Study Guide and Practice for the Semester 1 Exam- PreCalculus

Start Preparing Early:

Go back and review your notes from Modules 1 – 5.

Go back and review your old exams. Rework problems that you missed and make sure you understand the mistake you made.

Each chapter in your book has “Study Aides”. Use these as extra practice.

Review the Graded HW from each lesson.

Review the Practice Exercises that were assigned in each lesson.

Think about EVERYTHING you know about the following topics. We don't claim that this list is exhaustive

but it will give you a start in your review process:

1. Domain and Range of a function 2. Symmetry of functions 3. Inverse of a function 4. Composition Functions 5. Shifting Functions 6. Vertical Asymptotes, Horizontal Asymptotes, Slant Asymptotes, Holes 7. Zero of a function, Root of a function, x-intercepts 8. Imaginary Roots 9. Maximum or Minimum of a Function 10. Possible Rational Roots 11. Complex Conjugate 12. Descartes' Rule 13. Change of Base Formula for Logarithms 14. Properties used to expand or condense logarithmic expressions 15. Solving a logarithmic equation 16. Solving an exponential equation 17. Domain/Range of Logarithmic Functions 18. Domain/Range of Exponential Functions 19. The Unit Circle 20. Domain and range of the trigonometric functions 21. The acronym ASTC 22. The acronym SOH-CAH-TOA 23. The reciprocal Trig functions 24. Amplitude, Period, Phase and Vertical Shift of Trigonometric functions 25. Vertical asymptotes of a trig function 26. Finding the general and then specific solutions of a trigonometric equation 27. Finding solutions graphically for a trigonometric equation 28. Proving trigonometric identities 29. Applying Sum and Difference Identities 30. Applying multi-angle Trig Identities 31. Inverse Trigonometric Function: Quadrants where defined 32. Find the EXACT value of an inverse trig function. 33. Solve Applied Problems with Trigonometric functions

Practice Exam Questions – Check Answers below.

The following questions will be helpful to you in your preparation for the Segment One Exam. You should do them without the use of any calculator.

We don't claim that we have given an example of every topic tested on the exam or that doing only these questions will adequately prepare you for the exam! Check your answers at the end of the document and contact your instructor if you don’t understand any of the solutions.

1. Given: x

f(x)x 7

a) Find the domain of f(x)

b) Find the vertical asymptotes of f(x)

c) Find the horizontal asymptote of f(x)

d) Find the range of f(x)

e) Find the inverse of f(x).

f) What is the relationship between the domain, range and asymptotes for f and f inverse?

g) Write a new function g(x) so that g(x) has the same vertical and horizontal asymptotes as f(x) but so that it also has a hole at x=2

h) Write a new function h(x) so that h(x) has the same vertical asymptote as f(x) but has a slant asymptote.

2. Given 5 4f(x) x 2x x 2

a) Algebraically show why f(x) is NOT an odd function

b) List the possible rational roots of f(x)

c) What does Descartes' Rule tell you about the number of positive and negative real roots and the number of imaginary roots?

d) Write f(x) in completely factored form

e) List the zeros of f(x)

f) List the x-intercepts of f(x)

g) Find the y-intercept of f(x)

3. Given

2

2

x 2, x 1f(x)

2x 2, x 1

a) Find f(-2)

b) Find f(0)

c) Find f(1)

d) Find f(2)

e) Find the domain of f(x)

f) Find the range of f(x)

g) If 4g(x) x 1, find f[g(x)]

4. Given

2

2 2

f(x) log x 2

g(x) log x log 7

a) Find the inverse of f(x)

b) Find the domain and range of f(x)

c) Solve f(x) -g(x)=0

d) Rewrite f(x) as a base 3 logarithm

e) Solve f inverse = 5

5. Given 5 3

sin(u) ,withu in Quad III and cos(v) , with v in Quad IV13 5

a) Find the exact value of tan(v)

b) Find the exact value of sec (u)

c) Find the exact value of sin (u+v)

d) Find the exact value of cos(2v)

e) Find the exact value of tan(u-v)

f) Find the exact value of sin(u/2)

g) Find cos(Arcsin (-5/13))

h) Find 1sin 2cos 3 / 5

6. Prove 2sinx cot x cos x tanx cos x sin x

7a) Find the general solution for the equation 22sin x 2 cosx

b) Find the specific solution for the equation 22sin x 2 cosx in the interval 0, π

8a) Find the general solution for the equation 2sin 2θ 3 0

8b) Find the specific solution for the equation 2sin 2θ 3 0 in the interval 0, 2π

9. At a certain distance, the angle of elevation to the top of a building is 60 degrees. From 40 feet further back the angle of elevation is 45 degrees. How high is the building?

