Draft 190 Final - Dept of Math, CCNY



M19500 Sample Final with Solutions

Part I. Answer any five complete questions from this part. Cross out the question that you wish to omit. Show all work.

1. a) Rewrite the inequality [pic] using interval notation.

b) Rewrite the interval [pic] using inequality notation.

Solution: Interval and inequality notation both are based on the natural ordering of the number line: numbers increase as you go from left to right.

Solution to a)

[pic] [pic] is less than (but not equal to) [pic] and [pic] is less than or equal to $20$.

[pic] is to the left of (but not equal to) [pic] and [pic] is to the left of or is equal to $20$.

In interval notation we write

Answer to a): [pic].

The parenthesis “(“ says: “leave out[pic]” and the bracket “]” says include $20$.

Solution to b) [pic]

2. Rewrite [pic]as a reduced fraction.

Solution: Plan your work: [pic] will involve Exponents, then Multiply, then Add.

These are three of the steps in the order of operations acronym PEMDAS, which stands for Parentheses, Exponents, Multiply/ Divide, Add/Subtract.

Major idea: A negative power of a base is 1 over the corresponding positive power of that base.

First we give basic ingredients in the calculation, and then we finish with a neat presentation.

First do [pic] Therefore [pic]

Next do [pic] Therefore [pic]

In practice, steps should be written in order underneath one another as follows

Warning: the power -1/3 below applies only to the fraction [pic] but not to the number 2 to its left.

Solution

[pic]

3. Rewrite [pic] as a reduced fraction.

Solution: When you add a non-fraction and a fraction, you need to rewrite the non-fraction as a fraction whose denominator is the fraction’s denominator.

This was a mouthful. It’s better to use an example with numbers:

[pic]

Now a slightly harder example with letters:

[pic]

Now let’s do the example in the question. We will write vertically.

This uses extra space, but it helps you to proceed through the example without making errors.

[pic]

Related idea: please remember that

[pic] . We can substitute a positive number for x to get the important example

[pic]

4. Factor completely: [pic].

Solution: The expression is a polynomial (with fractional exponents).

It is a sum of three monomials, separated for clarity below

[pic]

To factor a sum of monomials (in one variable)

a) Factor out the GCD of the coefficients . In this case, 3 is the GCD of 3, -9, and 6.

b) Factor out the LOWEST Power of the variable, which, in this case, is case [pic].

The basic idea about factoring a power of x is:

To factor [pic] out of [pic], write [pic].

For example, [pic].

If a is the factored-out power and b is the original power, then

[pic]

Therefore, to factor [pic]

• first factor out 3 , the largest whole number that goes into 3, -9, and 6;

• then factor out the lowest power of x, which is [pic], as follows:

[pic]

5. Let [pic]

a) What are the period and amplitude of the function f ?

b) Sketch the graph of f on the grid below. Include x- and y-axes with appropriate scales. Label all maximum and minimum points with their coordinates.

Solution: a) The general sine function has the form y = A sin (Bx + C), where

• the amplitude is |A| = the absolute value of A,

• the period is [pic], and

• the phase shift is the angle x which solves equation[pic] = 0, namely [pic].

To find the amplitude and phase shift we match the general formula

[pic] by writing underneath it the specific example

[pic] to see that

[pic], and so the amplitude is [pic]

[pic], and so the period is [pic]

C = 0, and so the phase shift is 0.

b) The graph is shown below. The secret is to plot one period carefully by dividing the interval

[phase shift, phase shift + period] = [pic] into four equal segments, each of length

period/ 4 = [pic]. Start at (0,-2sin(0)) = (0,0). Because A = -2 is negative, move down to ( [pic], up to [pic], up to [pic], then down to [pic].

That completes the graph of one period. Then draw three more periods to complete the graph.

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|2 | | | | | | | |

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6. Rewrite [pic] as a reduced fraction.

Solution: The fractions [pic] and [pic] are nested in the numerator of the given fraction.

