APPM 1235 Final Exam Fall 2020

INSTRUCTIONS: Simplify and box all your answers. Write neatly and show all work. A correct answer with incorrector no supporting work may receive no credit. Books, notes, electronic devices (such as calculator or unauthorized electronicresources) are not permitted. Give all answers in exact form.

1. Write the word “agree” as your answer to question 1 to indicate that you will abide by the University honor code for thisexam. An honor code violation on the final exam will result in an F in the course.

Agree

2. Simplify the following: (21 pts)

(a) (x− 1)2 − (3x+ 1)(2x− 7)

Solution: (x− 1)2 − (3x+ 1)(2x− 7) = x2 − 2x+ 1−(6x2 − 21x+ 2x− 7

)= −5x2 + 17x+ 8.

(b)8x2y3 − 6x3y

2x2y

Solution:8x2y3 − 6x3y

2x2y=

8x2y3

2x2y− 6x3y

2x2y= 4y2 − 3x.

(c)(8x4y−5

)(1

2x−5

(2y2)3z0)

Solution:(8x4y−5

)(1

2x−5

(2y2)3z0)

=(8x4y−5

)(1

2x−58y6

)= 32x−1y.

(d)

(a1/2c1/3

a−1/2c1/6

)−1

Solution:

(a1/2c1/3

a−1/2c1/6

)−1=(a1/2−(−1/2)c1/3−1/6

)−1=(ac1/6

)−1= a−1c−1/6.

(e)√8x2y8

Solution:√

8x2y8 =√4 · 2x2y8 = 2|x|y4

√2.

(f) log5(25)− e3 ln 2 + log(1)

Solution: log5(25)− e3 ln 2 + log(1) = log5(52)− eln(23) + log(1) = 2− 23 + 0 = −6.

(g)1

a−1 − a1

a−1 + 1

Solution:1

a−1 − a1

a−1 + 1=

1a−1 −

a(a−1)a−1

1a−1 + a−1

a−1=

1−a2+aa−1aa−1

=−a2 + a+ 1

a− 1· a− 1

a=−a2 + a+ 1

a.

3. Solve the following equations for x: (24 pts)

(a) (x− 3)2 = 7

Solution:

(x− 3)2 = 7 (1)

x− 3 = ±√7 (2)

x = 3±√7 (3)

(b)1

x− 2+

3

x+ 2= 1

Solution:

(x− 2)(x+ 2)

(1

x− 2+

3

x+ 2

)= 1(x− 2)(x+ 2) (4)

x+ 2 + 3(x− 2) = x2 − 4 (5)

4x− 4 = x2 − 4 (6)

0 = x2 − 4x (7)

0 = x(x− 4) (8)

This results in potential answers x = 0, 4. Checking these values in the original equation we see that both valuesdo solve the original equation.

(c) 3 =√17− x− x

Solution:

3 =√17− x− x (9)

3 + x =√17− x (10)

(3 + x)2 = 17− x (11)

9 + 6x+ x2 = 17− x (12)

x2 + 7x− 8 = 0 (13)

(x+ 8)(x− 1) = 0 (14)

This results in potential answers x = 1,−8. Checking these values in the original equation, we see that only x = 1solves the original equation.

(d) log3(x) = 1− log3(x− 2)

Solution:

log3(x) = 1− log3(x− 2) (15)

log3(x) + log3(x− 2) = 1 (16)

log3 (x(x− 2)) = 1 (17)

x(x− 2) = 3 (18)

x2 − 2x− 3 = 0 (19)

(x− 3)(x+ 1) = 0 (20)

This results in potential answers x = −1, 3. Checking these values in the original equation, we see that only x = 3solves the original equation.

