Results 1 to 12 of 12

Math Help - [solved] Easy Question about usage of notation

  1. #1
    ixo
    ixo is offline
    Junior Member
    Joined
    Feb 2009
    Posts
    42

    How to clearly state my answer?

    Ok i get the hang of how to solve derivatives, apply them, and integrate them. However i'm wondering if i'm using correct notation to state my answers.

    Problem:
    Find all critical points, label minimum or maximum, points of inflection, and graph.
    y= x^2-x-2

    My answer:"EDITED" see below for solution
    Last edited by ixo; February 13th 2009 at 08:15 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327
    Quote Originally Posted by ixo View Post
    Ok i get the hang of how to solve derivatives, apply them, and integrate them. However i'm wondering if i'm using correct notation to state my answers.

    Problem:
    Solve for all critical points, minimums/maximum points, points of inflection, and graph.
    y= x^2-x-2

    My answer:
    A) y= x^2-x-2 \Rightarrow y=(x-2)(x+1)

    \therefore \{x: x^-x-2=x\}= -1,2
    Since you title this "notation", I must confess that I have no idea what this is suppose to mean. I assume that you just dropped the "2" in x^2 but that should be "= 0" not "= x". Also the left side of the equation is a set, the right side is not. You should have "{-1, 2}" on the right side: \{x: x^2- x- 2= 0\}= \{-1, 2\}. Since you were not asked for the intercepts, what does this "answer"?

    [tex]B) y\prime= 2x-1

    \therefore \{x: 2x-1=0\}= \frac{1}{2}
    There is no "therefore" here. Yes, \{x: 2x-1= 0\}= \{\frac{1}{2}\} but that is not true because y'= 2x-1. If you want to say those are the critical numbers, say that.

    \{y:y=(\frac{1}{2})^2-(\frac{1}{2})-2\} = -2.25
    Again, you don't say what this is supposed to answer. If you mean that (1/2, -2.25) is the only critical point say so.

    C) y\prime\prime= 2

    D \therefore y= x^2-x-2 opens up.
    Well, strictly speaking the graph of y= x^2-x-2 opens upward, but that's fairly common "abuse of notation".

    I don't know how to graph on latex but i would put a points at   (-1,0),(\frac{1}{2},-2 \frac{1}{4}),(2,0) and draw an upward opening parabola.
    And, specifically, (\frac{1}{2}, -2\frac{1}{4}) is the vertex of the parabola.

    Am i using them correctly? Any suggestions on better usage?
    Thanks.
    Last edited by HallsofIvy; February 14th 2009 at 05:52 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    ixo
    ixo is offline
    Junior Member
    Joined
    Feb 2009
    Posts
    42
    Does this seem any better? I was counted off on my last test because i just solved and wrote down the answers. In hindsight i do see that the work and answers were not clearly labeled and broken down. So i am trying to avoid that next time.

    My answer:
    f(x)= x^2-x-2
    by factoring f(x)=(x-2)(x+1)
    so  \{x: x^2 - x-2=0\}= -1,2
    y intercepts =  (-1,0),(2,0)

    Finding critical points:
    f\prime(x)= 2x-1
    x value using f\prime(x): \{x: 2x-1=0\}= \frac{1}{2}
    y value using f(x): (\frac{1}{2})^2-(\frac{1}{2})-2\ = -2.25
    so the critical point is located graphically at (.5,-2.25) and is a minimum because -1<.5<2 and -2.25<0


    Solving for inflection:
    y\prime\prime= 2 so there is no inflection.

    Quote Originally Posted by HallsofIvy View Post
    Well, strictly speaking the graph of y= x^2-x-2 opens upward, but that's fairly common "abuse of notation".
    Is this better:


    Summery:
    Graphically f(x)= x^2-x-2 is concave upwards with the vertex at (.5,-2.25) and has no inflection point.

    I don't know how to graph on latex but i would put a points at   (-1,0),(\frac{1}{2},-2 \frac{1}{4}),(2,0) and draw an upward opening parabolic graph.
    Last edited by ixo; February 13th 2009 at 08:12 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by ixo View Post
    f(x)= x^2-x-2
    by factoring f(x)=(x-2)(x+1)
    For best presentation, you may want to word this as a complete sentence, or write it as an implication as you originally had.

    so  \{x: x^2 - x-2=0\}= -1,2
    As HallsofIvy tried to say, this should be \{x : x^2 - x - 2 = 0\} = \{-1, 2\}.

    y intercepts =  (-1,0),(2,0)
    I would make it clear that this follows from the work above; e.g., "So, the y-intercepts are (-1, 0) and (2, 0)\text{.}" Don't use the equals sign like that unless you're writing in an informal situation.

    f\prime(x)= 2x-1
    To get the proper prime mark used for differentiation in \text{\LaTeX}, put the \prime in superscript: f^\prime(x) gives f^\prime(x). However, you can get the same thing with less work by using f'(x) (a single quotation mark): f'(x).

    x value using f\prime(x): \{x: 2x-1=0\}= \frac{1}{2}
    y value using f(x): (\frac{1}{2})^2-(\frac{1}{2})-2\ = -2.25
    Again, for full professionalism, use complete sentences. On a test, something like this should be okay, but I would maybe write it as:

    x\text{-coordinate}:\;f'(x) = 0\Rightarrow 2x - 1 = 0\Rightarrow x = \frac12
    y\text{-coordinate}:\;y = f\!\left(\frac12\right) = \left(\frac12\right)^2 - \frac12 - 2 = -2.25

    If you use the set notation, you should have \{x:2x-1=0\}= \left\{\frac12\right\}, since the right side should also be a set.

