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

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    Re: MATH1131 help thread

    Quote Originally Posted by leehuan View Post
    Not sure if troll on spelling
    Drongoski rarely does sarcasm, but it is clear here.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

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    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?
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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    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
    B Arts / B Science (Advanced Mathematics), UNSW

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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    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).
    Last edited by InteGrand; 27 Mar 2016 at 6:55 PM. Reason: Typo

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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    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.

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    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
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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    Can someone do the working out for this inequality

    x >= 6/(x-1)

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











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    Re: MATH1131 help thread

    Quote Originally Posted by InteGrand View Post










    Thanks so much. So we cannot use the 'multiply by (x-1)^2' method?
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    Re: MATH1131 help thread






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    Re: MATH1131 help thread

    Quote Originally Posted by InteGrand View Post
    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

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    Re: MATH1131 help thread

    sweet!
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    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 ^^^?
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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    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 ^^^?



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    Ancient Orator leehuan's Avatar
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    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

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    Re: MATH1131 help thread

    Quote Originally Posted by leehuan View Post
    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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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    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.

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    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.





    Last edited by leehuan; 2 Apr 2016 at 3:58 PM.

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    Re: MATH1131 help thread

    Quote Originally Posted by InteGrand View Post
    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
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    Re: MATH1131 help thread

    Quote Originally Posted by leehuan View Post
    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
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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    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.
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    Re: MATH1131 help thread

    How do I show that f is continuous at 0... f(x) = |x|
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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    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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    Re: MATH1131 help thread

    Quote Originally Posted by InteGrand View Post
    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].
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    Re: MATH1131 help thread

    Quote Originally Posted by Flop21 View Post
    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.

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    Re: MATH1131 help thread

    Quote Originally Posted by InteGrand View Post
    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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