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?  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 endeavour to 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.

These are called composite functions.  f o g is f(g(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.

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.

You will need to use the natural log to help you find the inverse of this function.  Be sure to read my Logs and Exponentials review in Lesson 1.

Note that you can easily see the domain and range of f(x) because you have been given the graph of the function.  That also tells you the range and domain of the inverse function. 

• Domain of f = range of f-inverse.
• Range of f = domain of f-inverse.

That makes sense because you merely had x and y change places to get the inverse function, so you have simply changed the domain into the range, and vice-versa.

To sketch the graph of f-inverse, merely invert some points on the given graph and plot them.  For example, the point (0,3) will invert to (3,0).  The horizontal asymptote at y=2 becomes a vertical asymptote at x=2.

They are actually incorrect to refer to some of these graphs as functions.  Specifically, part (a) is not a function because it fails the vertical line test.  But that is irrelevant, they aren't asking if the graphs are functions, I am just quibbling.

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.

First, note the domains of the given functions.  Be careful! In part (c), they have restricted the domain of f(x).  They say x>2, that is telling you the domain for this problem is restricted to (2, infinity).  Now that you know the domains of the original functions, you also know the ranges of the inverse functions, as they request.

Now proceed to find the inverses of these functions using the method I taught you in Lesson 8, as I mentioned earlier in the tips to help with question 4 above.

Make sure that you establish the domain and range of the original function, f.  That then tells you the range and domain of the inverse function, respectively.  The domain of the inverse function must match the range of the original function.  If you solve the domain of the inverse algebraically, you will be misled as to what the correct domain is.

This is a similar graph reading exercise to what I do at the start of Lesson 2 in my book.

This is a good run-through of limits.  Study Lesson 2 thoroughly to prepare for this question.  Hint: if I let u = cube root of x (that is what x to the power of 1/3 really means), what would u-cubed be?  That might help you see how to factor part (c).

Read the Squeeze Theorem section at the end of the lecture in Lesson 2 of my book.  They have given you a pretty easy one.