Math - Final

Mx. Haynes @Trinity

Before you continue…

This test will cover all the concepts from Book 2, Problem 123 all the way to Book 3, problem 135.

This was from logarithmic equations to graphing and finding limits of complex graphs involving division (holes, vertical asymptotes, etc).

Test #4

OK

Sequence tip … ദ്ദി(ᵔᗜᵔ)

Arithmetic: a_n = a₁ + d(n-1)

Geometric: a_n = a₁r^n-1

Summation of sequence tip… ദ്ദി(ᵔᗜᵔ)

Arithmetic: S_n = (n / 2)(a₁ + a_n)

Geometric:

S_n = a₁(r)⁰ + a₁(r)¹ + … + a₁(r)^n-1

- (r S_n = a₁(r)¹ + a₁(r)² + … + a₁(r)ⁿ)

1 - r S_n = a₁(r)⁰ - a₁(r)ⁿ

S_n = (a₁ - a₁(r)ⁿ) / (1 - r)

4^2-log₄8

4² / 4^log₄8

16 / 8

2

3^log₂₇8

27^{(1/3)log₂₇8}

27^{log₂₇8^(1/3)}

8^1/3

2

log₈81 * log₃4

(log₃81 / log₃8) × log₃4

((4)log₃4 / log₃8)

log₃256 / log₃8

log₈256

8^x = 256

2^3x = 2⁸

3x = 8

x = 8/3

(log_1/5(625) / log₂(1/8))

log_1/5(1/5)^-4 / log₂(2^-3)

-4 / -3

4/3

log₂(x-2) + log₂(x+1) = 2

log₂((x-2)(x+1)) = 2

2² = x² - x - 2

0 = x² - x - 6

0 = (x-3)(x+2)

x = -2 → log₂(-4) [DNE]

x = 3

Remember to check back in the equation for these problems!

x^log₂x / x² = 8

x^log₂x * x^-2 = x^log_x8

log₂x - 2 = log_x8

log₂x - ((log₂8) / (log₂x)) - 2 = 0

log₂x - (3/log₂x) - 2 =0

(log₂x)² - 3 - 2 log₂x = 0

a² - 2a - 3 = 0

(a-3)(a+1) = 0

a = 3; a = -1

log₂x = 3; log₂x = -1

x = 2³; x = 2^-1

x = 8; x = 1/2

The function f(x) is graphed below.

a) Find an equation for f(x).

b) For what value of x does f(x) = 3?

a) Vertical asymp. → log; x = -5

f(x) = log_a(x+5)

1 = log_a(-2+5)

a = 3

f(x) = log₃(x+5)

b) 3 = log₃(x+5)

3³ = x + 5

27 = x + 5

22 = x

Given the equation b^x = 8b² / 100. Write a simplified expression for the log(b) in terms of x. The answer may contain a logarithm.

log(b^x) = log(8b²/100)

x log b = log(8b²) - log(100)

x log(b) = log 8 + 2 log b - 2

log(b) = (log 8 + 2 log b - 2) / x

Given that log 8 ≈ 0.9 and log₂7 = 2.8, find log 28.

log 7 / log 2 = 2.8

3 log 2 = 9/10

log 2 = 3/10

log(7 × 4) = log(28)

log(7) + 2 log(2) = log(28)

2.8 log(2) + 2 log(2) = log(28)

24/5 log(2) = log(28)

24/5 × 3/10 = 72/50 = 36/25 = log(28)

Test #5

OK

Given the series: -1/4 + 1/16 - 1/64 + …

a) Write the sum of the first 15 terms using sigma notation.

b) Find the sum of the first 15 terms of the series. Leave your answer as an unsimplified expression.

a) Geometric with rate of -1/4 and start term of -1/4; See image.

b) S₁₅ = -1/4(-1/4)⁰ - 1/4(-1/4)¹ - 1/4(-1/4)² - … -1/4(-1/4)¹⁴

- (-1/4 S₁₅ = -1/4(-1/4)¹ - 1/4(-1/4)² - … - 1/4(-1/4)¹⁵)

5/4 S₁₅ = -1/4 + 1/4(-1/4)¹⁵

S₁₅ = 4(-1/4+ 1/4(-1/4)¹⁵) / 5

Find the sum of the following series.

