AP Precalculus Guide (new version)

Unit 1

1.1 - Change in Tandem

A function is a mathematical relation that maps a set of input values to a set of output values such that each input value is mapped to exactly 1 output value

Positive Function - the output values are above 0

Negative Function - the output values are below 0

Increasing function if…

1. Verbally: as the input values increase, the output values always increase

2. Analytically: for all a and b in the interval ___, if a < b, then f(a) < f(b)

Decreasing function if…

1. Verbally: as the input values increase, the output value always decreases

2. Analytically: for all a and b in the interval ___, if a < b then f(a) > f(b)

Zero: when the output value is zero

Concave up: bowl facing UP → slope increasing

Concave down: bowl facing DOWN → slope decreasing

Point of Inflection: point where the concavity changes

Justification on stating whether or not the function is increasing or decreasing…

The function f is increasing on the interval 4 < x < 10 or (4, 10) because for all a and b values, if a < b, then f(a) < f(b)

1.2 - Rates of Change

rate of change = slope

Rate of Change of a point

Ex. Estimate the rate of change at x = 1 for the function f(x) = -½x² + 3x - ½

1. Get as close as you can to the point (at least 3 decimal places)

   • (1, 2) & (1.001, 2.0019995)

2. Now calculate the slope

   • (2.0019995 - 2) ÷ (1.001 - 1) = 1.9995 ≈ 2

Positive ROC - indicates that as one quantity increases or decreases the other quantity does the same (same as if it were to say a function is increasing)

Negative ROC - indicates that as one quantity increases, the other decreases (same as if it were to say a function is decreasing)

1.3 - Rates of Change in Linear and Quadratic Functions

Average Rate of Change = Slope of a SECANT Line (calculates overall change in a quantity across a given interval)

average rate of change = f(a) - f(b) / a - b

The average rate of change for a linear function is constant. Regardless of the input-value interval length, the average rate of change always stays the same

• Does not matter where you pick your points, the slope is always the same

NOTE: linear functions only have an increase/decrease in the y values, meanwhile quadratics have an increase/decrease in the average rate of change (the amount of differences is the degree)

• The amount of differences you get from this indicates the degree. 1 is linear, 2 is quadratic, 3 is cubic, 4 is quartic, 5 is quintic, etc.

The average rate of change for a quadratic doesn’t stay the same. For consecutive equal-length input-value intervals, the average rate of change of the rate of change of a quadratic function is constant.

here the x is going up by one (15 +1 = 16; 16 +1 = 17)

The rate of change of the rate of change: what is the change in the slope/ROC (This will always be 0 for linear functions)

1. here the bottom is varying (18 + 2 = 20; 20 + 0 = 20; 20 - 2 = 18; 18 - 4 = 14)

2. Then you would do that again to find a commonality (+2, 0, -2, -4) → ( 0 - 2 = -2; -2 - 0 = -2; -4 - (-2) = -2)

3. Which means the rate of change of the rate of change is -2

The function is concave down because the rate of change is decreasing over equal length input intervals

Example problem on quiz:

• The function p is given by p(x) = g(x + 1) - g(x). If p(x) = 2, which of the following statements must be true?

• g(x + 1) - g(x) is really the slope (y2 [which is f(x + 1)] - y1 [which is f(x)]) / 1

• p(x) = 2 means that the slope is 2

• This means the graph always has a positive slope and that because p is positive and constant, g is increasing

Confusing Terms

Increasing / Decreasing ROC → The graph is concave Up / Down

Positive / Negative ROC → The function is Increasing / Decreasing (± slope; imagine a positive or negative linear line)

1.4 - Polynomial Functions and Rates of Change

Polynomial

p(x) = a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + … + a₁x² + a₁x + a₀

a_nxⁿ is the leading term | n is the degree | a_n is the leading coefficient

Local / Relative Maximum - if the polynomial switches from increasing to decreasing

Local / Relative Minimum - if the polynomial switches from decreasing to increasing

an included endpoint of a polynomial with a restricted domain may also be the local minimum or maximum

Global Maximum - the greatest of the local maximums (If it goes to infinity, there is none)

Global Minimum - the least of the local minimums

Two Zeros = if you have 2 zeros of a polynomial, there must be at least 1 local extrema between the two

Even degree = absolute extreme

• Positive leading coefficient = absolute minimum

• Negative leading coefficient = absolute maximum

1.5A - Polynomial Functions and Complex Zeros & 1.5B - Even and Odd Polynomials

If the real zero, a, of (x - a), has an even multiplicity, then the graph will bounce off the x-axis

Example:

• Given the function: (x + 1)²(x - 3)³(x + 4)¹

Complex Zeros - non-real zeros that will always come in pairs (Ex. 3 ± 2i)

• a ± bi

Even functions = symmetrical over the y-axis

• has the property f(-x) = f(x)

• You can prove that the function is even by substituting in -x and see if you get the same original function

  • Ex. f(x) = x⁶ - 4x² → f(-x) = (-x)⁶ - 4(-x)² → f(-x) = x⁶ - 4x²

Odd functions = symmetrical over the origin

• has the property f(-x) = -f(x)

1.6 - Polynomial Functions and End Behavior

The left or right side of a polynomial will either go up or down.

The limit expresses the end behavior of a function.

The Left Side

• The x values are approaching negative ∞

• lim x → -∞ f(x)

The Right Side

• The x values are approaching positive ∞

• lim _{x → ∞} f(x)

Even degree means that the left and right side will behave the same

Odd degree means that the left and right side will behave opposite

1.7 - Rational Functions & End Behavior

Rational Function - the ratio of two polynomials where the polynomial in the denominator cannot equal 0

• Usually can be described as:

Labeling some of the properties of the function:

• Domain: (-∞, -3) U (-3, 3) U (3, ∞) There is a hole and vertical asymptote

• lim_{x → -∞} f(x) = 0 | lim_{x → ∞} f(x) = 0

  • 0 is the horizontal asymptote because both limits approach this number

For inputs with a larger magnitude, the polynomial in the denominator dominates the polynomial in the numerator

Ways to remember:

• BOBO BOTNA EATS DC

  • BOBO - Bigger on bottom = 0 (BOB0)

  • BOTNA - Bigger on top = No Asymptote (BotNA)

  • EATS DC - Exponents Are The Same = Divide Coefficients (EATSDC)

1.8 - Rational Functions & Zeros

When there is an unfactored polynomial in a rational function, try to factor both numerator and denominator to make it easier to see holes and asymptotes

This is a helpful equation when working with a polynomial that is both factored on the numerator and denominator: (x-int)(hole) / (hole)(vertical asymptote)

