1. Nick Gandhi

2. This is Barkevious

3. He is struggling to get an “A” in
Pre-Calc Functions

4. He wants you to help him study
for the final exam.

5. Problem #1
 Find the domain of:

6.  First we need to
separate the two parts
of the equation so we
have (remember to
change the sign of
729!):
 Next, find the greatest
common factor of each
of the two parts
(remember, it is
possible to factor out
an “x”):

7.  Combine the two values
you factored out and set
up your new equation:
 Next, find where “x”
equals 0 in each part.
 Note: There are three
possible “x”s for the
second part.
 When graphed, you end Domain:
up getting a „w‟ shaped
graph.

8. Barkevious appreciates your
help!

9. Problem #2
 Solve for „x‟
2
30x 28x 303 276

10.  First, we add 303 to 2
the -276 30x 28x 27
 Next, factor out the 30
 Now use the (b/2)^2
formula, remember to
multiply this value by
30 and add to the
other side to keep it
balanced.

11.  Now factor the equation
 Divide both sides by 30
 Square root each side.
 Finally, subtract 7/15 from
each side to find „x‟

12. Barkevious is starting to feel
more confident about Pre-Calc!

13. Problem #3
 Divide then solve for x:

14. You‟re crazy right!?

15. Don‟t worry, its not as hard as it
looks.
 First, notice that we
do not have an x^3
value, so we plug in
0 for it.
 Next, we need to
multiply x^2 by x^2
to get x^4. And then
multiply the x^2 by
5x and -24.

16.  Now, we find that we
multiply -5x by x^2 to get -
5x^2.
 Do the same but only
multiply -136 by x^2 to get
-136x^2.

17.  Now that we solved the
long division, we need to
solve for x, so we take the
part we divided into the
quartic and the part we
used to solve the division.
 You are able to factor out
the first part nicely, but we
need to complete the
square for the second part.
 After using what we did in
problem 2, we find:

18. Barkevious is almost ready for
the exam!

19. Problem #4
 Find the function f ₒ g ₒ hand its domain when:

20.  For the first part, we need
to plug in h(x) for the „x‟ in
g(x), and then take that and
put it in f(x). So we get:
 Now, we need to find the
domain. So we are going
to set the denominator ( x 6) 21 12 0
equal to zero.
 Were going to solve for „x‟
21
so we subtract 12 first. ( x 6) 12

21.  To get x-6 by itself, we
need to take the 21st
root of both sides.
 Add 6 to both sides.
 Knowing what „x‟
equals, we now know
that in starting function,
„x‟ cannot equal this.

22.  For the domain of the function, „x‟ can be equal to all real
numbers except the 21st square root of -12 so our domain is:
 Remember, „x‟ cannot equal that value so we do not include
it in our domain. It can approach that value however.

23. Hooray! Barkevious is ready to get an “A” on
his final exam! And can get to sleep a couple
hours earlier too! (Thanks fox47 for the
inspiration).