1. Originally Posted by ecMathGeek
There's a very simple example of this:

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

$\displaystyle f(g(x))=(\sqrt{x})^2=x$
$\displaystyle g(f(x))=\sqrt{x^2}=|x|$
Thank you, I added that at the end of Lesson 1.2.

2. Originally Posted by ecMathGeek
There's a very simple example of this:

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

$\displaystyle f(g(x))=(\sqrt{x})^2=x$
$\displaystyle g(f(x))=\sqrt{x^2}=|x|$
This is not a fair example.
$\displaystyle f: \mathbb{R}\mapsto \mathbb{R}_{\geq 0}$
$\displaystyle g: \mathbb{R}_{\geq 0}\mapsto \mathbb{R}_{\geq 0}$.
Therefore, $\displaystyle f\circ g: \mathbb{R}_{\geq 0}\mapsto \mathbb{R}_{\geq 0}$ and $\displaystyle g\circ f: \mathbb{R}\mapsto \mathbb{R}_{\geq 0}$.
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.

3. 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!

4. Originally Posted by ThePerfectHacker
This is not a fair example.
$\displaystyle f: \mathbb{R}\mapsto \mathbb{R}_{\geq 0}$
$\displaystyle g: \mathbb{R}_{\geq 0}\mapsto \mathbb{R}_{\geq 0}$.
Therefore, $\displaystyle f\circ g: \mathbb{R}_{\geq 0}\mapsto \mathbb{R}_{\geq 0}$ and $\displaystyle g\circ f: \mathbb{R}\mapsto \mathbb{R}_{\geq 0}$.
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.

5. Originally Posted by ecMathGeek
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 $\displaystyle \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 $\displaystyle g(x) = x/2$. Then $\displaystyle f(g(x)) = x$ but $\displaystyle g(f(x))\ne x$ if x>1/2.

6. Originally Posted by Opalg
Define functions on the unit interval by $\displaystyle \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 $\displaystyle g(x) = x/2$. Then $\displaystyle f(g(x)) = x$ but $\displaystyle g(f(x))\ne x$ if x>1/2.
Correction:

$\displaystyle 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.\;$

$\displaystyle 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 $\displaystyle 0 \leq x<1/2$, f(x) and g(x) are inverse functions.

7. Originally Posted by ecMathGeek
Correction:

$\displaystyle 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.\;$

$\displaystyle 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).

Originally Posted by ecMathGeek
Restricted to the domain $\displaystyle 0 \leq x<1/2$, 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.

8. Originally Posted by colby2152
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?

Originally Posted by ThePerfectHacker
...But what CaptainBlank was talking about was different, their it is more complicated to find a counter-example.

9. Originally Posted by Jhevon
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!

10. Originally Posted by colby2152
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

11. ## 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 $\displaystyle f(x)$ be a function defined on some interval $\displaystyle (a, \infty)$. Then

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

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

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

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

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

Both the numerator and the denominator of this function approach infinity as $\displaystyle x$ approaches infinity, so is the limit of this function infinity? The answer to such a question is NO. With high values of $\displaystyle x$, 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:

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

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

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

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.

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

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”.

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

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

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.

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

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

Example 2-4B

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

Example 2-4C

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

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 $\displaystyle f(x)=e^x$ “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 $\displaystyle y=-1$ and $\displaystyle y=1$, so it cannot “grow” beyond these horizontal asymptotes.

12. Originally Posted by colby2152
the links don't work. they just send us to your host site

13. Originally Posted by Jhevon
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!

14. Originally Posted by colby2152
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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