Multiply top and bottom of the given fraction by the common denominator of the nested fractions, which is [pic]. The official steps are:

[pic]

But you should be able to condense the work as follows: [pic]

7. On the grid below, draw x- and y-axes with appropriate scales, and sketch the graph of the func tion

[pic] Draw your sketch with domain [pic] .

Because the domain is given, we will draw [pic]

Solution: Before you start, you need to get a sense of what y values will be needed for each part of the graph. Just work from left to right. Note that each piece of the graph will be a part of a straight line. Included endpoints are drawn as solid dots. Omitted endpoints are drawn as hollow dots.

First, [pic] for [pic] This is the line segment from (-1,1) (drawn as a solid dot) to (but omitting) point (1,3) drawn as a hollow dot.

Next, [pic] for [pic] This is the line segment from (1,2) (drawn as a solid dot) to (but omitting)

point (2,1), drawn as a hollow dot.

Last, [pic] for [pic] This is the line segment from (1,2) (drawn as a solid dot) to and including

point (3,2), drawn as a solid dot.

The above three paragraphs show that y goes from y = 1 to y = 3 , and so it is convenient to let y go from 0 to 3 in the sketch below. Before you plot any points or line segments, label the X and Y axes and place scale numbers on them, as shown below. Then plot the points, then the line segments.

| | | | |(1,3) | | | |

| | |Y | | | | | |

| | | | | | | | |

| | | | (1,2) | | | | (3,2) |

| | | | | | | | |

| | | | | | (2,1) | | |

| | | | | | | | |

8. a) Find an equation of the line through (6,9) that is parallel to the line joining points (2,4) and (2,6).

b) Find an equation of the line through (6,9) that is parallel to the line joining points (2,4) and (3,6).

Solution to a)

Notice that the given points (2,4) and (2,6) have the same x-coordinate, namely [pic]. If two points, on a line have the same x-coordinate, then the line is vertical and all points on the line have that x-coordinate. Thus the equation of the line through those two given points is [pic].

However, the problem asked for an equation of the line through (6,9) that is parallel to the line we just found. A line parallel to a vertical line is vertical, and so we want the vertical line through (6,9).

Answer: [pic]

Solution to b)

The points [pic] and [pic] have different x-coordinates, and so the line joining them has equation

[pic], where

• [pic] is either one of the points, say [pic] and

• m is the slope of the line joining the two points.

Now call the other point [pic].

Then [pic] is the slope of the line joining (2,4) and (3,6).

The requested line passes through point (6,9) and is parallel to a line with slope [pic], and so its slope is also [pic].

Substitute the values [pic] and [pic] in the general equation [pic]

to obtain [pic]. This is an answer.

It is OK to rewrite this answer in the slope-intercept form[pic], but you are not asked to that in this problem. Going through an extra step gives you extra chances to make mistakes, for which your teacher may deduct points.

9. If you start with the graph of the equation [pic] and then

• move the graph 3 units left; and then

• reflect the graph through the y-axis; and then

• move the graph 2 units down;

what is the equation of the resulting graph? Do not sketch.

The steps are as follow.

• When you shift the graph of the equation [pic] 3 units left,

you change the equation by replacing x by x + 3 .

You get the new equation [pic]

• When you reflect the graph of the equation [pic] through the y-axis

you change the equation by replacing x by –x.

You obtain the new equation [pic]

• When you shift the graph of the equation [pic] 2 units down,

you change the equation by subtracting 2 from the right hand side.

You get the new equation [pic]. This is the answer.

10. Given[pic].

Solution:

First find the inside functions:

[pic] and [pic]=61

Then you can rewrite the problem

[pic]

Another way to do this problem is to show that [pic] and [pic] are inverse functions.

Then [pic] for all [pic] and [pic] for all [pic], and the computation is then easy.

11. a) Sketch the graph of [pic]. Include on your graph all intercepts and the vertex, each labeled with its coordinates.

b) What is the maximum value of the function in part a)?