(e) 3−x−2 = 27

Solution:

3−x−2 = 27 (21)

3−x−2 = 33 (22)

−x− 2 = 3 (23)

−x = 5 (24)

x = −5 (25)

(f) 2x−1 = 5x

Solution:

2x−1 = 5x (26)

log2(2x−1

)= log2 (5

x) (27)

x− 1 = x log2 5 (28)

Letting c = log2 5 we get:

x− 1 = x log2 5 (29)

x− 1 = xc (30)

x− xc = 1 (31)

x(1− c) = 1 (32)

x =1

1− c(33)

x =1

1− log2 5(34)

4. Consider the function f(x) =3x

2x3 − 8x. Answer the following: (16 pts)

(a) Find the domain of f(x)

Solution: The domain is all real numbers except when 2x3 − 8x = 0. Solving this equation we get:

2x3 − 8x = 0 (35)

2x(x2 − 4) = 0 (36)

2x(x− 2)(x+ 2) = 0 (37)

Resulting in values x = −2, 0, 2 that must be removed from the domain. The domain is:(−∞,−2) ∪ (−2, 0) ∪ (0, 2) ∪ (2,∞).

(b) Determine whether f(x) is odd, even, or neither. Justify your answer.

Solution: Beginning with f(−x) we get:

f(−x) = 3(−x)2(−x)3 − 8(−x)

=−3x

−2x3 + 8x=

−3x− (2x3 − 8x)

=3x

2x3 − 8x= f(x)

So f(x) is even.

(c) Find x, y-coordinates of any hole(s). If there are none write NONE.

Solution:3x

2x3 − 8x=

3x

2x (x2 − 4)=

3

2 (x2 − 4)so there is a hole at x = 0. The y-coordinate is found by

plugging x = 0 into3

2 (x2 − 4)and we get

3

2 (02 − 4)= −3

8. So the hole is located at

(0,−3

8

).

(d) Find any horizontal/slant asymptotes. If there are none write NONE.

Solution: There is a horizontal asymptote of y = 0 since3x

2x3 − 8x→ 3x

2x3=

3

2x2→ 0 as x→ ±∞.

5. The following graph has equation of the form y = a(x−h)2+k. Find the equation of the function whose graph is given.(5 pts)

Solution: From the graph the vertex is located at (−3, 1). The vertex of a parabola is found at (h, k) so we get y =a(x − (−3))2 + 1 = a(x + 3)2 + 1. To find a we can use one of the other given points. Plugging in (−1,−2) we get−2 = a(−1 + 3)2 + 1 and solving for a we find:

−2 = a(−1 + 3)2 + 1 (38)

−2 = 4a+ 1 (39)

−3 = 4a (40)

−3

4= a (41)

Resulting in equation: y = −3

4(x+ 3)2 + 1.

6. Sketch the graph of the following functions. Label all intercepts and asymptotes as appropriate. (16 pts)

(a) f(x) = ln(x− 1)

Solution:

(b) g(x) = −√x

Solution:

(c) h(x) = tanx on the restricted domain(−π2,π

2

)Solution:

(d) k(x) = tan−1 x

Solution:

7. Given the following graph, identify a polynomial function that results in the graph. (5 pts)

Solution: The graph crosses at the x-intercept (−1, 0) and touches, but does not cross, at (2, 0). The resulting polynomialhas a zero, x = −1, of odd multiplicity and zero, x = 2, of even multiplicity. A polynomial with these zeros and correctmultiplicity can be written as y = (x + 1)(x − 2)2. However, the end behavior does not match that of the graph. Topreserve the polynomial zeros and multiplicity but achieve the correct end behavior we write: P (x) = −(x+1)(x−2)2.Note this gives the correct y-intercept as well since P (0) = −(0 + 1)(0− 2)2 = −4.

8. You are standing 94 feet from the base of a building. You estimate that the angle of elevation to the top of the 86th floor(the observatory) is 72o. If the total height of the building is another 50 feet above the 86th floor, what is the height ofthe building? Leave your answer in exact form, do not attempt to approximate your answer. (6 pts)

Solution: Let h denote the height to the 86th floor. To find the height to the 86th floor we can use the equation

tan (72o) =h

94. So h = 94 tan (72o). The total height to the top of the building is 94 tan (72o) + 50 ft.

9. Find the exact value: (17 pts)

(a) sin

(11π

6

)(d) arcsin

(sin

(5π

4

))Solution:

sin

(11π

6

)= −1

2arcsin

(sin

(5π

4

))= arcsin

(−√2

2

)= −π

4.

(b) sec

(2π

3

)(e) csc

(cos−1

(1

2

))Solution:

sec

(2π

3

)= −2 csc

(cos−1

(1

2

))= csc

(π3

)=

2√3

.