    Also, if you are typing your assignments, make sure you use appropriately sized parentheses around things like fractions and other large type. You can ensure grouping symbols are the right size by preceding the left and right symbol by \left and \right, respectively. For example,

    \left(\sum_{i=1}^n i^3\right) gives \left(\sum_{i=1}^n i^3\right), which is much prettier than (\sum_{i=1}^n i^3).

    so the critical point is located graphically at (.5,-2.25) and is a minimum because its y value is less than the y intercepts.
    Yes, but you need to explain that the reason your argument is valid is because this is a quadratic function. Otherwise, you could not conclude that the point was a minimum (or a relative minimum). It would probably be best, and easiest, to use the first or second derivative test.

    Solving for inflection:
    y\prime\prime= 2 so there is no inflection.
    There are no points of inflection. You could also point out the reasoning by writing y''=2\neq0.

    Summery:
    "Summery" is an adjective that describes something that is fit for summer. The word you are looking for is "summary."
    Follow Math Help Forum on Facebook and Google+

  5. #5
    ixo
    ixo is offline
    Junior Member
    Joined
    Feb 2009
    Posts
    42
    Didn't get the {:}={} but do now, didn't think to use ', and yes i misspelled summary thanks.

    How is this:


    y intercept: f(x)= x^2-x-2 =(x-2)(x+1) \Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1,2
    So the y intercepts are  (-1,0),(2,0)

    Critical points:
    f(x)= x^2-x-2 \Rightarrow f^\prime(x)= 2x-1 \Rightarrow 2x-1=0 \Rightarrow x= \frac12

    y value of critical point: \left(\frac12\right)^2-\left(\frac12\right)-2\ = -2.25
    so the critical point is located graphically at (.5,-2.25) and is a minimum because -1<\frac12<2 and -2.25<0

    Inflection points:
    y\prime\prime= 2 \neq 0 so there are no inflection points.


    Summary:
    Graphically f(x)= x^2-x-2 is concave upwards with the vertex at (.5,-2.25) and has no inflection point.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by ixo View Post
    Critical points:
    f(x)= x^2-x-2 \Rightarrow f^\prime(x)= 2x-1 \Rightarrow 2x-1=0 \Rightarrow x= \frac12
    f'(x)=2x-1 does not imply that 2x - 1 = 0\text. You could write

    f(x)= x^2-x-2\Rightarrow f^\prime(x)= 2x-1,\text{ so }f'(x) = 0\Rightarrow 2x-1=0 \Rightarrow x= \frac12

    so the critical point is located graphically at (.5,-2.25) and is a minimum because -1<\frac12<2 and -2.25<0
    You still need to better explain why the critical point is a minimum. Either state that the point is a minimum because it is the vertex of a parabola which opens upward, or use one of the derivative tests.

    Everything else seems okay, more or less.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    ixo
    ixo is offline
    Junior Member
    Joined
    Feb 2009
    Posts
    42
    I think i understand how to use implication and set notation now. Would this be the correct usage of them? How is this explanation of relative minimum? I'm sorry to continue on about this but i lost 1 point on 7 of 10 questions because of how i stated my answer.

    Original problem:Find all critical points, label minimum or maximum, points of inflection, and graph f(x)=x^2-x-2

    y intercept: f(x)= 0 \Rightarrow \{x:x^2-x-2 =0\}=\{-1,2\}
    So the y intercepts are (-1,0),(2,0)

    f(x)= x^2-x-2 \Rightarrow f^\prime(x)= 2x-1

    Critical point:  f^\prime(x)= 0 \Rightarrow \{x:2x-1=0\} = \{.5\}
    corresponding y coordinate: (.5)^2-(.5)-2\ = -2.25
    So the critical point is (.5,-2.25) and is a minimum because
    .5 is a critical point withing [-1,2] and both f(-1) and f(2) are both greater than f(.5) therefore .5 is the minimum.

    f'(x)= 2x-1 \Rightarrow f''(x)=2

    Inflection points: f''(x)= 0 \Rightarrow \{x:2=0\} =\emptyset
    So there are no inflection points.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327
    My only criticism would be that you left out what you said before:

    The graph of y= x^2- x- 2 is a parabola with vertex at (0.5, -2.25) and crossing the x-axis at (-1, 0) and (2, 0).