• Example function: (x + 3)(x - 2) / (x + 4)(x - 2)

• X intercepts (zeros) will be any factors on the numerator that aren’t holes

  • (x + 3); so that means that there is an x-intercept at x = -3

• Holes will appear on both the numerator and denominator (This is the value you cannot plug into the equation)

  • (x - 2) / (x - 2); meaning that x = 2 does not exist

• Vertical asymptotes are any factors that aren’t holes left in the denominator

  • (x + 4); so that means that there is a vertical asymptote at x = -4

1.9 - Rational Functions & Vertical Asymptotes

Finding the limit as x approaches a number/asymptote

• 2^- means approaching the vertical asymptote on the left side from left to right

  • lim_{x → 2^-} f(x) = -∞

• 2⁺ means approaching the vertical asymptote on the right side from right to left

  • lim_{x → 2⁺} f(x) = ∞

Order of dominance

1. Hole > 0

   • If there is a zero with a hole, then there will be no zero, but a hole instead

     • Ex. (x + 1)(x + 1) / (x + 1)(x + 2) means that there will be a hole at x = -1 instead of a zero because there is a hole of (x + 1)

2. Vertical Asymptote > Hole

   • If there is a vertical asymptote the same as a hole, then that hole will be a vertical asymptote

     • Ex. (x + 1) / (x + 1)(x + 1) means that there is a vertical asymptote at x = -1 instead of a hole because there is a vertical asymptote of (x + 1)

1.10 - Rational Functions & Holes

Writing limits for holes:

• From the left:

  • As x approaches 3 from the left, f(x) approaches 2

  • lim_{x → 3^-} f(x) = 2

• From the right:

  • As x approaches 3 from the right, f(x) approaches 2

  • lim_{x → 3⁺} f(x) = 2

Using a Table to Determine Limits

As you can see, the closer we get to 3, the closer the values get to 2

1.11A - Equivalent Representations & Binomial Theorem & 1.11B - Polynomial Long Division & Slant Asymptotes

Using pascal’s triangle, you can expand binomials. The rows of pascal’s triangle start from 0 (the top) and so on.

For example, if you had (x + 3)⁴ then the row you would use would be → 1 4 6 4 1

Binomial Theorem:

• The formula is (x + y)ⁿ where n would be the row #

  • (a+b)ⁿ = C₁aⁿb⁰ + C₂a^{n - 1}b¹ + C₃a^{n - 2}b² + … + C_{n - 1}a¹b^{n - 1} + C_na⁰bⁿ

  • C is the coefficient corresponding to the row in pascal’s triangle

  • n is the power

• Using the numbers in pascal’s triangle, expand the polynomial

• As the power of x decreases, the power of y increases

• Ex. (x + 5)³

  • 1x³5⁰ + 3x²5¹ + 3x¹5² + 1x⁰5³ → x³ + 15x² + 75x + 125

Slant Asymptote - When the degree of the numerator is one higher than the degree of the denominator

End behavior of slant asymptote:

• Finding the asymptote of the slant asymptote with long division or synthetic division will give you the linear equation. The end behavior of that function (in this case, y = x + 1) is the end behavior of the rational function

• lim_{x → ∞} f(x) = ∞

• lim_{x → -∞} f(x) = -∞

Long Division

• Long division can be used to divide any polynomial by another one

Synthetic Division:

• Synthetic division can only be used when dividing a polynomial by a factor like (ax + b)

1.12 - Transformations of Functions

Vertical Translations (anything done outside of the function)

• f(x) + d → up by d units

• f(x) - d → down by d units

Vertical Dilations (anything done outside of the function)

• a × f(x)

  • When doing a vertical dilation, only the y-values change, the x-values do not

  • Multiply the y-values by a

  • a > 1: vertical stretch

  • a < 1: vertical shrink

    

Horizontal Translations (anything done inside the function)

• f(x - d) → right by d units

• f(x + d) → left by d units

Horizontal Dilations (anything done inside the function)

• f(bx)

  • b is inverted (if it was f(3x) that would be a horizontal dilation of 1/3)

  • When doing a horizontal dilation, only the x-values change, the y-values do not

  • Multiply the x-values by the inverse of b

  • 1/b > 1: horizontal shrink

  • 1/b < 1: horizontal stretch

    

Reflections

• -f(x) → reflect over x-axis → vertical change

• f(-x) → reflect over y-axis → horizontal change

Transformations Numerically

Ex. let g(x) = 3f(x - 2) + 1, find g(3)

1. g(x) = 3f(x - 2) + 1 → set up equation

2. g(3) = 3f(3 - 2) + 1 → plug in the 3 for x in the modified equation

3. g(3) = 3f(1) + 1 → simplify

4. g(3) = 3(-12) + 1 → find what f(1) is and substitute that in

5. g(3) = -35 → solve

Transformations through domain & range

Ex. Given the graph for f has a domain of [-4, 3] and a range of (3, 9]. Let g(x) = -f(x + 5) + 2.

f(x)

• Domain: [-4, 3]

• Range: (3, 9]

g(x)

• Domain: [-4 - 5, 3 - 5] → [-9, -2]

  • f(x + 5) → move left 5 → subtract 5 from the domain

• Range: (3 * -1, 9 * -1] → (-3 + 2, -9 + 2] → (-1, -7]

  • Do the inversions and dilations first then the translations

Transformations Algebraically

Ex. Given f(x) = x² - 3x + 2, let g(x) = f(x - 3) + 2, find g(x)

• g(x) = f(x - 3) + 2 → set up equation

• [(x - 3)² - 3(x - 3) + 2] + 2 → substitute the input values in g(x) for x in f(x)

• (x² - 6x + 9 - 3x + 9 + 2) + 2 → distribute

• (x² - 9x + 20) + 2 → combining like terms

• g(x) = x² - 9x + 22 → solved

1.13 - Function Model Selection

Perimeter will be linear

Area will be quadratic

Volume will be cubic

Restricted Domain & Range

• Restricted domain and range is give between brackets and is usually given in a word problem or between a piece of a function

Piecewise Functions

• Range of a piecewise function

  

1.14 Function Model Construction

Types of Regression

• Linear

• Quadratic

• Cubic

• Quartic

• Exponential

• Logarithmic

• Logistic

• Sine

Inversely Proportional

• When something is inversely proportional, the equation y = k / x is used

  • k is some constant

• Ex. The number of workers at a job is inversely proportional to the time it takes to complete a task. If there are 15 workers and it takes 20 minutes to do a task, then how many workers would it take to complete a task in 3 minutes?

