First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

iforgotmyname · Mar 18, 2016

Re: MATH1131 help thread

Just let 6/(m-2)= eposilon

Flop21 · Mar 18, 2016

Re: MATH1131 help thread

iforgotmyname said:
Just let 6/(m-2)= eposilon

ahhHHHH

that's how you do it lol

Shadowdude · Mar 18, 2016

Re: MATH1131 help thread

Flop21 said:
are they suppose to help with actual questions or just how to use maple?

Either

I'm sure most consultants don't mind helping with actual questions

Drongoski · Mar 18, 2016

Re: MATH1131 help thread

First time I've heard of this Greek letter "eposilon". What does it look like?

leehuan · Mar 18, 2016

Re: MATH1131 help thread

Drongoski said:
First time I've heard of this Greek letter "eposilon". What does it look like?

Not sure if troll on spelling

Paradoxica · Mar 18, 2016

Re: MATH1131 help thread

leehuan said:
Not sure if troll on spelling

Drongoski rarely does sarcasm, but it is clear here.

Flop21 · Mar 27, 2016

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?

Shadowdude · Mar 27, 2016

Re: MATH1131 help thread

Flop21 said:
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

InteGrand · Mar 27, 2016

Re: MATH1131 help thread

Flop21 said:
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).

leehuan · Mar 27, 2016

Re: MATH1131 help thread

Flop21 said:
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.

$$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$

Flop21 · Mar 28, 2016

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

InteGrand · Mar 28, 2016

Re: MATH1131 help thread

Flop21 said:
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 $-2 \leq x < 1$.$

Flop21 · Mar 28, 2016

Re: MATH1131 help thread

InteGrand said:
$\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?

leehuan · Mar 28, 2016

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$

leehuan · Mar 28, 2016

Re: MATH1131 help thread

InteGrand said:
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

Flop21 · Mar 28, 2016

Re: MATH1131 help thread

sweet!

Flop21 · Apr 2, 2016

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 ^^^?

InteGrand · Apr 2, 2016

Re: MATH1131 help thread

Flop21 said:
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 $-1$''.$

leehuan · Apr 2, 2016

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

Flop21 · Apr 2, 2016

Re: MATH1131 help thread

leehuan said:
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?)?

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