-5 - 2 + 1 + … + 112

Arithmetic with start term of -5 and difference of 3

-5 + 3(n-1) = 112

3(n-1) = 117

n - 1 = 39

n = 40

S_n = (n / 2)(a₁ + a_n)

S₄₀ = (40/2)(-5+112)

S₄₀ = (20)(107)

S₄₀ = 2140

log₉(x²) + log₃(x-2) = 1

( log₃(x²) / log₃(9)) + log₃(x-2) = 1

(2 log₃(x) / 2) + log₃(x-2) = 1

log₃((x)(x-2)) = 1

3 = x² - 2x

0 = x² - 2x - 3

0 = (x-3)(x+1)

x = -1 → log₃(-3) [DNE]

x = 3

b^x-2 = 8b³ / 100^x

log(b^x-2) = log(8) + log(b³) - log(100^x)

(x-2) log b = 3 log(2) + 3 log(b) - 2x

(x-2) log b + 2x = 3 log 2 + 3 log b

a = log b

ax - 2a + 2x = 3 log 2 + 3a

(a+2)x = 3 log 2 + 5a

x = (3 log 2 + 5a) / a + 2

x = (3 log 2 + 5 log b) / log b + 2

Sigma notation tip… ദ്ദി(ᵔᗜᵔ)

Top - bottom + 1 = # of terms

Complete the statement to make the equation correct. (You do not need to find the sum)

5 terms!

x - (-3) + 1 = 5

x + 4 = 5

x = 1 (upper limit)

-4/5 = first term

-5 = rate

k+3 → needs to = 0 when n = -3

-4/5(-5)^k+3

Calculate the following sum. Do not simplify.

3 - 0 + 1 = 4 → 4 terms! (0, 1, 2, 3)

(2(0) - 1) * 2⁰ = -1

(2(1) - 1) * 2^-1 = 1/2

(2(2) - 1) * 2^-2 = 3/4

(2(3) - 1) * 2^-3 = 5/8

5/8 + 6/8 + 4/8 - 8/8 = 7/8

Evaluate the following sum.

Hint: Write out the first few terms of the series and the last few terms to see a pattern.

40 terms

log₃(27 * (1/9)⁰) = 3

log₃(27 * (1/9)¹) = 1

log₃(27 * (1/9)²) = -1

Arithmetic, difference of -2

3 - 2(40 - 1) = -75 (last term)

S_n = (a₁ + a_n)(n/2)

S₄₀ = (-75+3)(40/2)

S₄₀ = (-72)(20)

S₄₀ = -1440

Quiz #3

OK

Write the equation in standard form for the polynomial p(x) that has rational coefficients. p(x) has roots x = 3/2, and 3 + √2, with p(3) = 2.

x = 3 + √2

x - 3 = √2

(x-3)² = 2

x² - 6x + 9 - 2 = 0

x² - 6x + 7 = 0

p(x) = a(2x-3)(x²-6x+7)

2 = a(2(3)-3)(3² - 6(3) + 7)

2 = a(3)(-2)

2 = -6a

-1/3 = a

p(x) = -1/3(2x-3)(x²-6x+7)

p(x) = -1/3(2x³-12x²+14x-3x²+18x-21)

p(x) = -1/3(2x³-15x²+32x-21)

p(x) = -2/3x³ + 5x² - 32/3x + 7

Is x - 3 a factor of h(x) = 3x⁵ - 2x⁴ + x² - 13x + 14? Justify your answer.

No, because 3 is not a factor of 14.

If x²-2x+3 is a factor of the polynomial g(x) = 2x⁴-7x³+3x²+9x-27, find all roots of g(x).

Highest degree → 4 roots for g(x)

Divide g(x) by x²-2x+3 to get 2x²-3x-9

2x²-3x-9 = (2x+3)(x-3)

2x+3 = 0; x = -3/2

x-3 = 0; x = 3

x²-2x+3 = 0

x²-2x+1 = -2

(x-1)² = -2

x - 1 = ±√-2

x = 1 ± √-1 * √2

x = 1 ± i√2

f(x) is a polynomial with integer coefficients and roots of x = 5, x =4, and x = 1 - i√7. What is the lowest possible degree of f(x)? What are the x-intercepts of f(x)? Justify your answer.

4 because degree = # of roots

X-intercepts = roots, excluding imaginary → (5, 0); (4,0)

Match the graph with one of the following equations.

A) -(x-1)²(x+2)(x²+1)

B) -x(x-1)(x+2)

C) (x-1)³(x+2)

D) -(x-1)²(x+2)

Has a root at x = 0

B

Match the graph with one of the following equations.

