AP CALC AB

Unit 1

Estimate of the rate of change → b(32)-b(31.999) / (32-31.999)

How should you leave answers in decimal form? → round to 3 decimal places, truncate or cut off

Squeeze theorem → f is in between g and h, ( g(x) ≤ f(x) ≤ h(x) ), if lim x→a g(x) = L and lim x→a h(x) = L then lim x→a f(x) = L

Types of discontinuties →

Hole (removeable) = found when you cancel out factors in both denominator and numerator
VA (non-removeable) = found when denominator = 0 and is not canceled out as a hole
Jump (non-removeable)
Oscillate (non-removeable)
(x^2 +1) as a factor is not a discontinuity

Formal definition of continuity → if f(x) is continuous at x = c then,

f(c) is defined / in domain
Lim x→c f(x) exists
f(c) = lim x→c f(x)

Limit of infinity theorem → lim x→∞ (1/x^r) = 0 and lim x→-∞ (1/x^r) = 0 (where r is a rational number)

Slant asymtote → non-remainders when h(x) = f(x) / g(x)

Find the domain →

Demoinators
Even roots (radicals) = √x+3 → x+3 = 0
Logarithms = ln(x-1) → x-1 = 0

Basic rules for HA →

Denominator bigger means HA = 0
Numerator and denominator equal means HA = leading coefficient fraction
Numerator is bigger means HA = DNE

IVT →

f(x) is continuous on a,b
f(a) ≠ f(b)
f(c) is between f(a) and f(b)

Conclusion = according to the IVT there is a value c such that f(c) = k and a ≤ k ≤ b

Rule for limits of composite functions → if lim x→a g(x) = k and f(x) if continuous at x = k, then lim x→a f(g(x)) = f(lim x→a g(x)) ; the initial values sign depends on if it was approached from the top or bottom, positive or negative

Quadratic formula → x = (-b plus minus √(b^2 -4ac)) / 2a

a^3 + b^3 = (a + b)(a^2 -ab + b^2)

Unit 2

Average rate of change (secant slope) = f(a+h) - f(a) / ((a+h) - a)

Instantaneous rate of change (tangent slope) = lim h→0 f(a+h) - f(a) / h

Equation of tangent line → y-y1 = m(x-x1)

Differentiability fails to exist at →

Discontinuity
Corner or cusp
Vertical tangent
Oscillate

Power rule → d/dx = nx^(n-1)

Derivatives →

d/dx sinx = cosx
d/dx cosx = -sinx
d/dx a^x = a^x (lna)
d/dx e^x = e^x
d/dx loga x = 1/x (1/lna)
d/dx lnx = 1/x
d/dx tanx = sec^2 x
d/dx cotx = -csc^2 x
d/dx cscx = -cscxcotx
d/dx secx = secxtanx

Product rule → h(x) = fg, then dh/dx = df/dx (g) + (f) dg/dx

Quotient rule → h(x) = f/g, then dh/dx = df/dx (g) - (f) dg/dx / (g^2)

If f is differentiable at x = c → f is continuous at x = c

General derivative → d/dx = lim changex→0 (f(x+changex)-f(x)) / changex

Derivative at a point → d/dx at c = lim h→0 f(c+h)-f(c) / h

Unit 3

Chain rule → f(g(x) = dy/dx( g(x) ) (dg/dx)

Implicit differentiation → d/dx y^2 = 2y (dy/dx)

Horizontal tangent line exist when dy/dx = 0
Vertical tangent line exist when dy/dx is undefined

Derivative of an inverse function → d/dx (f^-1)^prime (x) = 1/ f^prime ( f^-1 (x) )

Inverse trig derivatives →

d/dx sin^-1 x = 1 / √(1 - x^2)
d/dx cos^-1 x = -1 / √(1 - x^2)
d/dx sec^-1 x = 1 / IxI √(x^2 - 1)
d/dx csc^-1 x = -1 / IxI √(x^2 - 1)
d/dx cot^-1 x = -1 / x^2 + 1
d/dx tan^-1 x = 1 / x^2 + 1

Log properties →

ln(1) = 0
ln(ab) = lna + lnb
ln(a/b) = lna - lnb
ln(a^n) = nlna

Higher power derivatives → d^4y/dx^4

Linear approx → L(x) - f(a) = dy/dx (a) (x-a)

Unit 4

Position → s(t)

Velocity → ds/dt = v(t)

Acceleration → d^2s/dt^2 = dv/dt = a(t)

v(t) < 0, → particle is moving left or down

v(t) > 0, → particle is moving right or up

v(t) = 0, → particle is at rest

Average velocity (a,b) → s(b) - s(a) / b - a

Speed = I velocity I

Speeding up → velocity and acceleration have same sign

Slowing down → velocity and acceleration have different signs

Displacement → distance from star to end ( net change )

Related rate problems → in respect to time d/dt

Concave up → under estimate

Concave down → over estimate

L Hosptials rule → lim x→a f(x) / g(x) = 0/0, then take the derivate of both numerator and denominator separately

Interperating a first derivative in context → the ______situation context______ is ____increasing/decreasing____ at a rate of ____I df/dx I____ at t = ___within correct units___

Interperating a second derivative in context → the rate at which ______ is ____increaing/decreasing____ at a rate of _______ at t = ______

Motion of acceleration →

a(t) > 0 then object is accelerating right or up
a(t) < 0 then object is accelerating left or down
a(t) = 0 constant velocity

Unit 5

MVT → d/dx (c) = f(b) - f(a) / b - a if f is continuous and differentiable over (a,b)

Extreme value theorem → if a function is continuous over the interval (a,b) then f has at least one minimum value and at least one maximum value

Critical point →

d/dx does not exist
d/dx = 0

The first derivative test →

Minimum value of x = c occurs when d/dx changes sign from negative to positive
Maximum value of x = c occurs when d/dx changes sign from positive to negative

Candidates for absolute extrema →

Critical points
End points of the domain

F double prime > 0 → concave up

F double prime < 0 → concave down

Point of inflection → when f double prime changes sign, when concavity changes

2nd derivative test → if d/dx (c) = 0 then,

When f double prime of c > 0 there is a relative minimum at x = c
When f double prime of c < 0 there is a relative maximum at x = c

Sketching graphs of derivatives → the slope of f is the y - value of d/dx

An object is speeding up when → velocity and acceleration have the same sign

An object is slowing down when → velocity and acceleration have different signs

Rolles theorem → if f is continuous and differentiable over (a,b) then f(a) = f(b), there is at least one number (c) in (a,b) such that d/dx (c) = 0

Relative extremas only occur at → critical points

If d/dx > 0 then → f is increasing

If d/dx < 0 then → f is decreasing

When d/dx is increasing → it is concave down

When d/dx is decreasing → it is concave up

When f double prime is > 0 → f is concave up

When f double prime is < 0 → f is concave down

Point of inflection → changes in concavity

Point of inflection theorem → if (c, f(c)) is a point of inflection, then either f double prime = 0 or f is not differentiable at x = c