Review for Final Exam
Review the study guide for the midterm.
Sections covered since the midterm:
1.4 Complex Numbers
1.6 Inequalities
2.1 Coordinate plane – midpoint formula, distance formula
2.2 Graphs of Equations in 2 variables, intercepts and circles (not symmetry)
2.4 Lines, slope, point-slope form, parallel, perpendicular
3.1 What is a function?
3.2 Graphs of Functions Vertical Line Test Domain Multi-part definitions
3.6 Composition of Functions
5.1 Exponential Functions
5.2 Logarithmic Functions
5.3 Laws of Logarithms
Sample problems
1. Do the calculation below and write the result as a + b i
a. 6  5i 2  3i



** Use FOIL: 6  5i2  3i  12  18i 10i  15  27  8i


b.

** Multiply numerator and denominator by the conjugate of the denominator:
3  5i 1 2i  3  6i  5i 10  7 11i  7  11 i
1 4
5
5 5
1 2i 1 2i
c.

3  5i
1  2i
36
2 9
** Recall you must immediately change to i notation and then simplify:
36
i 36
6i
2i
2i 2

 2


 i 2
2
2 9 i 2  i 9 3i 2  2


2. Solve the quadratic equation:
x2  4 x  5  0
** Use the quadratic formula a=1, b=4, c=5
4  4 2 
4 1 5  4  4
x
2
2
**
4  2i 22 i


 2  i
2
2
3. Solve the inequalities:


a. 6  2x  2x  9
**Solution:
15  4 x
15
x
4
b. 

1 4  3x 1


2
5
4
**Solution:





10  4(4  3x)  5 by multiplying through by LCD = 20
10  16 12x  5 multiplying though by 4
26  12x  11 subtracting 16 from each
13
11
dividing each by -12, reversing the sense and simplifying the fraction on the
x
6
12
left.
c.
x2  x  6
** x



2
 x  6  0 so factor:
x  3x  2  0 so set up a number line (not shown) marking x=3 and x=-2 as the
only points where a sign for the factored product could change sign and figure out the signs
for each of the 3 regions to get: (,2] [3,)
d
6
6
 1
x 1 x
** Move everything
to one side and get the denominators to the LCD=x(x-1)

6
6
 1  0
x 1 x
6x
6(x 1) x(x 1)


0
x(x 1) x(x 1) x(x 1)
6x  6x  6  x 2  x
0
x(x 1)
x 2  x  6
0
x(x 1)
(x  3)(x  2)
0
x(x 1)
So make a number line (not shown) with the 4 numbers on it that make any factor of the
numerator or denominator equal to zero. x=-2, x=0, x=1, x=3
This breaks the number line into 5 regions. The signs alternate -+-+-

You should get
(,2] (1,3]
4. Consider the points (2,-3) and (-4,7).
 the midpoint of the segment connecting them.
a. Find
Solution – average the x and y coordinates:
x

2  4 2
73

 1 and y 
 2 So the midpoint is at (-1,2)
2
2
2

b. Find the distance between them.
Solution: Use the distance formula
d  (2  (4)) 2  (7  (3)) 2  36 100  136  2  2  2 17  2 34
c. Find the slope and equation of the line connecting them.

Solution: m 
7  (3) 10
5

  So the equation, using point slope form is
4  2 6
3
5
y  3   (x  2)
3

d. Find the equation of the perpendicular bisector of the line between them.

Solution: This line goes through the midpoint (from a above) with slope the reciprocal of
that of c above So the equation, using point slope, is
3
y  2  (x 1)
5
e. Find the equation of the circle centered at the midpoint going through both these points.

Solution: Using the standard equation of a circle centered at the midpoint (from a above)
with radius set to half of the diameter calculated in b, we get:
(x 1)2  (y  2)2  34
5. Find the intercepts of:

a. y  x  2
2
Solution: Set y = 0 to get x intercepts x   2 . Set x=0 to get y intercept y=-2

b. xy  5  x
Solution: Set y = 0 to get x intercept
x=5. Set x = 0 to try to find y intercept, but this gives

0=5 which has no solution, so there are no y intercepts for this graph.