  • n = k / t → number of workers = k / time

  • 15 = k / 20 → set up equation to solve for k

  • k = 15 × 20 = 300 → solve for k

  • n = 300 / 3 → input k into new equation to solve for workers

  • n = 100 workers → it would take 100 workers to complete a task in 3 minutes

Piecewise Functions Algebraically

Piecewise functions can be shown like the image above. The function on the left is in the bounds of the inequality on the right.

Unit 2

2.1 - Change in Arithmetic and Geometric Sequences

Sequence → an ordered list of numbers; each listed number is a term

• Ex. {1, 3, 5, 7, 9, …} (… means it keeps going continuously)

  • 1 would be the first term not the zero term

Arithmetic Sequences

• Linear

• If each successive term in a sequence has a common difference (constant ROC), the sequence is arithmetic

• Equation for the nth term → a_n = a₀ + dn

  • a₀ is the initial value (zero term) and d is the common difference

    

• Equation for any term → a_n = a_k + d(n - k)

  • a_k is the k^th term of the sequence

  

Geometric Sequence

• Exponential

• If each successive term in a sequence has a common ratio (constant proportional change) the sequence is geometric

  • If dividing is involved (for example, {4, 2, 1, …} is ÷2 each time), then the ratio will be multiplied by a fraction (÷2 → ×1/2)

• Equation for the nth term → g_n = g₀rⁿ

  • g₀ is the initial value (zero term) and r is the common ratio

    

• Equation for any term → g_n = g_kr^{(n - k)}

  • g_k is the k^th term of the sequence

    

2.2 - Change in Linear and Exponential Functions

Linear Functions vs Arithmetic Sequences

• They are basically the same, except when using a sequence vs a function, the domain and points are discrete, meaning the domain only includes specific points and not all numbers in between

• The point (x₁, y₁) of a linear function is similar to (k, a_k) of an arithmetic sequence

Exponential Functions vs Geometric Sequences

• Again, they are basically the same except with discrete points for the sequence

• The point (x₁, y₁) of an exponential function is similar to (k, g_k) of a geometric sequence

How each output value changes analytically

• Arithmetic → Over equal-length input value intervals, if the output values of a function change at a constant rate, then the function is linear (adding the slope)

• Geometric → Over equal-length input value intervals, if the output values of a function change proportionally, then the function is exponential (multiplying the ratio)

• “Over equal-length input intervals” means that the x-values of the function or a table are increasing by a constant amount (ex. 1, 2, 3, 4 or 2, 4, 6, 8)

Finding whether a set of points is linear or exponential

1. Make sure the function is increasing over equal-length input intervals

2. Test to see whether the function is linear or exponential

   1. Linear functions change by adding or subtracting a constant amount

   2. Exponential functions change by multiplying or dividing a constant amount

3. This example is increasing x2 from 8-16 and 16-32

4. So this function is exponential because over equal-length intervals of 1, f(x) proportionally changes by a ratio of 2

   1. If this was linear you could say, the function is linear because over equal-length input intervals of ___, f(x) has a constant rate of change of ___

Finding a common difference/ratio from a set of points

• Common Difference

  • Given the points (a, b) & (c, d), you can find the common difference using the slope formula →

    

• Common Ratio

1. Find how many times r would have to be multiplied to get from 4 to 16 (in this case it is 2)

2. This means we would have to multiply 4 ⋅ r ⋅ r to get 16 → 4r² = 16

3. Then, you solve for r → r² = 16/4 = 4 → r = √4 = 2

4. So r = 2

• OR… given the points (a, b) & (c, d), you can find the common ratio using this equation →

2.3 - Exponential Functions

General form of an exponential function → f(x) = ab^x

• a = initial value

• a ≠ 0 & b > 0

Exponential growth → when a > 0 & b > 1

Exponential decay → when a > 0 & 0 < b < 1

Determining exponential function based on output values

• Ex. 3 × 3 × 3 × 3 × 6

  • a, the input value, will be the odd one out: 6

  • the rest are the common ratio multiplied by however many times

  • f(x) = 6(3)^x where x = 4

Exponential Functions depending on parameters

a > 0; b > 1 Exponential growth Increasing Ex. f(x) = 1(2)x	a > 0; 0 < b < 1 Exponential decay Decreasing Ex. f(x) = 1(0.5)x
a < 0; b > 1 a < 0, so the function remains below 0 Decreasing Ex. f(x) = -2x	a < 0; 0 < b < 1 a < 0, so the function remains below 0 Increasing Ex. f(x) = -(0.5)x

Characteristics of Exponential Functions

• They are always increasing or decreasing

• There is no extreme (except on closed intervals)

• Their graphs are always concave up or concave down

• They have no inflection points

• If the input values increase or decrease without bound, the end-behavior can be expressed as

  • lim_{x → ±∞} ab^x = 0

  • lim_{x → ±∞} ab^x = ±∞

Additive Transformation of an Exponential Function

• g(x) = f(x) + k

• If the output values of g are proportional over equal-length input-value intervals, then f is exponential

  • Meaning that if g(x) has proportional output values, then f(x) will always be exponential

  • But if f(x) is proportional, that doesn’t mean g(x) will be proportional

    

2.4 - Exponential Function Manipulation

Negative Exponent Property → b^-n = 1/bⁿ

Product Property → bⁿ * b^m = b^{n + m}

Product Property → (b^m)ⁿ = b^mn

Transformations of Exponential functions

Horizontal Translations & Vertical Dilations

ac(b)^x = b^kb^x = b^{(x + k)}

• Through the product property, when doing a horizontal translation, it is the same as doing a vertical dilation

• Every horizontal translation of an exponential function, f(x) = b^{(x + k)}, is equivalent to a vertical dilation

• Ex. 1(2)^{(x + 3)} → 1 × 2³(2^x) → 8(2)^x

  • Horizontal translation left by 3 units / Vertical dilation of 8 units

Horizontal Dilations

• Every horizontal dilation of an exponential function, f(x) = b^(cx), is equivalent to a change of base

  • (b^c)^x → b^c and c ≠ 0

  • Ex. 2^3x → (2³)^x → 8^x

The kth Root of b

2.5 - Exponential Function Context & Data Modeling

Finding the exponential function in the form f(x) = ab^x + k through a table

1. Find the ratio (b) (in this case b=2)

   1. Find the differences

      • 9-6 = +3; 15-9= +6; 27-15= +12; 51-27= +24

   2. Then, you can find the ratio

      1. 3 × 2 = 6 × 2 = 12 × 2 = 24

2. Then find a and k by setting up a system of equations

   • 6 = a(2)⁰ + k → 6 = a + k

   • 9 = a(2)¹ + k → 9 = 2a + k

     • 9 = 2a + k

     • -6 = -a - k

     • a = 3

   • 6 = 3 + k → k = 3

The number e

• e ≈ 2.718…

• The natural base e is often used as the base in exponential functions that model contextual scenarios