A) -(x-1)²(x+2)(x²+1)

B) -x(x-1)(x+2)

C) (x-1)³(x+2)

D) -(x-1)²(x+2)

Passes through x = 1 cubically

C

Match the graph with one of the following equations.

A) -(x-1)²(x+2)(x²+1)

B) -x(x-1)(x+2)

C) (x-1)³(x+2)

D) -(x-1)²(x+2)

Point (-1,-4)

-(-1-1)²(-1+2) = (-4)(1) = -4

D

Match the graph with one of the following equations.

A) -(x-1)²(x+2)(x²+1)

B) -x(x-1)(x+2)

C) (x-1)³(x+2)

D) -(x-1)²(x+2)

Process of elimination

A

Test #6

OK

Exponentiated i tip… ദ്ദി(ᵔᗜᵔ)

i¹⁰⁶⁷ = ?

You might be blanking, oh no! Never fear, just divide.

See, i¹ = i, i² = -1, i³ = -i and i⁴ = 1; remainder will give you your answer.

1067 / 4 = 266 R3

i¹⁰⁶⁷ = i³ = -i

As simple as that!

Simplify the following fully. (Any complex answers should be written in standard a + bi form).

i¹ + i² + i³ + … + i³¹⁴

S_n = i(i)⁰ + i(i)¹ + i(i)² + … + i(i)³¹³

- (i S_n = i(i)¹ + i(i)² + i(i)³ + … + i(i)³¹⁴)

(1 - i) S_n = i - i³¹⁵

S_n = (i - i³¹⁵) / 1 - i

315 / 4 = 78 R3; i³¹⁵ = i³ = -i

S_n = 2i / (1-i)

S_n = 2i(1+i) / (1-i)(1+i)

S_n = (-2 + 2i) / 2

S_n = -1 + i

Find the other five trigonometric ratios if csc(x) = -2, π < x < 3π/2

Q3 (sin negative, cos negative)

sin x = -1/2 (reciprocal of csc x)

cos x = -√3/2 (30-60-90)

tan x = 1/2 * 2/√3 (sinx/cosx)

tan x = √3/3

sec x = -(2√3) / 3 (reciprocal of cos x)

cot x = 3√3 / 3

cot x = √3 (reciprocal of tan x)

Write the equation in factored form for the polynomial graphed below. Assume the smallest degree for each root.

Cubically @x=-4, Bounce @x=2, Linear @x=0

ax(x+4)³(x-2)²

-16 = -2a(-2+4)³(-2-2)²

-16 = -2a(8)(16)

-16 / (-16)(16) = a

1/16 = a

y = 1/16x(x+4)³(x-2)²

Verify the following identity: (sin(x) + cos(x)) / (sec(x) + csc(x)) = sin(2x) / 2

(sinx+cosx)(secx-cscx) / (secx+cscx)(secx-cscx)

(sinx+cosx)(1/cosx-1/sinx) / (secx+cscx)(secx-cscx)

(sinx/cosx - cosx/sinx) / (sec²x - csc²x)

(sinx/cosx - cosx/sinx) / (1/cos²x - 1/sin²x)

(sin²x - cos²x / sinxcosx) / (sin²x-cos²x / sin²xcos²x)

sin²xcos²x / sinxcosx = sinxcosx

2sinxcosx = sin(2x) → sinxcosx = sin(2x) / 2

Sketching the reciprocal of a graph

x → -∞, y → 3

x → -4, y → 0

x-intercepts of the graph are __ of the reciprocal

vertical asymptotes of the graph with both sides going same way are __ of the reciprocal

What is important to note about select points?

x → -∞, y → 1/3

x → -4, y → ∞ (y gets very small in graph, so its reciprocal gets very big)

vertical asymptotes

x-intercepts (usually ends up forming a parabola as a result

Points at y = 1 or y = -1 on the graph keep the same position in the reciprocal graph! This is because the reciprocal of 1 or -1 is still 1 or -1. :)

Give an equation of a polynomial p(x) that satisfies all of the following properties.

• p(x) is an odd function

• lim_x→∞p(x) = ∞

• x = -2 is a root

• x = -i is a root

Odd function → -p(x) = p(-x)

if p(-2) = 0, then -p(2) = 0… so p(2) = 0; same with i and -i

-p(0) = p(-0); (0,0) must be included

p(x) = ax(x-2)(x+2)(x-i)(x+i)

p(x) = ax(x²-4)(x²+1)

p(x) = ax(x⁴-3x²-4)

p(x)=a(x⁵-3x³-4x)

x+2 is rightmost root, so test with 3?

