Download Final Exam

Math 120
Selected Solutions to the Final Exam Study Guide
1. Review applications problems from Exam 1.
2. Solve for x.
1
2
4

 2
a.
x  6 x  5 x  x  30
b.
Updated 12/8/16
ln( x  3)  ln( x  2)  ln 85
3.
Simplify each expression.
1
1
a.
If g ( x)  2
and r ( x)  2
, simplify g ( x)  r ( x)
x  3x
x 9
b.
Refer to exam III adding, subtracting, multiplying, and dividing fractions.
c.
Refer to exam IV problems involving simplifying radicals and radical expressions.
d.
Refer to exam IV problems involving simplifying complex number expressions.
4.
Know the difference between an equation, an expression and a function. Know which instruction might
be associated with each, i.e., simplify, solve and graph.
5. Be able to graph any of the following base functions with left/right and/or up/down shifts.
Be able to list the domain and range of all these functions.
a.
b.
c.
d.
Linear
Absolute value
Cubic
Radical
e)
f)
g)
h)
Rational
Quadratic
Exponential
Logarithm
6. Be able to use interval notation to solve absolute value and quadratic inequalities by examining the graph
of its associated function. (see also section 8.7 problems 7 – 12)
a. Refer to Exam II (problems like 7 x  2  5 )
b. Refer to Exam V (problems like
7.
 2x  5 x  4  0
)
Be able to recognize the difference between rational, irrational, and imaginary numbers. Refer to Exam IV.
Examples of:
8.
Rational numbers:
6, 73 ,  15 ,0, 121, 3 8
Irrational numbers:
 , e, 2, 3 25, 3 25
Imaginary numbers:
3i, 4, 18
Functions and their inverses.
a. Know the significance of a one-to-one function. What does it tell you about the inverse of f(x)?
If the function is one-to-one, then its inverse is a function.
b. Given a function, find its inverse. For example, if k ( x)  log 3 x , what is k 1 ( x) ?
k ( x)  log 3 x
Step 1: switch x and y.
x  log3 y
Step 2: solve for y =
log3 y  x
3x  y
k 1 ( x)  3x
Use the following functions to respond to questions 5 through 10. Rationalize all denominators. Use “i”
notation with imaginary solutions.
9.
1
3x
f ( x)  x  5
g ( x)  ( x  3) 2  1
h( x)  9x
q( x) 
p ( x)   x  1
k ( x)  log 3 x
r ( x)  3 x  5
w( x)  x 2  2 x  6
Use interval notation to identify the domain and range of any of the function listed above.
g ( x)  ( x  3) 2  1
Domain:
All reals.
 ,  
Range: Since the vertex is at  3, 1 and the parabola opens ‘up’, the
range is all y-values greater than or equal to -1.
1, 
f ( x)  x  5
Domain:
The value in the square root must not be negative.
x5  0
x  5
5, 
h( x )  3 x
Domain:
All reals.
Range:
The ‘y’ values are all positives.
k ( x)  log 3 x
 ,  
 0,  
This is the inverse of h(x). Domain and range are ‘switched’.
Domain:
 0,  
Range:
 ,  
10. Evaluate
3
3
3
3
7 3 7





7
f (2)
25
7
7 7
11. Find all values of x for which
a)
x  5  x 1
Square both sides…
x  5  x2  2 x  1
Now it is quadratic.
f ( x)  p( x)
0  x 2  3x  4
x  5  3
Zero product property. Check your solutions.
b) r ( x)  3
3
c) g ( x)  21
( x  3)2  1  21
Add one to both sides.
( x  3)2  20
Square root rule. (two solutions)
x2  2x  6   x  1
set = 0
x 2  3x  5  0
Doesn’t factor. Use another method.
e) w( x)  0
x2  2 x  6  0
Doesn’t factor.
Try completing the square.
f) h( x)  3(3 x  2)
32 x  3(3 x 2)
Same bases, therefore exponents are equal.
2x  3x  2
Now solve for x.
log3 x  2
Translate into an exponent equation.
d) w( x)  p ( x)
g) k ( x)  2
32  x
x 9
‘Cube’ both sides…
h) k ( x)  log 3 (2 x  1)
11. Find k 1 ( x)
log3 x  log3 (2 x  1)
Same bases. ‘Arguments’ are equal.
x  2x  1
Now solve for x.
To find the inverse, switch “x and y” and then solve for y = .
You will need to translate from logarithm to exponent to succeed.


3  1 or 1  3 .
p( f (2))  p
13. Rewrite w(x ) in vertex form.
Change to the form of the function that indicates where the vertex is.
Solution:
14. Use long division or synthetic division to perform
2  5  p
 3  
12. Find p( f (2))
w( x)   x  1  5
2
w( x)
.
 p ( x)
Note:  p ( x) is    x  1  x  1
w( x) x 2  2 x  6
Then

 p ( x)
x 1
Long division:
Synthetic division:
x 1 x2  2 x  6
1 1
2
6
___________
15. Write an equation for the graph that satisfies the given requirement.
a. The horizontal line through (-2,3)
b. The line through (6,2) and is perpendicular to 3x+5y = 11.
16. Solve the following systems of equations using substitution or elimination
6x  5 y  2
1
y   x3
(a)
(b)
2
1
4
x y 
4 x  8 y  24
3
3
17. Solve the system of equations. Write the solution as an ordered triple (x, y, z).
2 x  y  3z  1
3x  y  z  8
x  2 y  4z  3
18. Suppose that a flare is launched upward with an initial velocity of 48 ft/sec from a height of 128ft. Its
height in feet, h(t) after t seconds is given by:
h(t )  16t 2  48t  128 .
(show all your work)
a. How high will the flare be after 1 second?
h(1) 
b. When will the flare hit the ground?
Set h(t )  0
c. How high will the flare go?
The height is the “y” coordinate for the vertex.
d. When will it reach its maximum height?
The time is the “x” coordinate for the vertex.
solve the quadratic equation.
Use only the positive solution.