Percent Increase:

• Start with the initial amount a and grow with a % increase

• y = a(1 + %increase)^x

Percent Decrease:

• Start with the initial amount a and decay with a % decrease

• y = a(1 - %decrease)^x

Half-Life:

• An initial amount a that shrinks by half every h

• f(t) = a(1/2)^t/h

Doubling time:

• An initial amount a that doubles every d

• f(t) = a(2)^t/d

Equivalent Forms

• If t represents the number of weeks, then the base of f(t) = (1.01)^t indicates that the quantity increases by a factor of 1.01 every week

  • If you want the equation for every year, then that equation would look like this → (1.01⁵²)^t/52

    • This is because every 52 weeks, t = 52/52, means that it would account for 1 year

    • since we’re counting years, 1.01⁵² means that in 1 year, we’d have (1.01 × 1.01 × 1.01 … 1.01) 52 times per year

    • So when t = 52 weeks / 52 (one year), 1.01 would’ve increased 52 times

  • If you wanted an equation for every day, then it would look like this → (1.01^1/7)^t

    • This is because once t is > 7, the function will equal 1.01 which is the same as if it where (1.01)¹ weeks

2.6 - Competing Function Model Validation

The only way to know what model fits an equation best is to use context clues

Residual Value - a measure of how much a regression curve vertically misses a data point

Residuals → Actual - Predicted (AP)

The difference between the predicted and actual values is the error

Error → Predicted - Actual (PA)

• A positive error means that the model overestimated (you can see the y=0 line is above the point)

• A negative error means that the model underestimated (you can see the y=0 line is below the point

Residual Plot Patterns

If there is a clear pattern in the residual plot, it is a bad regression model

2.7 - Composition of Functions

The composition of functions means we take one function and substitute it into the other function

• f(g(x) or f ∘ g(x)

• Ex. f(x) = 2x + 3; g(x) = 5x²; find f(g(5))

  • 5(5)² → 5(25) → 125

  • 2(125) + 3 = 253

Identity Function

• When f(x) = x

• then f(g(x)) = g(x)

Finding a Composed Function

• f(x) = x² - 1 & g(x) = √x ; find f(g(x))

  • h(x) = (√x)² -1

  • h(x) = x - 1

Domain of a Composed Function

• When finding the domain of a function h(x) = f(g(x)), there are 2 things to consider

  1. because g(x) is the input, we can only use x-values in the domain of g

  2. any restrictions on the domain of f(g(x)) must be included

• Ex. problem above in Finding a Composed Function

  • g(x) is the input, so it can only be from [0, ∞)

  • and there are no restrictions in x - 1, so the domain is [0, ∞)

Decomposition of Functions

• if there is a radical, one function can simply be √x (this applies with 1/x as well)

• Ex. h(x) = √x³ + 1 ; find f(g(x))

  • f(x) = √x

  • g(x) = x³ + 1

• Ex2. h(x) = 1/(x² + 1) ; find g(f(x))

  • f(x) = x² + 1

  • g(x) = 1/x

2.8 - Inverse Functions

The inverse of a function is a reverse mapping of a function where (x, y) → (y, x)

They should reflect across the line y=x

If the graph turns (has a max/min) then it is no longer invertible because there will be output values that are the same for different input values

Find the Inverse of a function

1. instead of (for example) y = mx + b, switch the x and y so it is → x = my +b

2. solve for y

Finding the domain and range

• The domain and range swap for inverse functions

  

Composition of f and f^-1

• f(f^-1(x)) = f^-1(f(x)) = x

• If both functions, when composed with each other, cancel out, then they are inverses of each other

2.9 - Logarithmic Expressions

A logarithm is a way to find an unknown power using the base and a number

• Ex.

  • log₂(64) = 6 | 2⁶ = 64

2.10 - Inverses of Exponential Functions

The inverse of an exponential function can be modeled through a logarithmic function.

Exponential vs. Logarithmic

Power Rule → 2log₅(x) = log₅(x²)

Raising to a log power

• a^c(log_b(x))

• if a and b are the same and c = 1, then you can cancel out to get x

• 2^log₂(x) = x

You can also cancel out to get x if the bases are the same in this scenario → log₂(2^x) = x

Using these two, you can find if an exponential function and a logarithm function are inverses by plugging each function into each other (ex. f(g(x)) = x = g(f(x)))

2.11 - Logarithmic Functions

Logarithmic functions have inverse properties as exponential functions

Exponential

• x-intercept: NA

• y-intercept: (0, 1)

• asymptote: y = 0

• increasing: (-∞, ∞)

• decreasing: NA (only when 0 < b < 1)

• domain: (-∞, ∞)

• range: (0, ∞)

Logarithmic

• x-intercept: (0, 1)

• y-intercept: NA

• asymptote: x = 0

• increasing: (-∞, ∞)

• decreasing: NA (only when 0 < b < 1)

• domain: (0, ∞)

• range: (-∞, ∞)

Transformations of Logarithm Graphs

• Horizontal Translations

  • log_b(x ± h)

    

  • CHANGES DOMAIN

• Vertical Translations

  • log_b(x) + k

    

• Horizontal Dilation

  • log_b(cx)

  • Compresses x-values by 1/c

    

• Vertical Dilation

  • alog_b(x)

  • Extends y-values by a

    

• Horizontal Reflection

  • log_b(-x)

  • Reflection over y-axis

    

  • CHANGES DOMAIN

• Vertical Reflection

  • -log_b(x)

  • Reflection over x-axis

    

2.12 - Logarithmic Function Manipulation

Product Property → log_b(xy) = log_bx + log_by

Quotient Property → log_b(x ÷ y) = log_bx - log_by

Power Property → log_bx^k = klog_bx

Change Base Formula

Finding transformations through manipulation

1. f(x) = log₂(4x)

2. f(x) = log₂4 + log₂x

3. f(x) = log₂(2²) + log₂x

4. f(x) = 2 + log₂x

   1. This means that the function f(x) has a vertical translation of 2

2.13 - Exponential & Logarithmic Equations & Inequalities

Logarithmic Equations

• If you have logs with the same base on each side, you can cancel them

  • log₂(x + 3) = log₂(5) → x + 3 = 5

• You can also use logs to solve exponential equations

  • 2(3)^{x - 3} = 18 → 3^{x - 3} = 9 → 3^{x - 3} = 3² → x - 3 = 2

Inequalities

• First, you need to find out what restrictions there are in the inequality (making sure they are positive)