Solution: The function [pic]is a quadratic polynomial. Its graph [pic] is a parabola.

To find x-intercepts, set y = 0: [pic] . To find y-intercept, set [pic] to get [pic]

.

The x-coordinate of the vertex is [pic] . To find the y-coordinate of the vertex, plug x = 9/2 into

[pic]

and so the vertex is [pic] .

To show the vertex and intercepts, we need to have the domain include x-values 1 and 8, while the range must include y = -8 and y = 49/4 = 12 ¼.

b) Maximum value = y at the vertex = 49/4

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12. Graph the polynomial[pic]. Show clearly on your graph all intercepts, labeled with their coordinates, as well as the intervals in which y is positive, the intervals in which y is negative, and end behavior.

Solution: Intercepts: when [pic], then [pic] is the y-intercept. When [pic], then [pic] and [pic] are the x-intercepts. These are shown on the graph below.

The domain needs to include [pic] and x = 1/2 in order to show the x-intercepts, while the range needs to include [pic] and [pic] to show the y-intercepts.

The two x-intercepts x = -3 and x = 1/2 split the x-axis into three intervals, and you need to pick an x-value inside each interval to decide whether y is positive or negative in that interval.

In [pic] choose x = -4 to get [pic] =[pic] is positive in [pic]

In [pic] choose x = 0 to get [pic] is positive in [pic] .

In [pic] choose x = 1 to get [pic] is negative in [pic] .

Make sure that the graph does not include any sharp corners.

200

| | | | Y | | | | |

| | | |(0,152) | | | | |

| | | | | | | | |

| | | | | | | | |

| | | | | | | | |

| | | | | | | | |

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| | | | |-200 | | | |

Part III. Answer any five complete questions from this part. Cross out the question that you wish to omit. Show all work.

Trigonometry Background: Students have to know

• Basic right triangles, (45-45-90) degrees and (30-60-90) degrees, which tell us the trig functions of 30, 45, and 60 degrees, using the old SOHCAHTOA rule;

• Conversion between radians and degrees

o 180 degrees = [pic] radians ;

o 90 degrees = [pic]/2 radians

o 60 degrees = [pic]/3 radians ;

o 45 degrees = [pic]/4 radians ;

o 30 degrees = [pic]/6 radians .

• Definition: The reference angle of an angle θ (with initial line on the positive x-axis) is the acute angle between the terminal line of angle θ and the x-axis. In each of the pictures below, the angle θ is shown in black, its terminal line in red, and its reference angle in blue

θ θ θ

[pic]

Make sure to understand how the following statements follow from the pictures above.

• How to find the reference angle of a given angle [pic] between 0 and 2[pic]

o if the terminal line of [pic]is in Quadrant I, the reference angle of [pic] is [pic];

o if the terminal line of [pic]is in Quadrant II, the reference angle of [pic] is [pic]- [pic];

o if the terminal line of [pic]is in Quadrant III, the reference angle of [pic] is [pic]+ [pic];

o if the terminal line of [pic]is in Quadrant IV, the reference angle of [pic] is 2[pic] - [pic].

• How to find the trig functions of general angles: Suppose a point on the terminal line of the angle [pic] has coordinates P (x,y). Let [pic] = distance from P to (0,0). Then

o [pic] ; θ

o [pic] ;

o [pic] ;

o [pic] ; r

o [pic] ; (x,y)

o [pic] .

o Note that these rules tell you the trig functions of all angles, including multiples of 90 degrees. In particular, sine and cosine are defined for all angles, but tangent and secant are undefined for angles with terminal lines on the y-axis (since x = 0), while cosecant and cotangent are undefined for angles with terminal lines on the x-axis (since y = 0).

