Final Review With Solutions

1) 7.5.1, 7.6.1~7.6.4

• Work Application Problem People work at different rates. Let the letters r ,t, and A represent the rate at which the work is done, the time required, and the amount of work done, respectively. Then A = rt . Notice the similarity to the distance formula, d = rt . The amount of work is often measured in terms of jobs accomplished. Thus, if 1 job is completed, A = 1, and the formula gives 1 = rt 1 r= t as the rate. If a job can be accomplished in t units of time, then the rate of work is 1 job per unit of time. t In solving a work problem, we begin by using this fact to express all rates of work. EXAMPLE: John and Sejong are working on a neighborhood cleanup. Sejong can clean up all the trash in the area in 7 hours, while John can do the same job in 5 hours. How long will it take them if they work together? Let x = the number of hours it will take the two people working together. We can 1 make a chart here for A = rt , with A = 1 . Since A = 1 , the rate for each person will be , t where t is the time it takes each person to complete the job alone. For example, since Sejong can clean up all the trash in 7 hours, her rate is 1/7 of the job per hour. Similarly, John’s rate is 1/5 of the job per hour. Work

Rate

Sejong

1 7 1 5

John

Time working together x x

Fractional Part of the job done 1 x 7 1 x 5

Since together they complete 1 job, the sum of the fractional parts accomplished by each of them should equal 1. Part done by Sejong + Part done by John = 1 whole job

1 1 x + x = 1 7 5 Solve this equation. The least common denominator is 35. 1  1 35  x + x  = 35 ⋅1 5  7 5 x + 7x = 35

35 12 Working together, Sejong and John can do the entire job in 35/12 hours, or 2 hours and 55 minutes. Check this result in the original problem. x=

• Variation Direct variation: y varies directly as x if there exists some constant k such that y = kx This can be worded also “y is proportional to x. The number k is called constant of variation. In dir ectly variation, for k>0, as the value of x increases, the value of y also increase. Similarly, as x decreases, y also decreases. EXAMPLE: Supposes that y varies directly as z, and y=50 when z=100. Find k and the equation connecting y and z. Since y varies directly as z, y = kz for some constant k. Use the fact that y=50 when z=100 to find k. Substitute these values into the equation y = kz . 5 = k ⋅100 Now, solve for k. 50 1 k= = 100 2 The variable y and z are related by the equation: 1 y= z 2

Inverse variation: y varies inversely as x if there exists a real number k such that k y= . x

Also, y varies inversely as the nth power of x if there exists a real number k such that y=

k . n x

For inverse variation, for k>0, as x increases, y decreases, and as x decreases, y increases. An example of inverse variation can be found by looking at the formula for the area of a parallelogram. In its usual form, the formula is

A = bh Dividing both sides by b gives h=

A . b

Here, h(height) varies inversely as b(base), with A(the area) serving as the constant of variation. EXAMPLE: The weight of an object above the earth varies inversely as the square of its distance from the center of the earth. A space vehicle is an elliptical orbit has a maximum distance from the center of the earth (apogee) of 6700 miles. Its minimum distance from the center of the earth (perigee) is 4090 miles. If an astronaut in the vehicle weighs 57 pounds at its apogee, what does the astronaut weigh at its perigee? If w is the weight and d is the distance from the center of the earth, then k 2 d

w=

for some constant k. At the apogee the astronaut weighs 57 pounds and the distance from the center of the earth is 6700 miles. Use these values to find k.

57 =

k (6700) 2

k = 57(6700) 2 Then the weight at the perigee with d=4090 miles is

w=

57(6700)2 ≈ 153 Pounds. (4090)2

Joint variation: It is common for one variable to depend on several others. For example, if one variable varies as the product of several other variables (perhaps raised to powers), the first variable is said to vary jointly as the others. EXAMPLE: Strength of a rectangular beam varies jointly as its width and the square of its depth. If the strength of a beam 2 inches wide by 10 inches deep is 1000 pounds per square inch, what is the strength of a beam 4 inches wide and 8 inches deep? If S represents the strength, w the width, and d the depth, then S = kwd 2

for some constant, k. Since S=1000 if w=2 and d=10, 1000 = k (2)(10) 2 Solving this equation for k gives 1000 = k ⋅ 2⋅ 100 1000 = 200k k = 5,

so S = 5wd 2

Find S when w=4 and d=8 by substitution in S = 5wd 2 . S = 5(4)(8) 2 = 1280 The strength of the beam is 1280 pounds per square inch.

