Download 2. g(x)

Math 163 Final Exam Study Guide
State the domain (D) for 1 and 2.
Since division by zero is undefined,
x–4≠0
x≠4
So the domain is all real numbers, x ≠ 4
3x
1. f(x) =
x4
2. g(x) =
x6
We can take the square root of positive numbers and zero. So, what’s
inside the radical cannot be negative. Therefore
x + 6 ≥ 0 (meaning positive or zero)
x ≥ -6
3. Find the slope and y-intercept for the line 4x – 2y = 7.
slope = _______________
Solve for y:
4x – 2y = 7
4x – 7 = 2y
2x – 7/2 = y
y-intercept = _______________
slope = 2, y-intercept = -7/2
4. Write the equation of the line passing through the points A(3, -2) and B(-5, 4).
remember: m 
y 2  y1
x 2  x1
, and y = mx + b.
m = (-2 – 4)/(3 – (-5)) so, y = -3/4 x + b
= -6/8
-2 = -3/4(3) + b (plug in one of the points to solve for b)
= -3/4
-2 = -9/4 + b
-8/4 + 9/4 = b
¼=b
Therefore: y = -3/4 x + ¼
or using point/slope form: y +2 = -3/4(x – 3) or y – 4 = -3/4(x + 5)
For problems 5-8 , if f(x) = 4x – 5 and g(x) = x2 – 6, find:
5. f(2) – g(-3)
=
=
=
=
4(2) – 5 – [(-3)2 – 6] (plug 2 in f(x) and plug -3 in g(x))
8 – 5 – (9 – 6)
8 – 5 – (3)
0
6. f(x)•g(x)
= (4x – 5)(x2 – 6)
= 4x3 – 24x – 5x2 + 30
= 4x3 – 5x2 – 24x + 30
7. f(g(x))
Put the g(x) function inside f(x):
= 4(x2 – 6) – 5
= 4x2 – 24 – 5
= 4x2 - 29
8. f -1(x)
This means the inverse of f(x)
Switch the x and y values
so y = 4x – 5 becomes x = 4y – 5
Now solve for y
x + 5 = 4y
(x + 5)/4 = y = f -1(x)
9. (5 +
=
=
=
=
 9 ) – (4 –
 36 )
(5 + 3i) – (4 – 6i)
(5 + 3i) – 4 + 6i
5 + 3i – 4 + 6i
1 + 9i
10. (4 – 3i)(2 + 5i)
FOIL:
= 8 +20i – 6i – 15i2
= 8 + 14i – 15(-1) remember i2 = -1
= 8 + 14i + 15
= 23 + 14i
11. Given that x = -2 is a zero of f(x) = x4 + 2x3 – 5x2 – 10x, determine all of the real
zeros.
Use synthetic division
-2
1
2
-5
-10
0
-2
0
10
0
1
0
-5
0
0
The answer represents x3 – 5x
x3 – 5x = 0 set this equal to zero and factor
x(x2 – 5) = 0
x = 0 x2 – 5 = 0
x2 = 5
x = ±√5
so x = 0, x = -2, x = ±√5
12. Determine all the rational zeros for f(x) = 2x3 – 5x2 – 28x + 15. Use the Rational
Zero Theorem to make a list of possible zeros if necessary.
To make a list of the possible rational zeros, divide the factors of the constant term by the factors of the
leading coefficient:
±1, ±3, ±5, ±15
±1, ±2
= ±1, ±3, ±5, ±15, ±½, ±3/2, ±5/2, ±15/2
Use synthetic division:
5
2
-5
-28
15
10
25
-15
2
5
-3
0
2
5
-3
-6
3
Divide again
-3
2
-1
0
The result represents 2x – 1 = 0
solving for x gives x = ½
so the zeros are ½, -3, 5
Solve:
13. 2x4 – 15x3 + 18x2 = 0
x2(2x2 – 15x + 18) = 0
Factor out x2
2
x (2x – 3)(x – 6) = 0 Factor 2x2 – 15x + 18
x2 = 0, x = 3/2, x = 6
x = 0 is a double root, so there are “4” zeros.
14. |x + 5| = 12
If you think about | ? | = 12, the question mark could be either 12 or -12
This is why we solve both x + 5 = 12 and x + 5 = -12
so x = 7 , x = -17
15. 3x2 – 2x – 4 = 0
This does not factor so we must use the quadratic formula which gives:
x = 2 ± √4 – 4(3)(-4)
2(3)
x = 2 ± √52
6
x = 2 ± 2√13
6
x = 1 ± √13
3
16.
2x  7 + 4 = x
√2x + 7 = x – 4
2x + 7 = (x – 4)2 square both sides
2x + 7 = (x – 4)(x – 4)
2x + 7 = x2 – 8x + 16
0 = x2 – 10x + 9 combine terms on one side
0 = (x – 9)(x – 1) factor
x = 9, x = 1
Solve for x exactly:
17. log81 x =
3
4
Write in exponential form, start with the BASE
813/4 = x
27 = x
18. e3x = 12
Write in logarithm form
ln 12 = 3x
ln 12 = x
3
(base e is understood) ln is loge
x ≈ 0.828
19. ln (x + 3) – ln x = 2ln 2
Combine logarithms into a single expression
x3
ln
= ln 22
x
x3
= 4 (one to one property)
x
x + 3 = 4x (multiply both sides by x)
3 = 3x
1=x
20. log (x – 15) + log x = 2
Combine logarithms into a single expression
log x(x – 15) = 2
102 = x(x – 15)
write in exponential form
100 = x2 – 15x
0 = x2 – 15x – 100
0 = (x – 20)(x + 5)
x = 20, x = -5
x = 20 is the only solution because taking the log of a negative number is undefined.
Perform the indicated operations:
 3 4
 3 1
 0  3
21. 
= 
 + 