3⁵-3(3)³-4(3) = 243 - 81 - 12 = 150

^ Since it’s positive, a just needs to be greater than 0.

p(x) = x⁵-3x³-4x

Quiz #4

OK

Horizontal asymptote tip… ദ്ദി(ᵔᗜᵔ)

If degree_num > degree_denom, most likely a slant asymptote. Divide to find.

If degree_num = degree_denom, horizontal asymptote @y = a/b, where a is the coeff of num highest degree and b is the coeff of denom highest degree.

If degree_num < degree_denom, horizontal asymptote @y = 0.

A

Fill in tomorrow when you can ig

Tricky Book Problems

OK

Interest tip… ദ്ദി(ᵔᗜᵔ)

Regular Compounding

P(1 + r/t))^nt

Compounding Continuously

Pe^rt

t = amount of compounding

r = rate of change

n = number of years

p = principal amount

Suppose you invest $100 at 12% interest compounded semi-annually. This means that each half-year, your money grows by 6%. What happens if the interest is compounded quarterly?

P(1 + r/t))^nt

t = amount of compounding

r = rate of change

n = number of years

p = principal amount

100(1 + 0.12/4)⁴⁽¹⁾

100(1.03)⁴ = $112.55

Simplify log 2 + log 3 + log 4 + … + log n.

log(1) = 0, so…

log(n!)

If a = log₂3 and b = log₂5, what is log₂(32/25) in terms of a and b.

log₂(2⁵) - log₂(5²)

5 - 2 log₂(5)

5 - 2b

If log₉5 = x and log₉4 = y, write the following in terms of x and y.

a) log₉(10)

b) log₉(0.2)

a) log₉5 + log₉2

2 log₉2 = y; y = log₉2 / 2

x + y/2

b) log₉(1/5)

log₉(5^-1) = -log₉(5)

-x

Rewrite as a sum or difference of logarithms. Leave no exponents or radicals.

log(a³√(b²/c))

log(a³) + log((b²/c)^1/2)

log(a³) + log(b/c^1/2)

3 log a + log b - ½ log c

Suppose you invest $10,000 at an annual rate of 9% and that interest is compounded continuously. How much money would you have at the end of the year?

Pe^rt

t = amount of compounding

r = rate of change

n = number of years

p = principal amount

10000(e)^{0.09 × 1}

$10941.74

Solve for x (this applies to the next couple of problems).

log₆(log₂(x)) = 1

log₂(x) = 6

x = 2⁶

x = 64

x = log(log 10¹⁰⁰)

x = log 100

x = 2

log_x(0.0016) = -4

1/x⁴ = 16/10000

1/x⁴ = 4/2500

1/x⁴ = 1/625

x⁴ = 625

x = 5

x = log₇(log₅(log₂32))

x = log₇(log₅(5))

x = log₇(1)

x = 0

(log₈x)² - 3log₈x + 2 = 0

a² - 3a + 2 = 0

(a-2)(a-1) = 0

log₈x = 2; log₈x = 1

x = 8²; x = 8¹

x = 64; x = 8

log₄(x+2) + log₄(x-4) = 2

log₄((x+2)(x-4)) = 2

16 = x²-2x-8

0 = x²-2x-24

0 = (x-6)(x+4)

x = -4 [log₄(-2) → DNE]

x = 6

6^log₆x² = x + 30

x² - x - 30 = 0

(x-6)(x+5) = 0

x = 6; x = -5

log₃(2x+1) = log₃(3x-6)

2x + 1 = 3x - 6

-x = -7

x = 7

log₉x = (1/2) log₉144 - (1/3) log₉8

log₉x = log₉12 - log₉2

log₉x = log₉(12/2)

log₉x = log₉6

x = 6

What is the sum of 1 + 2 + 3 + … + 100?

