Over the course of the next few months - whenever I have the time, I will post some of my old Calclulus notes that I have found handy while I tutored the course through college. Feel free to criticize and offer your opinions for improvement!
Over the course of the next few months - whenever I have the time, I will post some of my old Calclulus notes that I have found handy while I tutored the course through college. Feel free to criticize and offer your opinions for improvement!
Lesson 1.1 - Functions: Basic Properties
A function f(x) is defined as a set of all ordered pairs (x, y), such that for each element x, there corresponds exactly one element y.
Figure 1.1-A
The domain of f(x) is the set x. In Figure 1.1-A, the domain is the set of numbers x that have a corresponding value for f(x). Therefore, the domain in this case is x: (-∞,∞)
The range of f(x) is the set y. In Figure 1.1-A, the range is the set of numbers y that have a value. Therefore, the range in this case is y: [0,∞)
How do you know when a graph is not a function?
Functions may not have more than one value of f(x) for an element x. This is why a circle or ellipse is not a function. When these equations are graphed by a calculator, or computer, only the top half is shown. If the bottom was shown, it would not be a proper function because it would fail the vertical line test.
Figure 1.1-B
The circle in Figure 1.1-B is proved to be not a function by applying the vertical line test. Draw a vertical line on a graph, and it is not a function if it crosses more than one point on that graph.
May Functions be combined?
Combinations of functions occur frequently in Calculus. This can be extremely useful whether you’re a Civil Engineer adding up multiple forces in one dimension or a financial analyst taking the present value of an asset that
depends on the stock marker.
Here are the basic combinations of functions in algebraic terms…
If $\displaystyle f(x) = 4x + 1$ and $\displaystyle g(x) = 1 - x^2$
Sum: $\displaystyle f(x) + g(x) = 4x + 1 + 1 - x^2 \Rightarrow - x^2
+ 4x + 2$
Difference: $\displaystyle f(x) - g(x) = 4x + 1 - (1 - x^2) \Rightarrow x^2 + 4x$
Product: $\displaystyle f(x)g(x) = (4x + 1)( 1 - x^2) \Rightarrow 4x - 4x^3 + 1 - x^2 = - 4x^3- x^2+ 4x + 1$
Quotient: $\displaystyle \frac{f(x)}{g(x)} = \frac{4x + 1}{ 1 - x^2}$
Composite: $\displaystyle f(g(x)) = 4(1 - x^2) + 1 \Rightarrow -4x^2
+ 5$
Often students become confused due to wide variety of notation. Here is the other common notation of function combinations:
$\displaystyle f(x) + g(x) \rightarrow (f + g)(x)$
$\displaystyle f(x) - g(x) \rightarrow (f - g)(x)$
$\displaystyle f(x)g(x) \rightarrow (fg)(x)$
$\displaystyle \frac{f(x)}{g(x)} \rightarrow \left ( \frac{f}{g} \right ) (x)$
$\displaystyle f(g(x)) \rightarrow (f \circ g)(x)$
I would like to point out another notation which crops up from time to time and ends up confusing people:
$\displaystyle f(x) + g(x) \rightarrow (f + g)(x)$
$\displaystyle f(x) + g(x) \rightarrow (f - g)(x)$
$\displaystyle f(x)g(x) \rightarrow (fg)(x)$
$\displaystyle \frac{f(x)}{g(x)} \rightarrow \left ( \frac{f}{g} \right ) (x)$
and the typical notation for composites:
$\displaystyle f(g(x)) \rightarrow (f \circ g)(x)$
-Dan
nice work, starting from basic principles huh? this thread will be very useful. as long as we're going this low, we might as well be comprehensive. i think a discussion of "relations" is in order. as well as discussing what the domain and range are in general, in terms of relations. sure, in single variable calculus, the domain will usually be the set of x-values and the range the set of y-values, but we both know that this is not always the case. expound on this. i think the use of mapping diagrams would be useful as well.
small typo here. the range for $\displaystyle y = x^2$ is $\displaystyle [0, \infty)$Therefore, the range in this case is y: (-∞,∞)
since you tutored i know you've realized that students usually have a lot of trouble with composite functions when they are seeing it for the first time...and yes, the hundredth time. i think expounding on this would be good. don't just show an example, but explain what you did exactly.Composite: $\displaystyle f(g(x)) = 4(1 - x^2) + 1 \Rightarrow -4x^2
+ 5$
Your advice and critiquing is what I was looking for. I am not the best at finding typos and small math mistakes. As far as explanations on domain and range as well as general relations, I think I can divulge more into that. I never thought composite functions were tough hence my lack of explanation. It always seemed like plug and chug to me! I'll definitely consider more details in that department as well. Thanks Jhevon!
i never did either. it always seemed straight forward to me..."whatever you see in brackets, just replace x with it." but i tell you, i've tutored myself, and trust me, explaining composite functions is often a pain for me, people just don't get it, and i can't see why as hard as i try.
one trick i use is to not only use numbers, like f(2), or functions, like f(g(x)), but other things--just for people to get at the concept, and not to make the math so tedious and serious.
