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Did you read my tips on how to study and learn Math 1500? If not, here is a link to those important suggestions: These are tips for the first assignment in the Distance/Online Math 1500 course, but I strongly recommend that you do this assignment as homework even if you are taking the classroom lecture section of the course. These assignments are very good (and challenging) practice.
The first assignment is a great way to build and review key skills that will be helpful for this course. Here is a link to the actual assignment, in case you don't have it: Note that you need to study Lesson 1 (Skills Review) and Lesson 2 (Limits) from my Intro Calculus book to prepare for this assignment. Don't have my book or audio? You can download a free sample of my book and audio lectures containing Lessons 1 and 2: The assignment also delves pretty deep into some high school algebra that I have not discussed in my book, so I will give you some more pointers when
relevant.
This question is applying the Vertical Line Test. As you were taught in high school, if you can draw a vertical line through more than one point anywhere on a graph, that proves the graph is NOT a function. Otherwise, it is a function. Merely visualize drawing a
series of vertical lines through each graph, as you scan the graph from left to right.
You don't have to show much work here. Either say, "Yes, it is a function because it passes the Vertical Line Test," or "No, it is not a function, because it fails the Vertical Line Test."
Did you read my tips I sent earlier on making sign diagrams and finding range?
Here are the links:
Don't worry about how long you may be spending making a table of values and sketching these curves. The questions on this assignment are intentionally challenging because they know you have the time and access to a calculator. If you have to spend ten minutes plotting ten or so points and drawing a graph, go ahead. That's what mathematicians do. Don't ever think you are supposed to just know what the graphs look
like. Nor, will you try to memorize such things. Rest-assured, you will learn some very useful skills as this course progresses to sketch complex graphs more quickly, but there is no shame in making a table of values to learn about the behaviour of a function.
Remember, if a bottom zero causes k/0, the graph has a vertical asymptote. Compute the limit as x approaches the bottom zero to determine if the graph rises up to infinity or
down to negative infinity.
Make sure you also compute limits as x approaches infinity or negative infinity any time they are part of the domain to identify if there is a horizontal asymptote on the graph. See my Lesson 2, question 15 for discussion of asymptotes.
These are called composite functions. f o g is f(g(x)) or f at g at x, telling you to sub g(x) in place of x in the f function. Conversely, g o f is g(f(x)). When you sub one function into another, please simplify as much as you possibly can.
However, that simplification may cause you to have a misleading idea of what the domains of the composite functions are.
Note the domain of f o g can be no better than the domain of the "inside" function (g in that case). Establish the domain of g, then establish the domain of "f o g" as well and whichever domain is smaller is the domain of f o g.
Similarly, the domain of "g o f" is the smaller of the domains of f and the domain of "g o f".
Simply
put, when you start with a function, say f, with a restricted domain, your domain can't get better, if you then sub f into g. However, your domain can get worse.
Part (a): I show you how to find the inverse of a function in Lesson 8 of my book (just after question 4 in the lecture, page 230). I like to immediately have x and y change places, then proceed to isolate
y.
Hint: to isolate y, collect the terms that have y on the left side of the equation, and move the other terms to the right. Then factor y out. Once you have isolated y, make sure you replace y with f^-1 (x), the f-inverse symbol.
Part (b): Once you have found the inverse function, the usual methods will determine the domain of that function.
Part (c):
To prove a function is one-to-one algebraically, you do the following:
Step 1: Define f(x1) by subbing x1 in place of x and define f(x2) by subbing x2 in place of x. Step 2: Write this definition down in your proof you will submit: "A function is one-to-one if f(x1) = f(x2) if and only if x1 =
x2." Step 3: Set f(x1) = f(x2) and use algebra (like cross-multiplication and multiplying everything out to get rid of brackets, etc.) to collect all the terms with x1 on one side of the equation and all the terms with x2 on the other side of the equation.
Step 4: If everything goes according to plan, all the terms disappear, leaving you with
"x1 = x2". Step 5: Assuming that does happen, you can declare you have proven f(x) is one-to-one.
Since inverse functions change x and y around, the x-axis becomes the y-axis and the y-axis becomes the x-axis. Merely turn the given graphs 90 degrees to get a rough approximation of what the graph of the inverse would look like (the graph is reading right to left instead of
left to right, so it is not a perfect representation of the inverse graph). If the rotated graph passes the vertical line test, the inverse is a function.
Put another way, if the orginal graph passes the Horizontal Line Test, its inverse would be a function.
I briefly discuss even and odd functions in Lesson 9 of my book, question 3(a).
An even function is where you get the same answers for y whether you sub a positive x value in or a negative x value.
Which is to say, f(1) = f(-1), f(2) = f(-2), f(3) = f(-3), etc.
An odd function is where you get the same answers for y but with opposite signs when you sub a positive x value in or a negative x value. Which is to say, f(-1) has the same value but opposite sign to f(1), f(-2) has the same value but opposite sign to f(2), f(-3) has the same value but opposite sign to f(3), etc.
But you can't use
actual numbers to prove that a function is even or odd. You must literally compute and simplify f(-x). Which is to say, sub -x in place of every x and simplify.
- If f(-x) = f(x), the original function, then f is an even function.
- If f(-x) = - f(x), the original function multiplied by -1, then f is an odd function.
I suggest that you first compute f(1) and
f(-1). If both answers are identical, the function is probably even. Then compute f(2) and f(-2). If those two answers are also identical, you are pretty much certain the function is even. Now prove it by computing f(-x) and showing it is exactly the same as f(x).
On the other hand, if f(-1) is the same value as f(1), but with the opposite sign. And, if f(-2) is the same value as f(2), but with the opposite sign, then you expect the
function to be odd. Compute f(-x) and show that it is the negative of f(x).
Finally, if f(-1) has no resemblance to f(1), or if f(-2) has no resemblance to f(2), then the function is neither even nor odd, and that this is sufficient to prove that to be so.
This is a good run-through of limits. Study Lesson 2 thoroughly to prepare for this question. This is the most important question on this assignment, in my opinion. Any of these limits could appear on your final exam.
Read the Squeeze Theorem section at the end of the lecture in Lesson 2 of my book. My examples before I do question 17, should be quite helpful here.
This one is really pretty easy because you
have been given the boundaries already. It is a simple matter of computing the limits. Make sure you have looked at my trig review in Lesson 1, specifically so that you know what cos0 equals.
Do note it is highly unlikely that Squeeze Theorem will appear on your Final Exam.
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