  • log(x + 2) + log(3) > log(5x - 2)

    • x + 2 > 0 → x > -2 (won’t work because it is below zero)

    • 0 < 5x - 2 → x > 2/5

• Then, you need to solve the equation

  • log(3x + 6) > log(5x - 2) → 3x + 6 > 5x - 2 → 6 > 2x - 2 → 8 > 2x → x < 4

• Then you follow along with the restrictions and your output to get → 2/5 < x < 4

2.14 - Logarithmic Function Context and Data Modeling

Usually, when given datasets, if the x-values are proportional and the y-values are linear, then the table is logarithmic

You can use your calculator to use natural log regression to find the equation for these points

2.15 - Semi-Log Plots

Taking these points…

… you get this exponential graph…

However, when you switch to a semi-log plot, the y-values appear as bases of 10 (10⁰, 10¹, 10²) and the graph appears linear…

Taking the log the equation gives this straight, linear line:

• y = 1(2.5)^x → log(y) = log(1) + log(2.5^x) → log(y) = 0 + xlog(2.5) → y = 0.3979x

• This is the same as taking the log of each of the y-values of the original function.

Unit 3

3.1 - Periodic Phenomena

A periodic function demonstrates a repeating pattern over successive equal-length intervals.

Period - the smallest change in x-values it takes for the function to repeat itself

• the smallest positive value of k such that f(x + k) = f(x)

3.2A - Radians & 3.2B - Sine, Cosine, and Tangent

An angle in standard position has one ray on the x-axis (called the initial side) and another called the terminal side

Radian → a unit of angle, equal to an angle at the center of a circle whose arc is equal in length to the radius

• The angle of one radian is when the angle creates an arc length on a circle that is equal to the radius of the circle

• This means that s = rand θ = 1

Having a negative angle means that you go clockwise around the circle

The rays of an angle that are touching the circle means that the arc is subtended by the angle

• The angle (θ) in radians = Arc Length (s) / Radius (r)

Unit Circle

• The unit circle is a circle with a radius of 1

• θ = s (because r = 1)

• ONE REVOLUTION → 2π

• ½ REVOLUTION → π

• ¼ REVOLUTION → π/2

• REMEMBER THE UNIT CIRCLE (It will greatly help you later!)

  

SOH-CAH-TOAH

• sin(θ)

  • the ratio of the vertical displacement of a point (P) from the x-axis to the distance between the origin and point P

  • opposite / hypotenuse → y / r → y

  • In the unit circle, sin(x) = y

• cos(θ)

  • the ratio of the horizontal displacement of P from the y-axis to the distance between the origin and point P

  • adjacent / hypotenuse → x / r → x

  • In the unit circle, cos(x) = x

• tan(θ)

  • slope of terminal ray

  • opposite / adjacent → y / x → sin/cos

• An example in the unit circle →

  

3.3 - Sine and Cosine Function Values

Given the unit circle, when taking the cosine or sine of the radian, we can get the x and y coordinate of the point

• cos(π/4) = √2/2

• sin(π/4) = √2/2

We can also get the coordinate by using the radius and cos/sin value

• Because cos(θ) = x / r → x = rcos(θ)

• Because sin(θ) = y / r → y = rsin(θ)

By knowing the unit circle, you can determine where a point is as well

• Example → (-3√3/2, -3/2)

• This point has both -x and -y, so it’s in Quadrant III, meaning the points it could be is 7π/6, 5π/4, or 4π/3

• The y-coordinate looks like ½, which means that the θ value is most likely 7π/6

• The r value is 3 because the points have 3 multiplied onto them

3.4 - Sine and Cosine Function Graphs

The Sine Function

• y = sin(θ)

  

• sin(θ) increases on the right side of the unit circle (0, π/2) & (3π/2, 2π)

• sin(θ) decreases on the left side of the unit circle (π/2, 3π/2)

• sin(θ) is concave up on the bottom side of the unit circle (π, 2π)

• sin(θ) is concave down on the top side of the unit circle (0, π)

The Cosine Function

• y = cos(θ)

  

• cos(θ) increases on the bottom side of the unit circle (π, 2π)

• cos(θ) decreases on the top side of the unit circle (0, π)

• cos(θ) is concave up on the left side of the unit circle (π/2, 3π/2)

• cos(θ) is concave down on the right side of the unit circle (0, π/2) & (3π/2, 2π)

3.5 - Sinusoidal Functions

The cosine function is just the sine function -π/2 and vice versa.

Cosine is an even function and has reflective symmetry over the y-axis

Sine is an odd function and has rotational symmetry about the origin.

y = a cos θ | y = a sin θ

Amplitude: | a | → half the difference between the maximum and minimum values

Midline: A horizontal line halfway between the maximum and minimum values. It is determined by finding the average of the maximum and minimum values (usually expressed as y = something)

Period (cycle): 2π → The reciprocal of frequency. The change in θ values required for the function to complete one full cycle

Frequency: 1/2π → The reciprocal of period. The number of cycles the graph completes per one radian.

3.6 - Sinusoidal Function Transformations

Standard Equations

• y = a × cos(b(x ± c)) + d

• y = a × sin(b(x ± c)) + d

• Amplitude (|a|) → the amplitude / half the distance between the max and min values / ALWAYS absolute value of a / negative flips over x-axis

• Midline (y = d) → A horizontal line halfway between the max and min values (also the vertical shift ±d)

• Period (\frac{2π}{|b|}) → The change in x values required for the function to complete one cycle (b is the horizontal stretch/shrink / negative reflects over y-axis)

• Frequency (\frac{|b|}{2π}) → The number of cycles the graph completes per one radian

• Phase Shift (±c) → A shift left or right (-c = right / +c = left)

Example

• y = -3\cos(2x) + 1

• y = 3\sin(2{x-\frac{\pi}{4}}) + 1

  

3.7 - Sinusoidal Function Context & Data Modeling

Let’s see how we can determine a sinusoidal function manually from points given to us

1. First, we need to find out where are maximum and minimum values are

   • Minimum → (2, -62), (6, -65), (10, -59), (14, -60)

   • Maximum → (0, 97), (4, 104), (8, 101), (12, 98)

2. We need to find the period and the frequency

   • For period, we just take wherever 2 maximums or 2 minimums are and subtract the x-values. We have 2 maximums at (0, 97) and (4, 104) so → 4 - 0 = 4