Combined variation: Combination of direct and inverse variation. EXAMPLE: The maximum load that a cylindrical column with a circular cross section can hold varies directly as the fourth power of the diameter of the cross section and inversely as the square of the height. A 9-meter column 1 meter in diameter will support 8 metric tons. How many metric tons can be supported by a column 12 meters high and 2/3 meter in diameter? Let L represent the load, d the diameter, and h the height. Then L=

kd 4 . h2

Now, find k. Let h=9, d=1, and l=8

8=

k (1) 4 92

Solve for k.

k = 648 Substitute 648 for k in the first equation. 648d 4 L= h2 Now, find L when h=12 and d=2/3 by substituting the values into the last equation. 648(2/3) 4 122 16 1 8 L = 648 ⋅ ⋅ = 81 144 9 L=

The maximum load is about 8/9 metric ton.

2) 8.4.1~8.4.4

• Radical Expressions Product Rule: n a ⋅ n b = n ab Product of two radicals is the radical of the product. 5⋅ 7

EXAMPLE:

= 5 ⋅ 7 = 35 4

8 y ⋅ 4 3r 2 = 4 24 yr 2

Quotient Rule:

n

a = b

EXAMPLE:

−

8 125

3

=

3

3 3

n

a The radical of a quotient is the quotient of the radical. n b

−8 −2 = 5 125

7 216 =

3

7 6

Simplified radical: 1. The radicand has no factor raised to a power greater than or equal to the index. 2. Exponents in the radicand and the index of the radical have no common factor (except 1) 3. The radicand has no fractions. 4. No denominator contains a radical.

EXAMPLE: Simplify 108

= 22 ⋅ 33 = 22 ⋅ 3 2 ⋅ 3 =6 3 Simplify 4 162

= 4 81 ⋅ 2 = 4 81 4 2 = 34 2

Multiplying Radicals:

kn

a km = n a m

• Equations with Radical Expressions Power rules for solving equations with radicals: If both sides of an equation are raised to the same power, all solutions of the original equation are also solutions of the new equation. Note: When the power rule is used to solve an equation, every solution of the new equation must be checked in the original equation.

Solving an equation with Radicals: Step1: Isolate the radical. Make sure that one radical term is along on one side of the equation Step2: Apply the power rule. Raise each side of the equation to a power that is the same as the index of the radical. Step3: Solve. Solve the rEsulting equation; if it still contains a radical, repeat step 1 and 2. Step4: Check. It is essential that all potential solutions be checked in the original equation.

EXAMPLE: Solve

4−x = x+ 2

x=0 or x=-5, but when you check you know that 0 is the true solution and -5 is extraneous.

Solve

m2 − 4m + 9 = m − 1

m=4

Solve

5m + 6 = 2 − 3m + 4

m=15 or m=-1

3)9.1.1~9.2.4 I advise you to complete the following Thinkwell exercises: 9.1.2, 9.1.3, 9.2.2, 9.2.3, 9.2.4

• Function A Function is a relation in which, for each value of the first component of the ordered pairs, there is exactly one value of the second component A Function is a set of ordered pairs in which no first component is repeated. A Function is a rule or correspondence that assigns exactly one range value to each domain value. A relation is a set of ordered pairs. Typically, in (x,y) form x is called independent variable and y is called dependent variable. Domain and Range: In a relation, the set of all values of the independent variable (x) is the domain; the set of all values of the dependent variable (y) is the range. The domain of a relation is assumed to be all real numbers that produce real numbers when substituted for the independent variable. Vertical line test: If a vertical line intersects the graph of a relation in more than one point, then the relation does not represent a function. Function notation: y = f ( x ) = expression

EXAMPLE: For the following, decide whether or not it is a function.

{(1,1) , (1, −1) , ( 2,4 ) , ( 2,−4 ) , ( 3,9 ) , ( 3, − 9)}

The set of ordered pairs above is not a function because the all three ordered pairs have same x-value, but different y-value.

