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Math Help - Calculus Notes

  1. #31
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    Quote Originally Posted by ecMathGeek View Post
    There's a very simple example of this:

    Let
    f(x)=x^2
    g(x)=\sqrt{x}

    f(g(x))=(\sqrt{x})^2=x
    g(f(x))=\sqrt{x^2}=|x|
    Thank you, I added that at the end of Lesson 1.2.
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  2. #32
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    Quote Originally Posted by ecMathGeek View Post
    There's a very simple example of this:

    Let
    f(x)=x^2
    g(x)=\sqrt{x}

    f(g(x))=(\sqrt{x})^2=x
    g(f(x))=\sqrt{x^2}=|x|
    This is not a fair example.
    f: \mathbb{R}\mapsto \mathbb{R}_{\geq 0}
    g: \mathbb{R}_{\geq 0}\mapsto \mathbb{R}_{\geq 0}.
    Therefore, f\circ g: \mathbb{R}_{\geq 0}\mapsto \mathbb{R}_{\geq 0} and g\circ f: \mathbb{R}\mapsto \mathbb{R}_{\geq 0}.
    Thus, the domains do not match up. We can construct more of these. And this example is excellent for pre-calculus and calculus students because they would not know the difference. But what CaptainBlank was talking about was different, their it is more complicated to find a counter-example.
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  3. #33
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    Okay, my websites have looked better in the past, but this was to be quick, simple, and easy: The Calculus Guru

    The name is set in stone (I hope MathGuru likes it). The slogan/catchphrase can change in some time. I will upload notes after they are fully critiqued on MHF (and I revise them accordingly). I may go back and add Section Zero to review algebra and trig.

    I would like to use the MHF logo, but I need permission to do so!

    Thank you for all the help so far!
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  4. #34
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    This is not a fair example.
    f: \mathbb{R}\mapsto \mathbb{R}_{\geq 0}
    g: \mathbb{R}_{\geq 0}\mapsto \mathbb{R}_{\geq 0}.
    Therefore, f\circ g: \mathbb{R}_{\geq 0}\mapsto \mathbb{R}_{\geq 0} and g\circ f: \mathbb{R}\mapsto \mathbb{R}_{\geq 0}.
    Thus, the domains do not match up. We can construct more of these. And this example is excellent for pre-calculus and calculus students because they would not know the difference. But what CaptainBlank was talking about was different, their it is more complicated to find a counter-example.
    I was thinking the same thing, it kind of seemed like a "cheap" example. However, it had never occured to me that if the domains match up and f(g(x))=x, g(f(x)) could somehow be something other than x. That's an example I would like to see.

    By the way, colby2152, these are excellent notes! They are straightforward, well organized, and very detailed.
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  5. #35
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    Quote Originally Posted by ecMathGeek View Post
    However, it had never occured to me that if the domains match up and f(g(x))=x, g(f(x)) could somehow be something other than x. That's an example I would like to see.
    Define functions on the unit interval by \textstyle f(x) = \left\{\begin{array}{ll}2x&\text{if $0\leqslant x<1/2$}\\1&\text{if $1/2\leqslant x\leqslant1$}\end{array}\right.\;, and g(x) = x/2. Then f(g(x)) = x but g(f(x))\ne x if x>1/2.
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  6. #36
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Opalg View Post
    Define functions on the unit interval by \textstyle f(x) = \left\{\begin{array}{ll}2x&\text{if $0\leqslant x<1/2$}\\1&\text{if $1/2\leqslant x\leqslant1$}\end{array}\right.\;, and g(x) = x/2. Then f(g(x)) = x but g(f(x))\ne x if x>1/2.
    Correction:

    f(g(x))=\left\{\begin{array}{ll}x&\text{if $0\leqslant x<1/2$}\\1&\text{if $1/2\leqslant x\leqslant1$}\end{array}\right.\;

    g(f(x))=\left\{\begin{array}{ll}x&\text{if $0\leqslant x<1/2$}\\ \frac{1}{2}&\text{if $1/2\leqslant x\leqslant1$}\end{array}\right.\;

    Restricted to the domain 0 \leq x<1/2, f(x) and g(x) are inverse functions.
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  7. #37
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    Quote Originally Posted by ecMathGeek View Post
    Correction:

    f(g(x))=\left\{\begin{array}{ll}x&\text{if $0\leqslant x<1/2$}\\1&\text{if $1/2\leqslant x\leqslant1$}\end{array}\right.\;

    g(f(x))=\left\{\begin{array}{ll}x&\text{if $0\leqslant x<1/2$}\\ \frac{1}{2}&\text{if $1/2\leqslant x\leqslant1$}\end{array}\right.\;
    Maybe I didn't make the definitions sufficiently clear. The definition of f(x) depends on whether x is less than or greater than 1/2. But the definition of g(x) is that g(x)=x/2 throughout the interval [0,1]. So for any x in [0,1], g(x)=x/2 and therefore f(g(x))=f(x/2)=x (because x/2 is necessarily less than 1/2).

    Quote Originally Posted by ecMathGeek View Post
    Restricted to the domain 0 \leq x<1/2, f(x) and g(x) are inverse functions.
    Not true: if you restrict the domain of both functions to the interval [0,1/2] then g(f(x)) will not be defined when x>1/4, because then f(x)=2x>1/2, which would then be outside the domain of g.
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  8. #38
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    I didn't know how to go about the trig section because students who take Calculus are to already have a good background in Trigonometry.
    as a math tutor, i disagree. yes, students should have strong backgrounds in trig before going on to calculus, but in the real world, they usually don't. i'll tell you this. if a student is very strong in precalc, calc 1 will be easy, and chances are, they wouldn't need a tutorial in the first place. besides, you started pretty basic when talking about functions, why treat trig any different?