(c) tan−1 (1)

Solution:

tan−1 (1) =π

4

10. If an angle θ is in the interval(−π2,π

2

)and we know sin θ < 0, what quadrant does θ lie in? (4 pts)

Solution: θ in(−π2,π

2

)means θ is in quadrant I or IV. sin θ < 0 means θ in in quadrants III or IV. To satisfy both

pieces of information θ must be in quadrant IV.

11. Verify the identity:1

1 + cot2 θ+ cos2 θ = 1 (6 pts)

One possible Solution: Starting with the left hand side:1

1 + cot2 θ+cos2 θ =

1

csc2 θ+cos2 θ = sin2 θ+cos2 θ = 1//

giving the right hand side.

12. Find all solutions to the following equations: (10 pts)

(a) 2 sin θ =√3

Solution: Isolating sin θ we get: sin θ =√3

2. This has solutions θ =

π

3+ k2π and θ =

2π

3+ k2π where k is any

integer.

(b) cos (3θ) tan θ − 1

2tan θ = 0

Solution:

cos (3θ) tan θ − 1

2tan θ = 0 (42)

tan θ

(cos (3θ)− 1

2

)= 0 (43)

We get two equations: tan θ = 0 and cos (3θ) =1

2. Let k be any integer. tan θ = 0 has answers θ = k2π

and θ = π + k2π. These can be combined into answers θ = kπ. We find the answers of cos (3θ) =1

2when

3θ =π

3+ k2π and 3θ =

5π

3+ k2π. Solving these for θ we get answers: θ =

π

9+ k

2π

3and θ =

5π

9+ k

2π

3.

13. Find the exact value: (8 pts)

(a) sin(10o) cos(20o) + sin(20o) cos(10o)

Solution: sin(10o) cos(20o) + sin(20o) cos(10o) = sin (20o + 10o) = sin (30o) =1

2.

(b) sin(π8

)

Solution: First note that sin(π8

)is positive since

π

8is in quadrant I. So sin

(π8

)= sin

( π4

2

)=

√1− cos

(π4

)2

=√1−

√22

2=

√2−√2

2

2=

√2−√2

2· 12=

√2−√2

4=

√2−√2

2.

14. For f(x) = 2 cos(πx+

π

2

)(12 pts)

(a) Identify the amplitude.

Solution: Comparing f(x) = 2 cos(πx+

π

2

)to y = a cos (bx+ c), amplitude = |a| = |2| = 2.

(b) Identify the period.

Solution: Comparing f(x) = 2 cos(πx+

π

2

)to y = a cos (bx+ c), period =

2π

|b|=

2π

|π|=

2π

π= 2.

(c) Identify the phase shift.

Solution: Comparing f(x) = 2 cos(πx+

π

2

)to y = a cos (bx+ c), phase shift = −c

b= −

π2

π= −1

2.

(d) Sketch two cycles of the curve.

Solution: The step size is calculated by usingperiod of f(x)

4=

2

4=

1

2. Utilizing the step size, amplitude, and

phase shift we get the graph:

Formulas that may be useful:

1. a3 − b3 = (a− b)(a2 + ab+ b2) 12. a3 + b3 = (a+ b)(a2 − ab+ b2)

2. The quadratic formula: x =−b±

√b2 − 4ac

2a13. Circle: (x− h)2 + (y − k)2 = r2

3. Arc length: s = rθ 14. Area of a sector: A =1

2r2θ

4. sin2 θ + cos2 θ = 1 15. 1 + cot2 θ = csc2 θ

5. tan2 θ + 1 = sec2 θ

6. sin(a− b) = sin a cos b− sin b cos a 16. sin(a+ b) = sin a cos b+ sin b cos a

7. cos(a− b) = cos a cos b+ sin a sin b 17. cos(a+ b) = cos a cos b− sin a sin b

8. cos(2θ) = cos2 θ − sin2 θ 18. sin(2θ) = 2 sin θ cos θ

9. cos(2θ) = 2 cos2 θ − 1 19. cos(2θ) = 1− 2 sin2 θ

10. sin

(θ

2

)= ±

√1− cos θ

220. cos

(θ

2

)= ±

√1 + cos θ

2

11. sin2 (θ) =1− cos(2θ)

221. cos2 (θ) =

cos(2θ) + 1

2