(First + last)(# terms / 2)

(50)(101) = 5050

Solve each equation or inequality (this applies for the next few problems).

log₂(x-1) ≤ 3

x-1 ≤ 8

a^x can never be negative or 0… so:

1 < x ≤ 9

2 log²(x) - log(x³) = 2

2 log²x - 3 log x - 2

a = log x

2a² - 3a - 2 = 0

(2a+1)(a-2) = 0

a = -1/2 → log x = -1/2 → x = 10^-1/2 → x = (√10)/10

a = 2 → log x = 2 → x = 10² → x = 100

√(log(x) - 1) = log √x

(log(x)-1)^1/2 = 1/2 log x

y = log x

(y - 1)^1/2 = 1/2y

y - 1 = 1/4y²

-1/4y² + y - 1 = 0

-y² + 4y - 4 = 0

(-y+2)(y-2) = 0

y = 2 → log(x) = 2 → x = 100

x^3-log₇x = 49

(7^log₇x)^3-log₇x = 7²

log₇x(3 - log₇x) = 2

3 log₇x - (log₇x)² -2 = 0

a = log₇x

-a² + 3a - 2 = 0

(-a+2)(a-1) = 0

a = 2 → log₇x = 2 → x = 49

a = 1 → log₇x = 1 → x = 7

(4^x-1) / (2^x + 1) = 5

(2^2x-1) / (2^x+1) = 5

(2^x+1)(2^x-1) / (2^x+1) = 5

2^x-1 = 5

2^x = 6

log₂(6) = x

log₃(x²-2x) ≤ 1

x²-2x ≤ 3

x²-2x-3 ≤ 0

(x-3)(x+1) ≤ 0

x-3 ≤ 0 → x ≤ 3

x+1 ≤ 0 → x ≥ -1

x²-2x ≠ 0

x(x-2) ≠ 0

x ≠ 0; x ≠ 2

[-1,0) U (2, 3]

Is (x-3) a factor of f(x) = x³ - 4x² + 5? Justify your answer.

No, because 3 is not a factor of 5.

Calculate the magnitude of |3+2i|.

√(3² + 2²) = √13

Graphing complex numbers

Regular numbers: x-axis, complex numbers: y-axis

3-2i → (3,-2)

Simplify i⁶⁷⁶⁹

i = i; i² = -1; i³ = -i; i⁴ = 1

6769 / 4 = 1692 R1

i³²⁴³ = i

Graphing tip… ദ്ദി(ᵔᗜᵔ)

You can use a sign chart! Find the x-intercepts and then plug in something less (or more) than that and see what the signs result to.

Multiplicity of 1: Passes through linearly

Multiplicity of 2: Bounces

Multiplicity of 3: Passes through cubically

Types of functions tip… ദ്ദി(ᵔᗜᵔ)

Even: f(x) = f(-x)

Odd: f(-x) = -f(x)

sin(2x) / (1 + sin(2x)) = 2/((1+tanx)(1+cotx))

2sinxcosx / (1 + 2sinxcosx)

2sinxcosx / sin²x + cos²x + 2sinxcosx

2sinxcosx / (sinx+cosx)²

2 / ((sinx+cosx)² / sinxcosx)

2/(sinx+cosx / sinx)(sinx+cosx/cosx)

2 / (1+cotx)(tanx+1)

Identities

OK

sin²x + cos²x = ?

1

tan x = ?

sin x / cos x

sin(2x) = ?

2sinxcosx

cos(2x) = ? (x3)

1 - 2 sin²x

2 cos²x - 1

cos²x - sin²x

Review Guide

OK

Given that sin x = (√5)/5 and cos x < 0, find the exact value in simplest radical form of the remaining five trigonometric functions of x.

sin²x + cos²x = 1

5/25 + cos²x = 1

cos²x = 20/25

cos x = ±(2√5)/5 … cos x < 0

tan x = sinx/cosx = √5/5 / (-2√5)/5 = -1/2

cos x = -(2√5)/5

tan x = -1/2

csc x = 5√5 / 5 = √5

sec x = -(√5)/ 2

cot x = -2

If sin x = 0.3, find the value of:

sin(-x) + sin x + sin(x + π) + sin(x + 2π) + sin(x + 3π)

Draw triangles.

sin(-x) = -0.3

sin(x+π) = (√91)/10

sin(x+2π) = -0.3

sin(x+3π) = -(√91)/10

-0.3 + 0.3 + (√91)/10 - (√91)/10 - 0.3 = -0.3

Find the exact values of x on the interval [0, 2π).

a) sec x = -2

b) 2 sin x = -√3

c) cot x = -1

a) cos x = -1/2

x = 2π/3, 4π/3

b) sin x = -(√3)/2

x = 4π/3, 5π/3

c) tan x = -1

x = 3π/4, 7π/4

Consider the trigonometric function f(x) = 3 sec(x/4) + 2.

