~Calculus AB!~

Okay so Calculus seems scary....and IT IS. We're kidding, once you get the hang of it, it'll be alright! Just try your best and don't give up.

When first starting off Calculus, you should know these important equations... Derivatives and Integrals!

Here is a link to all the Need Statements you should know, they're basic formulas and statements everyone should memorize when doing Calculus.

Also here are two other important links to use to review for the AP test! Theorems & Trignometry. Bound to get a 5 for sure.

These are all the notes we were able to type out so far, of course there is so much to Calculus AB so please do check out the documents pasted above which has a whole summary of the calculus that may come on the AP test! If youw ant more complex and extra explanations use resources like Organic Chemistry Tutor, Khan Academy, and Paul's Online Notes, they wil be your best friends.

Proving Symmetry

- symmetry to y-axis is an even function so f(-x) = f(x)
- symmetry to origin is an odd function so f(-x) = -f(x)
- symmetry to x-axis is neither even or odd so -f(x) = f(x)

ex. y = g(x) = cosx/(x⁴ + 1)
g(-x) = cos(-x)/((-x)⁴ + 1) = cosx/(x⁴ + 1) = g(x)
so g(x) is symmetric to the y-axis

Domain & Range of a function

domain: is the set of all possible x-value inputs
range: is the set of all possible outputs y-value

ex. y = 1/(x² - 3x + 2)
domain: denominator ≠ 0
(x-2)(x-1) ≠ 0
x ≠ 1, 2
(-∞, 1)∪(1, 2)∪(2, ∞)
range: (-∞, 0)∪(0, ∞)

Transformations of Functions

basic types of transformations (c > 0)
original graph f(x)
horizonal shift c units right f(x-c) *horizontal, c is inside the function*
c units left f(x+c)
vertical shift c units up f(x)+c *vertical, c is outside the function*
c units down f(x)-c
reflection x-axis -f(x)
reflection y-axis f(-x)

Intro to Limits

as x gets closer to 2 x → 2

f(x) gets closer to 3 f(x) → 3

∴ lim f(x) = 3

x → 2

say: limit as x approaches 2 of f(x) is 3

h(2) = does not exist

∴ lim h(x) = 3

x → 2

limit exists at holes

*limits DON'T exist at breaks/jumps & at vertical asymptotes*

∴ lim g(x) = -2 ∴ lim g(x) = 1

x → -2 x → -1 ⁺ *this means it's from the right*

∴ lim g(x) = DNE ∴ lim g(x) = -1

x → -1 x → -1 ^- *this means it's from the left*

*DNE: if the values of lim f(x) are not the same when you approach from the left & right*

∴ lim g(x) = +∞ ∴ lim g(x) = -∞

x → 3 ⁺ x → 3 ^- *this means it's from the right*

Evaluating limits

Rational Functions

def: a rational function is a function of the form r(x) = p(x)/q(x)

where p(x) & q(x) are polynomial functions & q(x) ≠ 0

to graph:

- find holes first, factor, & cancel

- zeros: let numerator = 0 solve for x

- y intercepts: let x = 0 solve for y

- vertical asymptotes: let denom = 0

- horizontal asymptotes: deg. num ≤ deg. of denom

- slant asymptote: deg num = 1 + deg. denom

- if deg. num > 1 + deg. denom there is no asymptote

- make sure to find holes before vertical asymptotes

- check for symmetry & then GRAPH! keep practicing how to graph...

Infinite Limits

to find infinte limits w/o graphing: lim when x → a ^± p(x)/q(x) = ±∞ if & only if

lim when x → a ^± p(x) = non-zero constant & lim when x → a ^± q(x) = 0

Limits at Infinity

to find infinte limits w/o graphing: lim when x → ±∞ C/xⁿ = 0

goal is to rewrite the function to produce expressions C/xⁿ

multiply numerator & denominator by 1/x^highest power of denom.

The Intermediate Value Theorem (IVT)

if f is contonuous on [a, b] and if f(a) ≠ f(b), then for any # k between f(a) & f(b) there exists at least one # c between a and b such that f(c) = k

ex. f(x) = x³ - x² + x -2 [0,3] f(x) = 4

✓ f is continuous on [0,3]

✓ f(0) ≠ f(3)

✓ f(0) < 4 < f(3)

c³ - c² + c - 2 = 4

c³ - c² + c - 6 = 0

(c-2)(c² + c² + 3) = 0

a = 0 < c = 2 < b = 3

∴ intermediate value thm applies

Derivatives & Tangent Lines

def: the derivative of f(x) at only pt x is

f'(x) = lim f(x + ∆x) - f(x) / ∆x

∆x → 0

alternate def: derative of f(x) at x = c

f'(x) = lim f(x) - f(c) / x - c

x → c

derative = slope of a tangent line = instantaneous rate of change

f'(x) *say f prime of x*

y' *say y prime*

dy/dx *say dee y dee x*

Differentiation Rules

constant rule: d/dex [c] = 0

power rule: d/dx [xⁿ] = nx^n-1

constant multiple rule: d/dx[cf(x)] = c ⋅ f'(x)

sum & difference rule: d/dx [f(x) ± g(x)] = f'(x) ± g'(x)