10. Given y 2 4sin(3x π) , find

a) domain

b) amplitude

c) period

d) phase shift

e) vertical shift

f) range

11. Given x

y 6sec2

, find

a) amplitude

b) period

c) phase shift

d) equations of at least 2 consecutive vertical asymptotes

Key to Segment One Exam Practice:

1. x

f(x)x 7

a) Doman: All reals, x not equal to 7 (no zero permitted in the denominator)

b) Vertical asymptote at x = 7

c) Horizontal asymptote at y=1 (degree of numerator=degree of denominator, HA is quotient of leading

coefficients)

d) Range: All reals y not equal to 1.

e)

1

xy

x 7

yx

y 7

xy 7x y

xy y 7x

y(x 1) 7x

7xy

x 1

7xf (x)

x 1

f) Domain of f = Range of f inverse, Range of f = domain of f inverse, Vertical asymptote of f = Horizontal

asymptote of f inverse. Horizontal asymptote of f = Vertical asymptote of f inverse.

g) In order for there to be a hole in the graph of the function g(x), there must be a common factor in the

numerator and denominator. Example: x x 2

g(x)x 7 x 2

produces a hole at x=2

h) In order for there to be a slant asymptote (instead of a horizontal asymptote) the degree of the

numerator must be 1 more than the degree of the denominator, with no common factor. Example: 2x

h(x)x 7

has a slant asymptote at y = x+7 (found by dividing denominator into numerator and

discarding remainder)

2. 5 4f(x) x 2x x 2

a) For a function to have odd symmetry, f (-x) = - f(x)

5 4

5 4

5 4

f( x) x 2 x x 2

x 2x x 2

x 2x x 2

f(x)

b) (+/-) (1, 2) [all factors of constant term divided by all factors of leading coefficient]

c) Since there are 2 sign changes in f(x) there must be 2 or 0 positive real roots.

Since there is 1 sign change in f(-x) there must be 1 negative real root

The Fundamental Theorem of Algebra states that there must be 5 roots possible so the possibilities are:

# Pos Real Roots # Neg Real Roots # Complex Conjugate Roots

Total Number of Roots

2 1 2 5

0 1 4 5

d) Found by factoring by grouping:

5 4f(x) x 2x x 2

4

4

2 2

f(x) x (x 2) (x 2)

x 1 x 2

x 1 x 1 x 2

x 1 x 1 x i x i x 2

x 1)(x 1)(x 2)(x i x i

but this can also be done by testing the possible rational roots (start with a negative possibility since we

know from Descartes' Rule that there must be 1 negative real root) and then continue until the

polynomial is reduced.

e) Zeros of f(x) are x = -1, 1, 2, -i, i

f) x-intercepts of f(x) are the real zeros of f(x): (1,0), (-1,0), (2,0)

g) y intercept is f(0) = 2

3.

2

2

x 2, x 1f(x)

2x 2, x 1

When evaluating a piece function it is important to use the correct piece of the function:

a) f(-2) = 6

b) f(0) = 2

c) f(1) = 4

d) f(2)=10

e) Domain is all reals

f) Range is y 2 (Check the graph or notice that the smallest y value produced is y=2)

g) 4g(x) x 1 1 since 4x 0 so we use the bottom piece of the function:

4

24

8 4

8 4

f g(x)

f x 1

2 x 1 2

2x 4x 2 2

2x 4x 4

4. 2

2 2

f(x) log x 2

g(x) log x log 7

a)

2

2

2

x

x

1 x

f(x) log x 2

y log x 2

x log y 2

y 2 2

y 2 2

f (x) 2 2

b) Domain: x > -2

Range: All reals

c)