     • So, our period is 4

     • And thus, our frequency is 1/4

3. Next, we need to estimate the vertical shift / midline

   • This can be done by taking the highest maximum and lowest minimum and finding the average between them

     • Highest max → (4, 104)

     • Lowest min → (6, -65)

   • Taking the average → (104 + (-65))/2 = 19.5

4. Now, we find the amplitude

   • To find the amplitude, we just find the distance between the midline and either the highest max or the lowest min

   • We can use the max because it’s easier → 104 - 19.5 = 84.5

   • If we wanted to use the min point, we could find it as so → 19.5 - (-65) = 84.5

5. Using the period, we can find the b value

   • 2π/|b| = 4 → 4b = 2π → b = 2π/4 → b = π/2

6. Finally, we need to find the phase shift

   • Cosine

     • The phase shift is easy with cosine because the maximum starts at 0 with a regular cosine function, meaning that our function is shifted by the nearest maximum to 0

       • In this case it is on zero, and therefore we don’t have to shift it, meaning our function looks like this

         

   • Sine

     • Sine is harder because the value of 0 on sine is in between a maximum and minimum

     • This means we have to subtract ½ the period from the max/min closest to 0 to get the min/max

       • 0 - 2 = -2

     • Next, we need to get the middle of those two points, so we just find the average (0 + -2)/2 = -1, meaning that we are shifted to the left by one which is our c value

       

This is an estimated function, by using a calculator and using sine regression, you can find what the calculator estimated

3.8 - The Tangent Function

• The tan(x) function gives the slope of the terminal ray

• It is equal to y/x or sin(x)/cos(x) as long as cos(x) ≠ 0

Example of Point on tan(x)an(x)

• If you had tan(π/6) you just divide the sin(π/6) by cos(π/6)

  • sin(π/6) = 1/2

  • cos(π/6) = √3/2

    

Properties of tan(x)tan(x)

• The slope starts at 0

• The slope gets larger until it approaches ∞

• At θ = π/2, the slope is undefined

• Then the slope is infinitely negative but starts to grow towards zero

• Then at π, the slope is zero again

• The values of tan(x) are the same in Quadrants I & III

• The values of tan(x) are the same in Quadrants II & IV

• Every ½ revolution of the circle, the tangent function repeats

• The period of tan(x) = π

Vertical Asymptotes

• For the graph tan(x), there is a vertical asymptote every

  

  • You start at π/2 and go for k periods, where k is an integer

• For the graph of tan(bx), the period is π/|b|

  • A vertical asymptote appears every

    

    • Where k is any integer

Transformations of Tan(x)

• y = a × tan(b(x + c)) + d

  • a → creates a vertical dilation by the factor of |a|; If a < 0, then reflect over x-axis

  • b → creates a horizontal dilation and changes the period by a factor of |1/b|; If b < 0, reflect over y-axis

    • By either widening (0 < b < 1) or shortening (b > 1) the vertical asymptotes around each line, you can find how wide it is

  • c → creates a horizontal shift by -c units

  • d → creates a vertical shift by d units

Example

• y = 2tan(1/2(x - π)) - 2

  

3.9 - Inverse Trigonometric Functions

Let’s say you needed to find an angle x

• sin(x) = 3/8

• Using the inverse sine function, you can get rid of the sine and find the x-value

• sin^-1(sin(x)) = sin^-1(3/8) → x = sin^-1(3/8)

REMEMBER → the inverse of a function is just switching the x and y values. When we do this for arcsin and arccos, we cannot have two values overlapping (vertical line test), so we have a restricted range

Inverse Sine / sin^-1(x) / arcsin(x)

• Domain: [-1, 1]

• Range: [-π/2, π/2]

  

  • Output values are only between [0, π/2] & [-π/2, 0] (right side of unit circle)

Inverse Cosine / cos^-1(x) / arccos(x)

• Domain: [-1, 1]

• Range: [0, π]

  

  • Output values are only between [0, π] (top half of unit circle)

Inverse Tangent / tan^-1(x) / arctan(x)

• Domain: (-∞, ∞)

• Range: (-π/2, π/2)

  

  • Output values are only between (0, π/2) & (-π/2, 0) (right side of unit circle excluding π/2)

Finding Inverse Function

• Example → f(x) = 2sin(x) + 1 for -π/2 ≤ x ≤ π/2

  1. We find it like any other inverse equation

     • x = 2\sin(y) + 1

     • x - 1 = 2\sin(y)

     • \frac{x-1}{2} = \sin(y)

     • \sin^{-1}(\frac{x-1}{2})=f(x)^{-1}

  2. Domain & Range of f(x)

     • Domain → [-π/2, π/2]

     • We can find the range of f(x) normally (amplitude of 2 and midline of 1)

       • Range → [-1, 3]

  3. Swap the domain and range for f(x)^-1

     • Domain → [-1, 3]

     • Range → [-π/2, π/2]

3.10 - Trig Equations & Inequalities

Solving Trigonometric Equations

• Restricted Domain (0 ≤ x ≤ 2π)

  • Exact Values

    • We can find exact values using the unit circle

    • Example equation → 2sin(x) + 5 = 4

      1. First, we get the sin by itself

         • 2sin(x) = -1 → sin(x) = -1/2

      2. Our answer will be in values where sin(x) = -1/2

         • If we know one, for sine, the answer will be across from it → (11π/6 & 7π/6). These answers are also in our restricted domain, so that’s it!

           

  • Approximate Values

    • Approximate values we can find using a calculator

    • Example → tan²x + 2tan(x) - 8 = 0 (think of this as x² + 2x - 8 = 0)

      1. We can factor this out to (tanx + 4)(tanx - 2) = 0; this gives us two equations

      2. Our first equation is x = tan^-1(-4)

         • This gives us approximately -1.326, which is our of our range, but we can get the same value by adding 2π (instead of going clockwise, we need to go fully around the circle counterclockwise to get to the same value) → -1.326 + 2π = 4.957

      3. Now we have to get the other solution to tan^-1(-4)

         • It’s helpful to plot the point on the unit circle. 4.957 would be in Quadrant IV and we can draw a line through the origin to help us see what number the other value would be

         • tan(x) reflects across the origin; since this is basically half a rotation (π), we just need to subtract π from our value above → 1.816

           

         • Both of these values are inside of the restricted domain, so they are are answer for the first equation → 4.957 & 1.816

      4. Do the same thing for the other equation

         • x = tan^-1(2)

         • x = 1.107

         • x = 1.107 + π = 4.248

• All Values

  • Exact Values

    • Here, we don’t have any restricted domain, so we will format our answer differently