like if $\displaystyle f(x) = x^2$ and $\displaystyle g(x) = x + 1$
i'd tell them:
$\displaystyle f(2) = 2^2$
$\displaystyle f(h) = h^2$
$\displaystyle f(g(x)) = (g(x))^2 = (x + 1)^2$
$\displaystyle f($$\displaystyle ) = $ $\displaystyle ^2$
(that last one always gets them, hehe)
but i do that just to show them: i don't care what is in the brackets, as long as it is in the domain of the function, just plug it in where x is! (is in the domain of x^2?)
it actually gets harder to explain when there are more than one terms. if i had $\displaystyle f(x) = 2x^2 + x + 1$ for instance, God help me explaining that to some people!
those are two tricks you should show us here. no need to be formal. whatever works and is always accurate is fine. be formal only in your definitions and theorems and proofs, but as far as explaining goes, whatever gets the job done
anyway, we're going off topic here. back to the notes!
Lesson 1.2 - Functions: Inverse Functions
Functions f and g are inverses of each other if:
$\displaystyle f[g(x)] = x$ for each x in the domain of g
$\displaystyle g[f(x)] = x$ for each x in the domain of f
The inverse of the function f is denoted $\displaystyle f^{-1}$.
To find $\displaystyle f^{-1}$, switch x and y in an equation and solve the equation in terms of x for y.
Example 1.2-A:
If $\displaystyle f(x) = 4x + 5$, then $\displaystyle f^{-1}$$\displaystyle (x)$ = ???
$\displaystyle y = 4x + 5$
Switch x and y
$\displaystyle x = 4y + 5$
$\displaystyle 4y = x -5$
$\displaystyle y=\frac{x-5}{4}$
A function and its inverse are reflections of each other about the line $\displaystyle y = x$. The following graph in Figure 2.1 shows what the functions from the prior example look like.
Figure 1.2-A
Existence of an Inverse
Not all functions have corresponding inverse functions.
For example, consider $\displaystyle f(x) = x^2$.
There are two numbers that $\displaystyle f(x)$ takes to 9, $\displaystyle f(3) = 9$ and $\displaystyle f(-3) = 9$. If $\displaystyle f(x)$ had an inverse, then $\displaystyle f(3) = 9$ would imply that the inverse of f takes 9 back to 3.
On the other hand, since $\displaystyle f(-3) = 9$, the inverse of f would have to take 9 to -3. Therefore, no function exists that an inverse of the function: $\displaystyle f(x) = x^2$.
Let’s look at this graphically. If $\displaystyle f(x)$ had an inverse, then its graph would be the reflection of the graph of $\displaystyle f(x)$ about the line $\displaystyle y = x$ like Figure 2.1 showed. The graph of $\displaystyle f(x) = x^2$ and its reflection about $\displaystyle y = x$ are drawn below in Figure 2.2.
Figure 1.2-B
Note that the reflected graph does not pass the vertical line test, so it is not the graph of a function. Instead, the proper graph of $\displaystyle f^{-1}(x)$ would be in Figure 1.2-A.
A function $\displaystyle f(x)$ has an inverse if and only if when its graph is reflected about the line $\displaystyle y = x$, and the result is the graph of a function. However, note that we can tell if the inverse is a function before we try graphing it…
Horizontal Line Test
Let $\displaystyle f(x)$ be a function.
If any horizontal line intersects the graph of $\displaystyle f(x)$ more than once, then $\displaystyle f(x)$ does not have an inverse.
One-to-one
If no horizontal line intersects the graph of $\displaystyle f(x)$ more than once, then $\displaystyle f(x)$ does have an inverse.
A function $\displaystyle f(x)$ is one-to-one if and only if $\displaystyle f(x)$ has an inverse. Another way to determine if a function $\displaystyle f(x)$ is one-to-one is to test values, therefore if for every a and b in its domain, $\displaystyle f(a) = f(b)$ implies $\displaystyle a = b$.
An All Too Common Mistake
Sometimes the student, or even the teacher, will assume that this identity follows through for inverses: If $\displaystyle f[g(x)]=x$ then $\displaystyle g[f(x)]=x$. This is certainly not true as is the case of:
Let
$\displaystyle f(x)=x^2$
$\displaystyle g(x)=\sqrt{x}$
$\displaystyle f(g(x))=(\sqrt{x})^2=x$
$\displaystyle g(f(x))=\sqrt{x^2}=|x|$
Not sure there can be any ....
$\displaystyle f(g(x)) = x$
$\displaystyle \therefore f^{-1}(f(g(x)) = f^{-1}(x)$
$\displaystyle \therefore g(x) = f^{-1}(x)$
$\displaystyle \therefore g(f(x)) = f^{-1}(f(x)) = x$
But perhaps for the right choice rules, if you defined the domains in a cunning way ......
maybe.
i was just asking to make a point. we always talk about two functions being the "inverses of each other." and i know a lot of professors require for you to show f(g(x)) = x, and g(f(x)) = x if you want to show two functions are inverses of each other. but why can't you just show one, and then it would imply the other? just say f is the inverse of g and then we would automatically know that g is the inverse of f, or say f(g(x)) = x and it would follow immediately that g(f(x)) = x as well. why the extra work? what do these professors know that we don't?