    Quote Originally Posted by ThePerfectHacker View Post
    ...But what CaptainBlank was talking about was different, their it is more complicated to find a counter-example.
    can someone please find this legendary counter-example made by CaptainBlack? i'm very curious about it now
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  9. #39
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    Quote Originally Posted by Jhevon View Post
    as a math tutor, i disagree. yes, students should have strong backgrounds in trig before going on to calculus, but in the real world, they usually don't. i'll tell you this. if a student is very strong in precalc, calc 1 will be easy, and chances are, they wouldn't need a tutorial in the first place. besides, you started pretty basic when talking about functions, why treat trig any different?
    Excellent point Jhevon. I will either expand that section or add a "lesson" in Chapter Zero!
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  10. #40
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    Excellent point Jhevon. I will either expand that section or add a "lesson" in Chapter Zero!
    Chapter Zero

    i guess books with chapter zero are for people who believe the natural numbers start at zero. sounds cool anyway
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  11. #41
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    Last edited by colby2152; February 29th 2008 at 12:51 PM.
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  12. #42
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    Lesson 2.4 - Limits: Infinite Limits

    Lesson 2.4 - Limits: Infinite Limits
    Infinite limits can be defined in a precise manner that is left out of these notes. The following definitions are meant for an introductory Calculus course rather than an Analysis course.

    The most common application of infinite limits is analyzing their nature at positive or negative infinity. Likewise, we would like to see if a horizontal asymptote exists.

    Definition – Let f(x) be a function defined on some interval (a, \infty). Then

    \lim_{x \rightarrow \infty} f(x) = L

    means that the values of f(x) can be made arbitrarily close to L by taking [tex]x sufficiently large.

    What this simply states is that if a function is defined on some interval (a, \infty) then it has a limit.
    Likewise, the same is true for negative values where an interval is (-\infty, a) and as x approaches negative infinity, it becomes close to a limiting value.

    If a function has a value for its infinite limit, then that value is a horizontal asymptote. Another way to analyze infinite limits besides the numerical, algebraic, and graphical methods that were discussed before is to analyze the behavior of the function.

    Example 2.4-A

    f(x)=\frac{2x^2-4x+3}{5x^2-7x+2}

    \lim_{x \rightarrow \infty}f(x)= \lim_{x \rightarrow \infty} \frac{2x^2-4x+3}{5x^2-7x+2}

    Both the numerator and the denominator of this function approach infinity as x approaches infinity, so is the limit of this function infinity? The answer to such a question is NO. With high values of x, only the highest degree of a polynomial creates significant change in the value of the function, so you can actually view the limit of this function like this:

    \lim_{x \rightarrow \infty} \frac{2x^2}{5x^2}

    This function would then reduce and would we have a value for this limit.

    \lim_{x \rightarrow \infty} f(x)=\frac{2}{5}

    This so happens to be a rule of thumb to follow. If there is a fraction where both the numerator and denominator are polynomials, then the highest power can be analyzed. If both powers in the numerator and denominator are equal, then the limit is just a ratio of the coefficients.


    \lim_{x \rightarrow \infty}  \frac{ax^n+...}{bx^n+...}=\frac{a}{b}

    This limiting value is a horizontal asymptote of the function. The function holds no values at this asymptote, and in this case it holds no values beyond the asymptote. This line can be thought of as a boundary.

    Now, if the highest powers are different, then there are two different scenarios. If the numerator has a higher power than the denominator, then the function tends to infinity. The “top” simply “grows” faster than the “bottom”.

    \lim_{x \rightarrow \infty}  \frac{(ax^{n+m}+...)}{(bx^n+...)}=\infty

    If the “bottom” has a higher power than the “top”, then the function tends to zero. Think of the function as x approaches infinity. Both the top and bottom will “grow”, but the bottom grows much faster, and the function imitates:
    f(x)=\frac{1}{x}

    When you divide a small number by a very large number, you will get an even smaller number. This process continues and the values approach zero.

    \lim_{x \rightarrow \infty}  \frac{ax^n+...}{bx^{n+m}+...}=0

    Basic properties that were learned in Lesson 2.1 are carried out here as well.

    Example 2-4B

    \lim_{x \rightarrow \infty}  \frac{-5x^4}{3x^4+ 42}=\frac{-5}{3}

    Example 2-4C

    \lim_{x \rightarrow \infty}  \frac{-2x^2}{x+ 7}=-\infty


    What happens when we need to evaluate a limit that is not exclusively polynomials? You need to know the nature of functions that are not polynomial. We could also compare the function to another one, but that involves using the Squeeze Theorem which will be explained in the next lesson.

    Exponential Functions
    Functions such as f(x)=e^x “grow” very fast and will always “grow” faster than a polynomial. Since this function increases at such a high rate, its natural log counterpart does not.

    Trigonometric Functions
    These are fairly trivial to analyze since they have boundaries. For example, the sine function is bounded by y=-1 and y=1, so it cannot “grow” beyond these horizontal asymptotes.
    Last edited by colby2152; March 11th 2008 at 06:16 AM.
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  13. #43
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    the links don't work. they just send us to your host site
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  14. #44
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    Quote Originally Posted by Jhevon View Post
    the links don't work. they just send us to your host site
    I am not sure why, because when I enter the url address in my web browser, it comes up just fine. Could you try doing that for me to see if it works? (just so I can figure out this problem?)

    Thanks!
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  15. #45
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by colby2152 View Post
    I am not sure why, because when I enter the url address in my web browser, it comes up just fine. Could you try doing that for me to see if it works? (just so I can figure out this problem?)

    Thanks!
    indeed. copying and pasting the url in your message takes us to various pdf files. but when we click on the link itself, it sends us to freehostia.com

    did you try clicking on it from your post?
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