2

2 2

2 2 2

2 2 2

2

0

f(x) log x 2

g(x) log x log 7

f(x) g(x) 0

log x 2 log x log 7 0

log x 2 log x log 7 0

x 2log 0

7x

x 22

7x

x 21

7x

x 2 7x

1x

3

d) Using change of base formula: cb

c

log alog a

log b

3

2

3

log x 2log x 2

log 2

e)

1 x

1

x

x

f x 2 2 (from part a)

Solve f x 5

2 2 5

2 7

xLn(2) Ln(7)

Ln(7)x

Ln(2)

5. 5 3

sin(u) ,withu in Quad III and cos(v) , with v in Quad IV13 5

a) Given cos(v)=3/5, means x=3, r=5. Using Pythagorean theorem, y=-4 (negative because in quadrant

IV. You can also draw a right triangle in quadrant IV to see this)

y 4tan v

x 3

b) Given sin(u)=-5/13, means y = -5, r = 13. Using Pythagorean theorem, x= -12 (negative because in

quad III. You can also draw a right triangle in quad III to see this)

r 13 13sec(u)

x 12 12

c) sin(u+v)=sin(u)cos(v)+cos(u)sin(v)

=5 3 12 4 33

13 5 13 5 65

(find each trig function value using same method from part a and b. sin = y/r, cos=x/r)

d)

2

2

cos 2v 2cos v 1

32 1

5

7

25

e)

tan u tan vtan u v

1 tan u tan v

5 4

12 3

5 41

12 3

63

16

f) u 1 cosu

sin2 2

Angle (u/2) is in quadrant II because u is in quadrant III and so sin(u/2) is positive

00

0 0

180 u 270

u90 135

2

121

u 13sin

2 2

25

26

5 26

26

g) cos(Arcsin(-5/13) is asking for the cosine of the angle whose sine is (-5/13). If the sine of the angle is -

5/13, then y=-5, r=13 and x=12.

The cosine value is therefore 12/13 (since cosine is defined as x/r). (note that the angle must be in

quadrant IV because the sine value is negative and the arcsine is only defined in quadrants I and IV)

h) This is asking for the sine of 2 times the angle whose cosine is 3/5. We are working in Quadrant I. We

expand the sine double angle identity and fill in, using x = 3, r = 5 and y = 3 from the definitions of sine

and cosine as y/r and x/r respectively.

1

1 1

sin 2θ 2sin θ cos θ

sin 2cos 3 / 5

2sin cos 3 / 5 cos cos 3 / 5

4 32

5 5

24

25

6.

2

2 2

sin x cot x cos x tan x cos x sin x

cos x sin xsin x cos x

sin x cos x

cos x sin x sin x

cos x sin x cos x sin x

7a)

2

2

2

2

2sin x 2 cos x

2(1 cos x) 2 cos x

2 2cos x 2 cos x

2cos x cos x 0

cos x(2cos x 1)

cos x 0

x n2

2cos x 1

1cos x

2

2x n 2

3

4x n 2

3

b) In the interval 0, , solutions are 2

,2 3

8a. 2sin 2θ 3 0

3sin(2 )

2

4 52 2n , 2n

3 3

2 5n , n

3 6

b) In the interval 0, 2 , 2 5 5 11

, , ,3 3 6 6

9.

0

0

ytan 45

x 40

y1

x 40

x 40 y

ytan 60

x

y3

x

yx

3

y40 y

3

yy 40

3

1y 1 40

3

40y

11

3

450

600

40 ft

y

x

10. y 2 4sin(3x π)

a) domain: all reals

b) amplitude = 4

c) period = 2

3

d) Phase shift:

3x π 0

πx

3

(or π

left3

)

e) vertical shift = 2 (up)

f) range:

y 2 4sin(3x π)

1 sin(3x π) 1

4 4sin(3x π) 4

2 2 4sin(3x π) 6

Range : 2, 6

11. x

y 6sec2

a) Amplitude: none (since secant has no maximum or minimum value there cannot be any amplitude)

b) period: 2π

4π12

c) phase shift: none (no constant added to the argument of the trig function so no phase shift)

d) Asymptotes for secant function occur normally at π 3π

and2 2

(because that is where the cosine

function is equal to 0).

New asymptotes:

π 3π

2 2and1 2 1 2

π and 3π