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.
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.
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 x1/3 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.
Question 9 obviously requires the Squeeze Theorem, which I teach in Lesson 2.