a) Draw a sketch of f(x) on the interval (-6π, 6π).

b) Evaluate f(4π).

c) Determine the exact value(s) of x for which f(x) = -4 on the interval (-6π, 6π).

a) 2π / 1/4 = 8π (Period)

1/cosx (Reciprocal graph: parabola down, parabola up, parabola down)

Amplitude of 3 (Multiply y-vals by 3)

Vertical Translation Up 2 (Add 2 to y-vals)

b) See graph…

f(4π) = -1

c) 3 sec(x/4) + 2 = -4

sec(x/4) = -2

cos(x/4) = -1/2

x/4 = -4π/3, -2π/3, 2π/3, 4π/3

x = -16π/3, -8π/3, 8π/3, 16π/3

Determine the domain of g(x) = (√3x+4) / (2x² - x -15).

3x+4 ≥ 0; x ≥ -4/3 … -8/6

(2x+5)(x-3); x ≠ 3, x ≠ -5/2 … -15/6

Domain: [-4/3, 3) U (3, ∞)

f(x) is an odd function and g(x) is an even function.

a) Complete the table.

x: -4, -2, 0, 2, 4

f(x): 2, _, 0, -2, _

g(x): 5, -6, -2, _, _

b) f(-2) + g(2)

c) g(f(-4))

d) (f(4)*g(0)) / (g(-2) - f(-2))

a) f(-x) = -f(x); g(x) = g(-x)

f(-2) = -f(2) = 2

f(-4) = -f(4) = 2; f(4) = -2

g(2) = g(-2) = -6

g(4) = g(-4) = 5

b) 2 - 6 = -4

c) g(2) = -6

d) -2 * -2 / -6 - 2 = 4/-8 = -1/2

Given f(x) = (2x+3) / (5x+2), find the inverse f^-1(x) and determine the domain and range of both f(x) and f^-1(x).

5x + 2 ≠ 0

x ≠ -2/5

Domain f(x) = Range f^-1(x): (-∞, -2/5) U (-2/5, ∞)

x = (2y+3) / (5y+2)

5xy+2x = 2y+3

5xy-2y = -2x+3

y(5x-2) = -2x+3

f^-1(x) = (-2x+3) / (5x-2)

5x - 2 ≠ 0

x ≠ 2/5

Domain f^-1(x) = Range f(x): (-∞, 2/5) U (2/5, ∞)

sin x + (csc x / sec²x) = csc x

sin x + (cos²x / sin x)

(sin²x + cos²x) / sin x

1/sin x

csc x

(cosx / 1+sinx) + (1+sinx / cosx) = 2 sec x

cos²x / cosx(1+sinx) + (1+sinx)(1+sinx) / cosx(1+sinx)

(cos²x + sin²x + 2sinx + 1) / cosx(1+sinx)

2 + 2 sinx / cosx(1+sinx)

2(1+sinx) / cosx(1+sinx)

2/cosx

2 sec x

(1+sinx / sinx) + (cotx - cosx / cosx) = 2 csc x

(cosx(1+sinx) / sinxcosx) + (sinx(cotx - cosx) / sinxcosx)

(cosx + sinxcosx / sinxcosx) + (cosx - sinxcosx / sinxcosx)

2 cosx / sinxcosx

2/sinx

2 csc x

lim_x→2(x²-x-2)/(x²-4)

(x-2)(x+1) / (x-2)(x+2)

hole @ (2, 3/4)

3/4

lim_x→2(x²+4x+4)/(x²-4)

(x+2)(x-2)/(x-2)(x+2)

hole @(-2,0)

1+2 / 1-2 = 3/-1 = -3

3+2 / 3-2 = 5/1 = 5

DNE [diff signs]

lim_x→∞(x²-4)/(3x²+x+1)

infinite limit, num & denom are = so highest degree coeffs

1/3

lim_x→1(x³+2^×2-2x-1)/(x²+2x-3)

(x-1)(x²+3x+1)/(x-1)(x+3)

hole @ (1, 5/4)

5/4

log_x64 = 2

x² = 64

x = ±8 (logarithms are positive only)

x = 8

log₃(4x-7) = 2

4x - 7 = 9

4x = 16

x = 4

log₅(x+6) + log₅(x+2) = 1

log₅((x+6)(x+2)) = 1

(x+6)(x+2) = 5

x² + 8x + 12 = 5

x² + 8x + 7 = 0

(x+7)(x+1) = 0

x = -7 [DNE]; x = -1

(Plug back into equation and see if logs remain positive.)