{( 2,5 ) , ( 3,7 ) , ( 4,9) , ( 5,11)} Above is a function with domain: {2, 3, 4, 5} and range: {5, 7, 9, 11} EXAMPLE: Decide whether the each given relation defines y as a function x. Give the domain. y = x2 x+ y <4 xy = 1 2 x −9 x = y2 y=

4) 10.2.3, 10.3.3, 10.4.1, 10.5.3~10.7.5 I advise you to complete the following Thinkwell exercises: 10.2.3, 10.3.3, 10.4.1, 10.6.1, 10.6.2, 10.7.1~10.7.5

• Quadratic Functions Quadratic: f ( x ) = ax 2 + bx + c for real numbers a, b, and c with a ≠ 0 Vertical Shifts: the graph of f ( x ) = x 2 + k is a parabola with the same shape as the graph of f ( x ) = x 2 . The parabola is shifted k units upward if k>0, and |k| units downward if k>0. The vertex is (0, k). Horizontal Shifts: The graph of f ( x ) = ( x − h) 2 is a parabola with the same shape as the graph of f ( x ) = x 2 . The parabola is shifted h units horizontally: h units to the right if h>0, and |h| units to the left if h<0. The vertex is (h,0) Vertex and Axis: The graph of f ( x ) = ( x − h) 2 + k is a parabola with the same shape as f ( x ) = x 2 and with vertex at (h, k). The axis is the vertical line x=h.

1. Graph of the quadratic function f ( x ) = a ( x − h) 2 + k, a ≠ 0 is a parabola with vertex at (h, k) and vertical line x=h as axis. 2. The graph opens upward if a is positive and downward if a is negative. 3. The graph is wider than f ( x ) = x 2 if 0<|a|<1. The graph is narrower than f ( x ) = x 2 if |a|>1.

• Completing square to find the vertex To derive a formula for the vertex of the graph of the quadratic function f ( x ) = ax 2 + bx + c , complete the square on the standard form of the equation.

2

  −b   4ac − b 2 y = a  x −   + 4a   2a   y = a [ x − h]

2

+k

This equation shows that the vertex (h, k) can be expressed in terms of a, b, and c. However, it is not necessary to remember this expression for k, since it can be found by b replacing x with − . Using function notation, if y=f(x), then y-value of the vertex is 2a b f (− ) . 2a  b Vertex formula:  − ,  2a

 b  f −   2a  

EXAMPLE: Complete the square of each following equation. f (x ) = x 2 − 4 x + 5 x2 − 4x + 4 + 1 ( x − 2) 2 + 1

y = −3 x + 6 x − 1 1 2 − ⋅ ( −3x + 6 x − 1) 3 1 2 x − 2x + 3 2 2 x − 2x +1 − 3 2 2 ( x −1) − 3 2

Discriminant: b 2 − 4ac is discriminant of f ( x ) = ax 2 + bx + c . Quadratic formula: x =

−b ±

b2 − 4ac 2a

By calculating the discriminant before solving a quadratic equation, we can predict whether the solutions will be rational numbers, irrational numbers, or imaginary numbers.

Discriminant

Type of Solution

Positive, and the square of an integer

Two different rational solutions

Positive, but not the square of an integer

Two different irrational solutions

Zero

One rational solution

Negative

Two different imaginary solution

EXAMPLE: Two mechanic take 4 hours to repair a car. If each worked along, one them could do the job in 1hour less time than the other. How long would it take the slower one to complete the job done? EXAMPLE: Predict the number and type of solutions for the following equations. a. 6 x 2 − x − 15 = 0 b. 3m 2 − 4m = 5 c. 4x 2 + x + 1 = 0

a. Dicriminant is 361. It is a perfect square of 19. Since a, b, and c are integers, the solutions will be two different rational numbers, and the equation can be solved by factoring. b. Discriminant is 76, not a square of an integer, 76 is irrational. From this and from the fact that a, b, and c are integers, the equation will have two different irrational solutions, one using 76 and one using − 76 . c. Discriminant is -15, a negative number and a, b, and c are integers; this quadratic equation will have two imaginary number solutions.