindent So overall, the solutions are x\geq 3 or -2 \leq x < 1.$

8. ## Re: MATH1131 help thread

Originally Posted by InteGrand
$\noindent We have x \geq \frac{6}{x-1} (note x\neq 1, since then we have a zero denominator). We'll do it in cases.$

$\underline{Case 1. x >1 }$

$\noindent Here, x-1>0, so when we multiply through the inequation by x-1, we don't change the sign. On doing so, we have x(x-1) \geq 6 \Longleftrightarrow x^2 - x-6 \geq 0 \Longleftrightarrow (x-3)(x+2) \geq 0 (and x>1 in this case). Sketching the parabola, the solution will be x \geq 3.$

$\underline{Case 2. x<1 }$

$\noindent Here, x-1<0, so when we multiply through the inequation by x-1, we \emph{do} change the sign. On doing so, we have x(1-x) \leq 6 \Longleftrightarrow x-x^2 \leq 6 \Longleftrightarrow x^2-x+6 \leq 0\Longleftrightarrow (x+2)(x-3) \leq 0 (and x<1 in this case). Sketching the parabola, the solution will require -2 \leq x \leq 3 and x<1. So -2 \leq x < 1 in this case.$

$\noindent So overall, the solutions are x\geq 3 or -2 \leq x < 1.$
Thanks so much. So we cannot use the 'multiply by (x-1)^2' method?

9. ## Re: MATH1131 help thread

$Yes we can and it will still work$

\\ \begin{align*}x&\ge \frac { 6 }{ x-1 } \\ x\left( x-1 \right) \left( x-1 \right)& \ge 6\left( x-1 \right) \\ \left( { x }^{ 2 }-x \right) \left( x-1 \right) -6\left( x-1 \right) &\ge 0\\ \left( { x }^{ 2 }-x-6 \right) \left( x-1 \right)& \ge 0\\ \left( x+2 \right) \left( x-1 \right) \left( x-3 \right)& \ge 0 \end{align*}

$Keeping in mind that x\neq 1\\ -2\le x < 1, x\ge 3$

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

sweet!

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 ^^^?

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 ^^^?
$\noindent The hyperbola thing you're thinking of is y=\frac{1}{3-x}. But this one is y= \frac{1}{\sqrt{3-x}}, which is different. Because of the square root in the denominator, we must always have y>0. Furthermore, it is easy to show that for any positive y_0, there exists a real x_0 such that y_0 = \frac{1}{\sqrt{3-x_0}}, namely x_0 = 3 - \frac{1}{y_0 ^2}. Hence the range is \mathbb{R}^+, the set of all positive reals.$

$\noindent The imaginary'' thing in the WolframAlpha plot is referring to the fact that for certain real values of x (namely x>3), the function will take on imaginary values. An imaginary number is one of the form yi, where y\neq 0 and y\in \mathbb{R}, and i is the square root of -1''.$

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

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?)?

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.

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.

$Note that \sqrt{1}\neq\pm1$

$Rather, -\sqrt{1}=-1 and \sqrt{1}=1 so as we have no negative sign in FRONT of y=\frac{1}{\sqrt{3-x}} it cannot ever be negative$

$The equation would have to be y=-\frac{1}{\sqrt{3-x}} to bring out a negative branch, but that gets rid of the positive branch$

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

19. ## Re: MATH1131 help thread

Originally Posted by leehuan

$The part pertaining to y\ge0, \, x\ge 0 tells me to use the first quadrant.\\ Then more specifically, the domain of the function is 0\le x \le 2. \\ Lastly, it's just a matter of sketching y \le 2x over these boundaries$
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

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.

21. ## Re: MATH1131 help thread

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

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.

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].

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.

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?

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