    • Example → 4cos²x = 2

      1. First, we get cos by itself → √cos²x = √1/2 = cosx = ±√2/2

      2. By knowing the unit circle, we know what these values are

         • cos^-1(√2/2) = π/4 and the other value (for cos) can be found on the other side of the y-axis → 7π/4 or -π/4

           

         • We find the other equation now →

         • cos^-1(-√2/2) = ±3π/4

      3. Format the answer

         • cos^-1(√2/2) = ±π/4 + 2πk where k is any integer

           • Start at ±π/4 and every k times period (2π), we hit an answer

         • cos^-1(-√2/2) = ±3π/4 + 2πk where k is any integer

  • Approximate Values

    • Example → 3sin²x = sinx

      1. Get everything = 0 & simplify → 3sin²x - sinx = 0 → sinx(3sinx - 1) = 0

      2. Solve for each equation

         • sin^-1(0) = 0 & π → 0 + πk where k is any integer (you hit an answer every π times around the circle)

         • sin^-1(1/3) = 0.34 + 2πk where k is any integer

           • To find the other value, we know that the other value for sin will appear across from it, meaning that we can add or subtract π to get the other value)

           • π - 0.34 = 2.81

           • 2.81 + 2πk where k is any integer

Solving Trigonometric Inequalities

• Restricted Domain w Exact Values (0 ≤ x ≤ 2π)

  • Equation → 4cos(x) - 1 ≤ -3

    1. Solve like normal → cos(x) ≤ -1/2

    2. Using the unit circle, find out the two answers

       • cos(2π/3) = -1/2

       • cos(4π/3) = -1/2

    3. Determine < or >

       • Because this equation is <, it would include everything to the left between the radians 2π/3 and 4π/3

       • Therefore, our answer is 2π/3 ≤ x ≤ 4π/3

• All Exact Values

  • Equation → cos(2x) + 4 = 4

    1. Solve like normal, however, when taking the cos^-1(x), don’t divide by 2

       • cos(2x) = 0 → cos^-1(0) = 2x

       • Cos equals zero at angles π/2 and 3π/2

       • Because we hit an answer every π periods, our answer would be π/2 + πk where k is any integer

       • However, we divide the entire answer by the two afterwards → π/4 + kπ/2 where k is any integer

3.11 - Secant, Cosecant, Cotangent Functions

Reciprocal Trig Functions

• Cosecant / csc(x) / 1/sin(x)

  • Cosecant is 1/sin(x) or 1/y

  • When drawing out the sin function csc(x) has maximums and minimums at the maximums and minimums of sin(x)

  • csc(x) also has vertical asymptotes where sin(x) has zeros → 0 + πk where k is any integer

    

• Secant / sec(x) / 1/cos(x)

  • Secant or sec(x) is 1/cos(x)

  • Sec(x) is like csc(x) where it has a max/min where cos(x) has a min/max

  • Sec(x) has vertical asymptotes where cos(x) has zeros → π/2 + πk where k is any integer

    

• Cotangent / cot(x) / 1/tan(x) / cos(x)/sin(x)

  • Cot(x) is cos(x)/sin(x)

  • Cot(x) has zeros where cos(x) has zeros

  • Cot(x) has vertical asymptotes where sin(x) has zeros → 0 + πk where k is any integer

    

3.12 - Equivalent Representations of Trig Functions

Pythagorean Identities

• Our first Pythagorean identity is sin² + cos² = 1

  • We can also get sin² = 1 - cos²

  • We can also get cos² = 1 - sin²

• By dividing by sin and cos, we can get our other identities

  • sin²/sin² + cos²/sin² = 1/sin² → 1 + cot² = csc² → cot² = csc² - 1

  • sin²/cos² + cos²/cos² = 1/cos² → tan² + 1 = sec² → tan² = sec² - 1

We can use these identities to simplify equations such as:

Solving Equations

• Given the equation 1 = sec²x + tanx, and we need to solve for x for values 0 ≤ x ≤ 2π

  1. By shifting everything to one side we get sec² - 1 + tan = 0

  2. Knowing our identities, we simplify it down to → tan² + tan = 0

  3. We cans simplify further and get two equations → tan(tan + 1) = 0

  4. Now, we solve as we have been

     • tan^-1(0) = 0, π

     • tan^-1(-1) = 3π/4, 7π/4

Sum/Difference Identities

• The sum and difference identities look confusing, but they are easy!

• Helpful Tricks

  • a and b, no matter the function, will be in the same place

  • sin keeps the same sign

  • cos always has cos with cos and sin with sin (coscos - sinsin)

  • sin has mismatched functions (sincos + cossin)

  

• Finding Values/Simplifying Using Sum/Difference Identities

  • Let’s say we’re given the equation → sin(π/4 + 5π/3)

    1. Use the sin identitiy → sin(π/4)cos(5π/3) + cos(π/4)sin(5π/3)

    2. Using the unit circle, we can get these values → (√2/2)(1/2) + (√2/2)(√3/2)

    3. We can now multiply across → (√2/4) + (√6/4)

    4. Because they have the same denominator, we can add them together, but make sure you know how to add radicals → (√2 + √6)/4

  • To simplify, use the same process; for example, if you have → sin(x - π/6)

    1. Use your identity → sin(x)cos(π/6) - cos(x)sin(π/6)

    2. Then we get the equation → (√3/2)sin(x) - (1/2)cos(x)

Double Angle Identities

• Double Angle Identities can be found using the sum/difference identities because f(2x) = f(x + x)

  

• We can find them like so

  • sin(2a) → sin(a + a) → sin(a)cos(a) + cos(a)sin(a) → 2sin(a)cos(a)

  • Cosine has 3 optional values

    1. Sine & Cosine

       • cos(2a) → cos(a + a) → cos(a)cos(a) - sin(a)sin(a) → cos²a - sin²a

    2. Just Cosine

       • Using the Pythagorean Identities, we can swap out -sin² as cos² - 1

       • cos²a + cos²a - 1 → 2cos²a - 1

    3. Just Sine

       • We can also swap out cos²a with 1 - sin²

       • 1 - sin² - sin² → 1 - 2sin²a

• Solving Equations

  • Given the equation 3cos(2x) = 2cos²x we can use our identities to solve

    1. Because we have only cos, it’d be easier to use the cos double identity → 3(2cos²x - 1) = 2cos²x

    2. Distribute the 3 → 6cos² - 3 = 2cos²

    3. Set the equation = 0 → 4cos² - 3 = 0

    4. Solve as normal!

       • cos^-1(+√3/2) = π/2, 3π/2

       • cos^-1(-√3/2) = π/6 , 5π/6

3.13 - Trigonometry & Polar Coordinates

Polar Graph → a graph of concentric circles of varying radius (r) where the origin is the pole