    And you could add the obvious point: the graph crosses the y-axis at (0, -2).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by ixo View Post
    y intercept: f(x)= 0 \Rightarrow \{x:x^2-x-2 =0\}=\{-1,2\}
    So the y intercepts are (-1,0),(2,0)
    The fact that the set of all x such that x^2 - x - 2 = 0 is the set \{-1,\,2\} has nothing to do with f(x) being zero (in this set builder notation, the variable x is just a dummy variable, and doesn't necessarily represent the same variable used in f). Get rid of the implication, or write it as f(x) = 0\Rightarrow x^2-x-2=0\Rightarrow x=-1,\,2 (the comma-separated values for x should be fine in an informal setting, but you could write out x = -1\text{ or }x = 2 if you like).

    Yes, your implication is logically true (since the consequent is always true), but it doesn't make sense to use it in this case since there's no obvious relationship between the two statements. The same goes for the two other uses of set builder notation with implications.

    So the critical point is (.5,-2.25) and is a minimum because
    .5 is a critical point withing [-1,2] and both f(-1) and f(2) are both greater than f(.5) therefore .5 is the minimum.
    The "therefore .5 is the minimum" phrase seems redundant since you've already said that it is a "minimum because..."

    And you still haven't fully explained why the point is a minimum.

    For example, consider the function f(x)=-32x^6+96x^5-44x^4-72x^3+76x^2-24x\text. There is a critical point at (0.5, -2.25), f(-1)=0 and f(2) = 0 are both greater than f(0.5), but f(0.5) is not a minimum (it is actually a relative maximum!).

    The only reason what you say is valid is because you are dealing with a quadratic function, so you need to explain this. Again, I suggest that you either mention that the point is a minimum because it is the vertex of a parabola opening upwards, or use the first derivative test (or second derivative test) to show that it is a relative minimum (and thus an absolute minimum since it is the only extremum on a curve that is everywhere continuous).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    ixo
    ixo is offline
    Junior Member
    Joined
    Feb 2009
    Posts
    42
    Ok so the in the set builder notation is still not correct. I'm going to stop using it.

    I left out the portions which were deemed "ok" as to reduce redundancy.

    Are you stating that i need to show f'(x)<0 and f'(x)>0 so i get in the habit for larger degree polynomials? The reason why I am concluding in that particular way is to show proof through rolle's theorem.

    so:
    y intercept: f(x)= 0 \Rightarrow x^2-x-2 =0 \Rightarrow x=-1,2
    So the y intercepts are (-1,0),(2,0)

    f(x)= x^2-x-2 \Rightarrow f^\prime(x)= 2x-1 \Rightarrow f''(x)=2

    Critical point:  f^\prime(x)= 0 \Rightarrow 2x-1=0 \Rightarrow x= .5

    corresponding y coordinate: (.5)^2-(.5)-2\ = -2.25
    So the critical point is (.5,-2.25)
    Inflection points: y\prime\prime= 2 \neq 0 no inflection points.

    .5 is a minimum critical point within [-1,2] where both f(-1) and f(2) are greater than f(.5). Also because f^\prime(x)<0 when x<.5 and f^\prime(x)>0 when x>.5

    Summary:
    Graphically f(x)= x^2-x-2 is concave upwards with the vertex at (.5,-2.25) and has no inflection point.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by ixo View Post
    Are you stating that i need to show f'(x)<0 and f'(x)>0 so i get in the habit for larger degree polynomials?
    No, I'm just saying that you need to explain why the critical point is at a minimum. One way to do it is to use the derivative test (like you did here), but you could also just say that the point is the vertex of the parabola, which opens upwards, and is therefore the minimum point. You just need to show some kind of reasoning.

    y intercept: f(x)= 0 \Rightarrow x^2-x-2 =0 \Rightarrow x=-1,2
    So the y intercepts are (-1,0),(2,0)
    These are x-intercepts, not y. I know, I completely missed it before. Sorry.

    Other than that, your work seems acceptable at this point.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    ixo
    ixo is offline
    Junior Member
    Joined
    Feb 2009
    Posts
    42
    In other words i need to always include the limits as f'(x) approaches the critical points from both sides to show my proof of min or max for each point?

    Yes those are my x intercepts not y, good eye, I didn't catch that either.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Easy Sig fig question
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: January 31st 2010, 05:36 PM
  2. [SOLVED] Easy Vectors Question
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: April 21st 2009, 09:59 PM
  3. [SOLVED] Easy Factoring Question?
    Posted in the Algebra Forum
    Replies: 5
    Last Post: February 2nd 2009, 04:45 PM
  4. [SOLVED] Easy trignoometry question
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: January 30th 2009, 02:24 PM
  5. another printer cartridge usage question..
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: January 15th 2009, 07:57 PM

Search Tags


/mathhelpforum @mathhelpforum