• ASTC: To figure out the sign (plus or minus) of the Cosine, Sine, or Tangent of angle [pic]: If the terminal line of the angle is in

o A: Quadrant I, then All three trig functions of [pic] are positive;

o S: Quadrant II, then only Sine is positive ( and cosine, tangent are negative);

o T: Quadrant III, then only Tangent is positive (and cosine, sine are negative);

o C: Quadrant IV, then only Cosine is positive (and tangent, sine are negative).

o

• Without using ASTC: To figure out the sign (plus or minus) of the Cosine, Sine, or Tangent of angle [pic]: Notice that

o [pic] = x/r has the same sign as x, since r is always positive. Thus [pic] is

▪ positive in Q1 and Q4, where x is positive and is

▪ negative in Q2 and Q3, where x is negative.

o [pic] = y/r has the same sign as y, since r is always positive. Therefore [pic] is

▪ positive in Q1 and Q2, where y is positive and is

▪ negative in Q3 and Q4, where y is negative.

o tan [pic] =y/x is

▪ positive in Q1 and Q3, where y and x have the same sign

▪ negative in Q2 and Q4, where y and x have opposite signs.

▪

• How to compute the trig function of a general angle as either plus or minus the trig function of its reference angle

o The magnitude of a trig function of [pic] equals the trig function of the reference angle of [pic] . In other words,

o A trig function of any angle [pic] equals plus or minus the trig function of the reference angle of [pic].

o The sign of the trig function of [pic] is obtained from the ASTC rule above.

• Definitions of inverse trigonometric functions.

o If [pic] , then [pic] is the unique angle [pic] with[pic]AND [pic]. If x > 1 or x < -1, then arcsin x is not defined.

o If [pic] , then [pic]is the unique angle [pic] with[pic] AND [pic] . If x > 1 or x < -1, then arccos x is not defined.

o If x is any real number, then [pic]is the unique angle [pic] with [pic]AND [pic] .

Now let’s work on some trig problems.

13. Simplify [pic] .

Remember that the value of arcsin (or arccos or arctan) must always be expressed in radians.

Solution: The angle [pic] equals 135 degrees, in Quadrant II. Therefore its sine is positive and is equal to the sine of the reference angle of 135 degrees, which is 180 – 135 = 45 degrees. Thus [pic] . If you have memorized conversions between radians and degrees, you don’t have to refer to degrees.

We want to find [pic] .

By definition, this is the angle [pic] with[pic] , but also satisfying the crucial requirement that [pic]. Since

• [pic] (think of a 45-45-90 right triangle) and

• [pic]

it follows that [pic].

Comment: The above example shows (at least ) two important ideas.

• arcsin is a function. Therefore it has at most one value. If you write something like arcsin(blah) = [pic], your teacher may decide to deduct full credit.

• arcsin(sin [pic]) does NOT have to equal [pic]. However, keep in mind that

• sin(arcsin x ) = x provided x is in [-1,1]. For any other x, sin(arcsin x ) is undefined.

14. Find[pic].

Solution: Let [pic]. From the definitions listed earlier,

we need to find angle [pic] satisfying TWO conditions: [pic] AND [pic]

In the standard 30-60-90 , [pic] . Now put the 30 ° angle in standard position in quadrant I, then draw the “X” diagram that extends the terminal line of that angle.

Clearly the two conditions [pic] and [pic] are satisfied only by an angle with terminal line in Quadrant IV, when tangent is negative. One such angle is obtained by moving counterclockwise from the positive x axis to get 360 ° - 30 ° =330 ° . However, this angle doesn’t satisfy [pic]. To fix this, do an about-face: rotate the positive x-axis clockwise 30 degrees to get to the red line in Quadrant IV as shown below. . This gives the solution [pic]

Remember that inverse trig function values are expressed in radians, not degrees.

Y

Q II: tan is Negative QI: sin, cos, and tan are All positive.

S A

[pic]

X

[pic]

T C

15. a) Convert[pic]radians to degrees. b) Find [pic] by using an addition formula for sine.

Solution to a) Remember that [pic] radians equals 180°.

Thus [pic]

Solution to b) Rewrite 105° as a sum: 105° =60° +45°

Now substitute A = 60° and B = 45° in the formula for sine of a sum and simplify:

[pic]

If you used cos 45° = sin 45° =[pic] you would get

[pic]

Either method is fine.