A coordinate on the polar coordinate system is (r, θ) where r is the radius (each concentric circle, counting out from the pole) and θ is the angle around the circle

• When plotting points, it helps to go to the angle first and then count out to the radius

If r is negative, then you start from the origin/pole and count backwards on the angle line

• For example, if you had (-2, π/6), you’d start from the origin and, on the line of π/6, count out 2 rings on the line of 7π/6

If θ is negative, then you go clockwise around the circle

Because of this, this means we can get 4 different ways to plot a point on the polar graph:

• (2, 7π/6)

• (-2, π/6)

• (2, -5π/6)

• (-2, -11π/6)

  

Conversions

• Polar → Rectangular

  • This is the easier of the transformations

    • x = r × cos(θ)

    • y = r × sin(θ)

• Rectangular → Polar

  • We can use the identities to find r (x² + y² = r²)

    • r = √x² + y²

  • Because tan(θ) = y/x, we can get the angle from both points by taking the inverse

    • θ = tan^-1(y/x)

  • Make sure to check for boundaries (for example 0 ≤ x ≤ 2π)

Imaginary/Complex Numbers (a + bi)

• Complex Numbers consist of a real part (a) and an imaginary part (bi)

• Conversions

  • The polar form looks like this

    

  • Rectangular Complex → Polar Complex

    • This is similar to the other conversions

    • r = √a² + b²

    • θ = tan^-1(b/a)

    • Example

      • 2 - i

        • r = √2² + (-1)² = √5

        • θ = tan^-1(-1/2) = -0.464

        • √5[cos(-0.464) + isin(-0.464)]

  • Polar Complex → Rectangular Complex

    • This is also similar to the other conversion

    • a = r × cos(θ)

    • b = r × sin(θ)

    • Example

      • 3cos(π/3) + 3isin(π/3) → 3(1/2) + 3i(√3/2) → 3/2 + i(3√3)/2

3.14 - Polar Function Graphs

Lines

• Lines are when θ = something like π/3

  

Circles

• Circles are when r = a × cos(θ)/sin(θ)

• a is the maximum r values

• Cosine

  • Example of r = 2cos(θ)

  • Cycle: [0, π]

  • Positive: open right

  • Negative: open left

    

• Sine

  • Example = r = 2sin(θ)

  • Cycle: [0, π]

  • Positive: open above

  • Negative: open below

    

Roses

• Cosine → petal on x-axis

• Sine → no petals on x-axis

• r = a × cos/sin(nθ)

  • a is the max distance from the pole

  • n is even → 2n petals

  • n is odd → n petals

• Odd n Cosine

  • Example → r = 2cos(3θ)

  • # of petals: 3

  • Cycle: [0, π]

    

• Even n Cosine

  • Example → r = 2cos(2θ)

  • # of petals: 4

  • Cycle: [0, 2π] (2n petals means that the cycle is double too)

    

• Odd n Sine

  • Example → r = 2sin(5θ)

  • # of petals: 5

  • Cycle: [0, π]

    

• Even n Sine

  • Example → r = 2sin(2θ)

  • # of petals: 4

  • Cycle: [0, 2π]

    

Find Endpoints with Restricted Domain

• For example, given the equation: r = 4cos(3θ), what are the endpoints of π/6 ≤ θ ≤ π/3

  1. The graph would look like this

     

  2. We then input both π/6 and π/3 into the equation to get our endpoints

     • r = 4cos(3(π/6)) → r = 4cos(π/2) = 0 → (r, θ) = (0, π/6)

     • r = 4cos(3(π/3)) → r = 4cos(π) = -4 →(r, θ) = (-4, π/3)

Limaçon

• r = a ± bcos(θ)

• r = a ± bsin(θ)

  • a + b → gives you the max radius

  • a - b → gives you the max radius in opposite direction

  • a > 0 and b > 0

• Different ways limacons can appear based on cos/sin and + or -

  

1. Cardioids (a = b)

   • Example → r = 3 + 3cos(θ)

   • Max r → 3 + 3 = 6

     

   • Cardioids look like hearts

     

2. Dimpled Cardioid (a > b | 1 < a/b < 2)

   • Example → r = 5 - 3cos(θ)

   • Max r → 5 + 3 = 8

     

   • Dimples/Convex Cardioids look like their name, they have a small dent or they appear flatter on one side

     

3. Inner Loop Limacon (a < b)

   • Example → 1 + 3cos(θ)

   • Max r → 1 + 3 = 4

     

   • Like the name, they have an inner loop

     

Archimedes’ Spiral

• where r = θ and θ is greater than 0

  

3.15 - Rates of Change in Polar Functions

The Rate of Change of r

• Given the equation r = 6sin(3θ), we get this graph

  

  • On the intervals

    • (0, π/6) → r is positive & increasing

    • (π/6, π/3) → r is positive & decreasing

    • (π/3, π/2) → r is negative & decreasing

    • (π/2, 2π/3) → r is negative & increasing

• The polar version of this graph looks like this

  

  • the rate of change is measured by the distance from the pole

Example

• Equation → r = f(θ) = 3 + 5sin(θ) on the interval π ≤ θ ≤ 2π

• Here is the data table →

  

  • Interval(s) where f is increasing

    • r is increasing from -2 to 3 so → (3π/2, 2π)

  • Interval(s) where f is decreasing

    • r is decreasing from 3 to -2 so → (π, 3π/2)

  • Is there any relative extrema?

    • Yes, at least one because f changes from decreasing to increasing

  • Is the distance between r/f(θ) and the pole is increasing/decreasing on the interval 5π/4 ≤ θ ≤ 3π/2

    • r is negative and r is decreasing → distance is increasing from the pole

  • What is the AROC of f between 3π/2 and 7π/4

    • r₂ - r₁ / θ₂ - θ₁

    • -2 - (-0.53) / 3π/2 - 7π/4 = 1.871 units per radian

  • What interval(s) must r = 0

    • When it changes from + → - or - → +

    • (7π/6, 5π/4) & (7π/4, 11π/6)

  • Using the aroc above, what is the estimated value of 5π/3?

    • Use point-slope formula and taking a value close to 5π/3:

      • r₁ - (-2) = 1.871(θ₁ - 3π/2)

      • f(5π//3) = 1.871(5π/3 - 3π/2) - 2 = -1.02

Good Luck on the AP Precalculus Exam!