Derivatives and Integrals

Product & Quotient Rules

product rule

if h(x) = f(x) ⋅ g(x)

then h'(x) = f'(x)g(x) + g'(x)f(x)

think it this way: deriv of 1st times 2nd plus deriv of 2nd times 1st

quotient rule

if h(x) = f(x) / g(x)

then h'(x) = f'(x)g(x) - g'(x)f(x) / [g(x)]²

think it this way: deriv of the top times the bottom minus deriv of bottom times the top all over the bottom squared

Higher Order Derivatives

original: f(x) 1st deriv: f'(x) 2nd deriv: f''(x)

Chain Rule

product rule

if h(x) = f(x) ⋅ g(x)

then h'(x) = f'(x)g(x) + g'(x)f(x)

think it this way: deriv of 1st times 2nd plus deriv of 2nd times 1st

quotient rule

if h(x) = f(x) / g(x)

then h'(x) = f'(x)g(x) - g'(x)f(x) / [g(x)]²

think it this way: deriv of the top times the bottom minus deriv of bottom times the top all over the bottom squared

PVA (Position - Velocity - Acceleration)

particle moving horizontally or vertically

position function: s(t) or x(t) or y(t)

velocity: v(t) = s'(t)

accerlation: a(t) = v'(t) = s'(t)

average velocity Δs/Δt = change in position/change in time

particle changes direction v(t) changes sign

total distance traveled from t = t₁ to t = t₂

D_T |s(t₁) - s(t_c)| + |s(t_c) - s(t₂)|

t_c = time particle changes direction

Implicit Differentiation

basically you solve for dy/dx

ex. derative of y² is 2y ⋅ dy/dx

Related Rates

related rates is to find the rates of chnage of 2 or more related variables that are changing with the respect to time

ex. when water drains out of a conical tank the volume V, the radius r, and the height of the water h are all functions of time

V = 1/3 ⋅ πr²h

dv/dt = 1/3 ⋅ π[2r ⋅ dr/dt ⋅ h + dh/dt ⋅ r²]

bsically determind what you are looking for, when this is occuring, and what you are given

then make a sketch, label variables, write an eq relating to them, differentiate implicty with respect to time

finally substitute known values!

Rolle's Theorem / Mean Value Theorem

Just check out the Theorems

Extrema (find the minimum and maximum), The First Derivative Test (for relative extrema), Concavity & Points of Inflection (POI), & Concavity & 2nd Derivative Test

Okay so honestly, if you take a look at the Needs Statements page, you will be able to solve each one of these problems

Logs & Exponentials

y = log_bx, b > 0 ⇔ x = b^y

y = log_b(xy) = log_b(x) + log_b(y)

y = log_b(x/y) = log_b(x) - log_b(y)

y = log_b(x^y) = ylog_b(x)

y = log_b(^ysqrt(x)) = 1/y ⋅ log_b(x)

ALSO:

e_lnx = x

ln(e_x) = x

Logarithmic Differentiation

use ln as an aid in differentation in non-ln properties

take ln of both sides

Inverse Function

the inverse function is denoted as f^-1 (f inverse)

ex. y = sqrt(2 - 3x)

x = sqrt(2 - 3y)

x² = 2 - 3y

3y = 2 - x²

y = 2 - x² / 3

f^-1(x) = 2 - x² / 3

L'Hopital's Rule

it's done to solve indeterminate form of 0/0 or ∞/∞

basic rule is you take the derivative of the top and bottom seperately to solve!

Integration

ex. y = sin(3x)

dy/dx = cos(3x) ⋅ 3

∫dy = ∫cos(3x)3dx

y = 1/3 ⋅ 3sin(3x) + C

Formal U-substitution

∫xsqrt(2x-1)dx

= ∫(u+1)/2 ⋅ u^1/2 du/2

= 1/4∫u^3/2 + u^1/2 du

= 1/4[2/3 ⋅ u^5/2 + 2/3 ⋅ u^3/2]

= 1/4[2/15 ⋅ u^3/2 (3u + 5)

= 1/30 ⋅ (2x-1)^3/2 (3(2x-1) + 5)

= 1/30 ⋅ (2x-1)^3/2 (6x-3) + 5)

u = 2x - 1

x = (u + 1) / 2

dx = 1/2 ⋅ du

Newton's Law of Cooling

the rate of change in temp of an object is proportional to the difference between the object's temp & the temp of the surrounding medium

y = temp of object & y_s temp of surrounding medium SO... dy/dt = k(y - y_s)

Area of Region between two curves

area between curves = ∫ (top curve) - (bottom curve) dx

the bottom bound is left

the top bound is right

Integration by parts

∫udv = uv - ∫vdu

repeated integration by parts: ∫xⁿsin(ax)dx, ∫xⁿcos(ax)dx, ∫xⁿe^axdx