6. Find the equation of the line through (3,2) with slope -4.
Solution: y=2 = -4(x-3) using point slope form.
7. Find the slope and intercepts of the line 2x – 3y -12 = 0
Solution: Solving for y, add 12 and -2x to both sides and then divide by -3 to get
y
2
2
x  4 so the y intercept is at y = -4, the slope is . To get the x intercept, go back to
3
3
the original equation and set y = 0. So 2x-12=0 So x=6.

8. If f (x) 

x
f (x 
h)  f (x)
find and simplify
x 1
h
xh
x

f (x  h)  f (x) x  h 1 x 1

** Substituting
a compound fraction.
h
h
The LCD of the included denominators is (x  h 1)(x 1) Multiply numerator and
denominator by this:

 x  h
x 


x  h 1(x 1)
x  h 1 x 1  

h(x  h 1)(x 1)
(x  h)(x 1)  x(x  h 1)


h(x  h 1)(x 1)
x 2  hx  x  h  x 2  xh  x

h(x  h 1)(x 1)
h
1


h(x  h 1)(x 1) (x  h 1)(x 1)

9. Find the domain of the functions:
x4
a. f (x)  2
x  x6


x4
x4
** f (x) 
So the domain is all real numbers except x=-3

x 2  x  6 (x  3)(x  2)
and x=2.
b. g(x) 
1
2x 1
** Since we cannot take square roots of negatives in our functions:

2x 1  0 So adding 1 to both sides and dividing by 2, x 


1
2

10. Plot


a.
2x  3 if
f (x)  
 3  x if
x  1
b.
 x
if

f (x)  9  x 2 if
 x 1 if

x0
0x3
x3
x  1
** We’ll try to do these in the review session
11. Write the following function in words:
f (x)  x 2  log(x)
** For any number find its square and subtract the log base 10 from that. Then take the
square root of the result.

12. Sketch the graph of

y  e2x
**
13. Sketch the graph of
y  ln(2  x)

**

14. Express in logarithmic form:
** Solution:
log 4 (0.125)  


3
2
4  0.125
3
2
15. Evaluate:
log 4 (1/2)
Solution: This asks what power you put on 4 to get (1/2). By inspection, this is -1/2

16. Use a calculator to find
ln(1 3)
**Solution: 1.00505 from the calculator
17. Use the laws
 of logarithms to combine:
2(log( x)  2 log( y)  log( z))
** Solution:
2(log(x)2log(y) log(z))  2log(x) 4 log(y) 2log(z)

 log(x 2 ) log(y 4 )  log(z2 ) using the 3rd law for each term

 log(x 2 y 4 )  log(z 2 ) using the first law

x 2 y 4 
 log 2 by the second law of logarithms.
 z 


18. Use the laws of logarithms to expand:
3x 2
ln(
)
10
(x 1)

3x 2
)  ln(3x 2 )  ln((x 1)10 ) by law 2
** Solution: ln(
10
(x 1)
= ln(3)  ln(x


2
)  ln((x 1)10 ) by law 1
 ln(3)  2 ln(x) 10 ln(x 1) by law 3 (twice)
19. Draw 2 examples of graphs that fail the vertical line test.
20. Suppose

f (x)  3x 2 and g(x)  ln(x 2  2) and h(x)  e2 x1
Find:
a.



g f (x)  g( f (x))  g(3x  2)



 ln((3x  2)2  2)
b.

g f
h g f
h g f (x)  h(g( f (x)))  h(g(3x  2))
 h(ln((3x  2)2  2))
 e2ln(( 3x2)
2
2)1
Word problems: p. 387 #67,71, 75, 77 p. 398 #79,80, p. 405 #62