16. Find all solutions of [pic] for [pic].

Solution:

Begin with the basic tricky review fact from algebra: The solution of [pic] is

[pic], NOT JUST x = 2.

In our case, the solutions of [pic]are [pic], NOT JUST [pic]

We know that one solution (in Quadrant I) is given by [pic]

We now invoke the basic principle:

If [pic] then x has reference angle [pic].

All such angles are shown by the diagram back in problem 14.

To get all solutions with [pic] take

[pic] in Quadrant I

[pic] in Quadrant II

[pic] in Quadrant III

[pic] in Quadrant IV

Answer: [pic]

17. Given [pic], rewrite [pic] as a simplified polynomial.

WARNING: The meaning of the word “substitute” in algebra class is not the same as in English class.

In English: Substitute Jack for Jill in the sentence “I really like Jill” means

replace ‘Jill’ by ‘’Jack’ to get “ I really like Jack”

In Math: substitute Jack for Jill in the sentence “I really like Jill” means

replace ‘Jill’ by ‘’Jack’ ENCLOSED IN PARENTHESES to get “ I really like (Jack)”

Problem solution: To find [pic], substitute [pic] for [pic] in [pic]:

This means: Replace every ‘x’ in the function definition by [pic] as follows:

[pic]

[pic] THE PARENTHESES ARE CRUCIAL!!!!

Now simplify the given expression

[pic] Substitute the values above into the numerator :

=[pic] This is the result. The first set of parentheses may be removed. To get rid of the second set, you need to distribute the minus sign as follows:

=[pic] Distribut the -3 in front of (x+h) to get

=[pic] Now collect terms to get:

=[pic] You know that [pic]=[pic] and so

you Substitute [pic] for [pic]:

=[pic] In truth, the parentheses weren’t needed here. Now collect terms =[pic] Next factor the numerator

=[pic] Cancel the common factor to get

[pic] = THE ANSWER.

18. Find the exact value of [pic] by using a half-angle formula.

The relevant half-angle formula is

[pic] Substitute [pic] to get [pic] .Thus

[pic]

Since the angle [pic] is in Quadrant IV, its sine is negative by ASTC. Therefore

[pic] .

This is an acceptable answer, since no instructions were given regarding simplification. If you want (or if you are asked) to simplify, the nicest result is obtained quickly if you multiply the numerator and denominator of the main fraction by 2.

[pic]

19. Sketch the graph of the equation[pic]. On your graph show all intercepts (each labeled with its coordinates) and any asymptote (labeled with its equation).

Solution: To find the y-intercept, set x = 0 to get [pic].

Since [pic] is defined only for [pic] , [pic] is undefined and so

there is no y-intercept.

The vertical asymptote is the line[pic], that is, the vertical line x = 2.

To find the x-intercept, set y = 0 to get

[pic] Play it safe: check the answer [pic]The x intercept is 10.

| | | | | | | | |

| |x=2 | | | | | | |

| | |optional | | | | | |

| | | | | | | | |

20. Find all solutions for x in the equation [pic].

Main idea: Rewrite so that log functions are on the left side and anything else is on the right side. Then rewrite the left side as the log of a single expression.

[pic]

Thus x = 8 or x = -4. These answers must be checked by substitution into the original equation.

Check x = 8:

[pic]

Check x = -4 :

[pic]

No, since [pic] is defined only for x > 0.

Answer: x = 8.

21. A sample of the radioactive element migrainium weighs 1000 pounds at 1:00 P.M. and decays exponentially. If it weighs 500 pounds at 5:00 PM, how much will it weigh at 7:00 PM?

Solution:

Let N(t) = the amount of migrainium at time t.

Then [pic] . Here N(0) is the amount at time 0, which is given as 1000 pounds.