16^x + 4^x+1 - 21 = 0

4^2x + 4^x * 4 - 21 = 0

a = 4^x

a² + 4a - 21 = 0

(a+7)(a-3) = 0

a = -7 → 4^x = -7 → x = log₄(-7) = DNE

a = 3 → 4^x = 3 → x = log₄(3)

x: -2, -1, 0, 1, 2

f(x) = 2^x: _, _, _, _, _

x: _, _, _, _, _

g(x) = log₂x: -2, -1, 0, 1, 2

x: _, _, _, _, _

y = 2f(x+4): _, _, _, _, _

1/4, 1/2, 1, 2, 4

1/4, 1/2, 1, 2, 4

x: -6, -5, -4, -3, -2

y = 2f(x+4): 1/2, 1, 2, 4, 8

x: 1, 2, 4, 8, 16

y = g(x/4)+3: 1, 2, 3, 4, 5

f(x) = (2(x+3)²(x-3)(x-1)²) / ((x+2)(x-2)²(x-1)²(x-5))

a) lim_x→-2^-f(x) = ?

b) lim_x→-2⁺f(x) = ?

c) lim_x→-2f(x) = ?

d) lim_x→2f(x) = ?

e) lim_x→5f(x) = ?

f) lim_x→1f(x) = ?

g) lim_x→∞f(x) = ?

h) lim_x→-∞f(x) = ?

a) (2(-4+3)²(-4-3)(-4-1)²) / ((-4+2)(-4-2)²(-4-1)²(-4-5))

2(1)(-7)(25)/(-2)(36)(25)(-9) = (-)/(+) = (-)

lim_x→-2^-f(x) = -∞

b) (2(-1+3)²(-1-3)(-1-1)²) / ((-1+2)(-1-2)²(-1-1)²(-1-5))

2(4)(-4)(4)/(1)(9)(4)(-6) = (-)/(-) = (+)

lim_x→-2⁺f(x) = ∞

c) two values

lim_x→-2f(x) = DNE

d) no hole, vertical asymptote

2(2+3)²(2-3)(2-1)² = 2(25)(-1)(1) = -50

2^-: (1.5+2)(1.5-2)²(1.5-1)²(1.5-5) = (+)(-)²(+)²(-) = (+)(-) = (-)

2⁺: (3+2)(3-2)²(3-1)²(3-5) = (5)(1)²(2)²(-2) = -40 → as it approaches 2, - gets smaller and smaller in the denominator; since both are -, this means that there IS a limit.

big negative / small negative = -∞

lim_x→2f(x) = ∞

e) no hole, vertical asymptote

2(5+3)²(5-3)(5-1)² = 2(64)(2)(16) = 4096

5^-: (4+2)(4-2)²(4-1)²(4-5) = (+)(+)²(+)²(-) = (-)

5⁺: (6+2)(6-2)²(6-1)²(6-5) = (+)(+)²(+)²(+) = (+)

Different directions

lim_x→5f(x) = DNE

f) hole

2(1+3)²(1-3) / (1+2)(1-2)²(1-5) = 2(4)²(-2) / (3)(-1)²(-4) = -64 / -12 = 16/3

lim_x→2f(x) = 16/3

g) 2(big+3)²(big-3)(big-1)² / (big+2)(big-2)²(big-1)²(big-5)

2(big-3) is WAY less than (big+2)(big-5)

small / big means it approaches zero

lim_x→∞f(x) = 0

h) 2(-big+3)²(-big-3)(-big-1)² / (-big+2)(-big-2)²(-big-1)²(-big-5)

Still… 2(-big-3) is WAY less than (-big+2)(-big-5)

small / big means approaching zero

lim_x→-∞f(x) = 0

Identify any x-intercepts, vertical asymptotes, holes, and end behavior of the graph h(x) = (x³-3x²-6x+8) / (x²-7x+12)

h(x) = (x-4)(x²+x-2) / (x-4)(x-3)

h(x) = (x-4)(x+2)(x-1) / (x-4)(x-3)

hole at (4, 18)

vertical asymptote at x=3

x-ints: (-2,0); (1,0)

y-int: (0, 2/3)

x³/x² → num>denom; slant asymptote… divide to find

slant asymptote at y = x + 4