2
  2 5
 6 7 
 4
5
 4

  0  3
0 
22.  2
=
4
2

 3  1
2
23. Find A-1 for A = 
1
3
 4
 

5
A-1 =  5
1
2


5
 5
3

4
 (4)(0)  (5)(4) (4)(3)  (5)(2)  20  2

 

6
(2)(0)  (0)(4) (2)(3)  (0)(2) =  0
 (3)(0)  (1)(4) (3)(3)  (1)(2)  4  11
Remember: A-1 =
 d  b
1
ad  bc  c
a
Solve each system of equations by any method:
24. 2x + 8y = -13
3x – 5y = 9
3(2x + 8y = -13)
-2(3x – 5y = 9)
 6x + 24y = -39
 -6x + 10y = -18
34y = -57
y = -57/34
3x
3x
3x
3x
3x
x
– 5(-57/34) = 9 plug y into the 2nd original equation
+ 285/34 = 9
= 9 – 285/34
= 306/34 – 285/34
= 21/34
= 7/34
(7/34, -57/34)
25. x – y = 3
x – y2 = 1
solve for x in the first equation: x = y + 3
y + 3 – y2 = 1
substitute for x in the second equation
0 = y2 – y – 2
gather all terms on one side
0 = (y – 2)(y + 1) factor
y = 2, y = -1
plug each y value in equation 1 to find the corresponding x
(5, 2) (2, -1)
26. eq 1 5x + 2y – z = -1
eq 2 2x – 3y – 2z = 3
eq 3
x + y – z = -6
eq 1 5x + 2y – z = -1
(-1)eq 3 -x – y + z = 6
eq 4 4x + y
= 5
eq 2 2x – 3y – 2z = 3
(-2)eq 3 -2x – 2y + 2z = 12
-5y
= 15
y = -3
plug y = -3 into eq 4
4x + (-3) = 5
4x = 8
x=2
plug x and y into eq 3
2 + (-3) – z = -6
-1 – z = -6
-z = -5
z=5
(2, -3, 5)
Draw the graph of each equation:
27. 3x – y = 6
27.
solve for y
y = 3x – 6
28. y = ln (x + 2)
the parent function
y = ln x
shifted two units to the left
28.
29. f(x) = x2 – 8x + 12
29.
factor to find the zeros
(x – 6)(x – 2) = 0
x = 6, x = 2
and/or find the vertex using
x = -b
2a
x = -(-8)
2(1)
x = 8/2
x=4
plug x = 4 into the equation to
find y
y
y
y
y
=
=
=
=
(4)2 – 8(4) + 12
16 – 32 + 12
-16 + 12
4
vertex (4, -4)
30. f(x) = (x – 4)3 + 2
parent function: y = x3
30.
parent function: y = x3
shifted 4 units to the right and
2 units up