Measure time in hours, beginning with t = 0 at 1:00 PM

At 1:00 PM, [pic] and we are given [pic]

Thus [pic]

At 5:00 PM, [pic], and [pic]

Solve for K:

[pic]

Thus [pic]

The question asks: Find the amount at 7:00, which corresponds to t = 6.

[pic]

The answer to a word problem must be a clear sentence:

At 7:00 P.M., the migrainium weighs [pic] pounds.

22. Solve the inequality[pic]. Write your answer using interval notation.

Background: |x| is the distance between x and 0.

Model example 1: Solve |x| < 5.

Solution: |x| < 5 means: the distance between x and 0 is less than 5. This happens provided -5 < x and x < 5 . See the diagram below. The ‘o’ is written to indicate that -5 and 5 are missing from the solution set.

The solution to [pic] is -5 < x < 5, written in interval notation as ( -5 , 5 )

Below is a picture of the interval (-5,5)

------------------------oxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxo----------------------------

-5 0 5

Similarly, the solution to[pic] is [pic], written in interval notation as [ -5 , 5 ]

Model example 2: Solve |x| > 5.

Solution: |x| > 5 says that the distance from x to 0 is greater than 5.

This happens when x < -5 or 5 < x ,. See the diagram below.

The solution to [pic] is x < -5 or 5 < x, written in interval notation as [pic]

This is shown below.

[pic]

xxxxxxxxxxxxxxxxxxxo--------------------------------------------------------oxxxxxxxxxxxxxxxxxxx

-5 0 5

Similarly, the solution to[pic] is [pic], written in interval notation as [pic]

Solution to original problem:

Since the solution to [pic]is [pic], we can replace x by 2 – 3x to obtain

the solution to [pic]is

[pic] Solve these two equations in parallel as follows.

[pic]

[pic]

Answer:It is preferable to write the intervals from left to right as [pic]

23. Suppose fencing material costs 20 dollars per yard. Write an expression (involving x) for the total cost of a fence that goes around a rectangular field if the field’s width is x yards and the field’s area is 40 square yards.

Solution:

Review: Rectangle area = length times width

Rectangle perimeter = 2(length) + 2(width).

Let A = rectangle area; we are given x = width; let L = length.

Then A = Lx. We are given that the area is 40 Therefore

40 = Lx. Divide both sides by x to obtain a formula for the length: 40/x= L

| L = 40/x |

| |

| |

| |

|x |

|x |

| |

| |

| |

|L = 40/x |

Next step:

If P is the perimeter of the rectangle, then [pic]

Finally, Fence cost = [fence cost per yard] times yards of fence

= [pic] dollars = cost of the fence, in dollars.

24. a) Find the inverse function [pic] of the function [pic]

b) For the functions f and g in part a), find the value of [pic] without using algebra. Explain your reasoning. [If you use algebra, you will waste time and get zero credit].

Solution to a)

Set y = f(x), solve for x

[pic]

which says that if the input to the inverse function is y, the output is [pic]

Switch variables. If the input to the inverse function is x, the output is [pic]

Answer to a) : The inverse function is [pic]

Solution to b): By definition of inverse function , [pic] for all x.

In particular, when [pic], we have [pic]

-----------------------

Y

(3À/4,2) 7À/4,2) (11À/4,2) (15À/4,2) (3π/4,2) 7π/4,2) (11π/4,2) (15π/4,2) (0,13π/4) ) (0,9π/4 (0,π/4)

0 π/2 π 3π/2 2π 5π/2 3π 7π/2 4π

X

-2 (π/4,-2) (5π/4,-2) (9 π/4,-2) (13π/4,-2) (0,9π/4 (0,π/4)

3

2

1

0

-1 0 1 2 X 3

15

Y

10

5

0

-5

-10

-15

(9/2,49/4)

(1,0)

(8,0)

0 1 2 3 4 5 X 6 7 8 3

(0,-8)

(-3,0)

(1/2,0)

-4 -3 -2 -1 0 1 2 3 X 4

6 Y

3

0

-3

-6

-9

-12

0 2 4 6 8 10 12 14 16

(10,0) X

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