Question 1 requires the Vertical Line Test.  If a vertical line can cross more than one point anywhere on the curve, the graph is not a function.  It is sufficient to say that a vertical line does/doesn't cross more than one point as your reason for why the graph isn't/is a function.

 

Question 2.  I show you how to find domain and range in Lesson 1 of my book.  Remember, any x value that causes the denominator to be zero (a Bottom Zero) must be excluded from the domain.  Also, if there is a square root, use a sign diagram to determine when the sign under the square root is negative since that, too, must be excluded from the domain.

 

Many students have trouble understanding my steps for making a sign diagram in that lesson.  Here is an easier method to remember (although it is slower for most students, it is perfectly adequate for this stage of the course).  Once you have found the Top and Bottom Zeros and marked them on the number line, pick a number from each region and substitute it into the entire function.  The sign of your answer will be the sign of that region.  For example, if you have marked the numbers 1 and 4 on the number line, choose a number less than 1 (such as 0) to sub into the function to find the sign of that region, choose a number between 1 and 4 (such as 2) to sub into the function to find the sign of that region, and choose a number greater than 4 (such as 5) to sub into the function to find the sign of that region.

 

A safer way to find the range of a function is to draw a graph of the function and then visualize the range that way.  Find the domain of the function first.  Put the endpoints of the domain on a Table of Values and add about three more points inbetween the endpoints as well.  Plot those points and connect the dots to get a reasonably reliable picture of the function.  If the domain includes infinity or negative infinity, compute the limit as x approaches infinity and negative infinity to see what is happening to the function (and so y) at that time.  Include your sketch of the curve in your solution to back up your conclusion about the domain and range.

 

Question 3 is 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)).   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 g o f .

 

Question 4.  I show you how to find the inverse of a function in Lesson 8 of my book (just before question 4 in the lecture).  Note that, since, to invert a function, x and y change places, that means the graph of f ^-1 is basically the graph of f "turned sideways" (the x-axis becomes the y-axis and vice-versa).  This also means the domain of f is identical to the range of f ^-1 and the domain of f ^-1 is the range of f. It is much easier to find domain than range.

 

A one-to-one function is one where each value of x has only one value of y (of course, that is true of all functions), but also each value of y has only one x.  Graphically speaking that means all functions pass the Vertical Line Test, but one-to-one functions also pass the Horizontal Line Test.  One way to answer part (a) in this question is to graph the function and show it passes both the Vertical and Horizontal Line Tests.  But the graph isn't easy and make take several points.  It also requires you check for Vertical and Horizontal Asymptotes by doing all k/0 limits and limits as x approaches negative and postive infinity.

 

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: A function is one-to-one if:

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.

 

Question 5.  Since f ^-1 is the f graph turned sideways, for a function to be invertible as a function also, it must pass the Horizontal Line Test as well as the Vertical Line Test.

 

Question 6 uses the same concepts I discussed above in question 4.  I strongly suggest you find the domain of the original function f(x) first.  That tells you the range of the inverse function.  Also, note in part (c) that they have restricted the domain of f(x) to x>4.  They have told you the domain of the graph! Do not consider values of x that are not greater than 4.  Yes, the function normally would include values of x<4, but you have been specifically told to restrict to x>4.  That therefore restricts the range of the inverse function accordingly.

 

I suggest that, for all these questions, get the domain of f(x) (and so the range of the inverse), then sketch the graph of f(x).  From the graph, you will then easily see the range of f(x) (and so the domain of the inverse). 

 

Question 7 is similar to my example right at the start of Lesson 2.

 

Question 8 requires the various tricks and tips I teach in Lesson 2.  Note that x^1/3 is the cube root of x.  Hint: Let u = cube root of x, then what would x itself be in terms of u?  I think you will find it easier to factor once you put u in place (if you have studied my factoring tips in lesson 1, especially how to factor a sum or difference of cubes).

 

Question 9 obviously requires the Squeeze Theorem, which I teach in Lesson 2, question 17.