AP Calculus Study Guide

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continuity
we can draw its graph without lifting a pencil from paper, has no holes, breaks, or jumps on interval

f(c) exists
lim x->c f(x) exists
lim x->c f(x)=f(c)
jump discontinuity
when the left and right hand limits exist, but are different
removable discontinuity
lim x->c f(x) not equal to f(2)
infinite discontinuity
when it has a vertical asymptote
extreme value theorem
if f is continuous on the closed interval [a,b], then f attains a minimum and maximum value somewhere in the interval
the intermediate value theorem
if f is continuous on the closed interval [a,b], and M is a number such that f(a)<M<f(b), then there exists a number c, the the interval
instantaneous rate of change
limit of the difference quotient
memorize derivatives
refer to page 114
Mean Value Theorem
if the function f(x) is continuous at each point on the closed interval a<x<b and has a derivative at each point on the open interval a<x<b, then there is at least one number c, a<c<b, such that f(b)-f(a)/b-a=fprimec
Rolle's Theorem
addition to hypotheses of the MVT, given that f(b)=f(a)=k then there is a number c, between a and b such that fprimec=0
intermediate forms
0/0, or infinity/infinity, 0xinfinity, infinity-infinity, o^0, 1^infinity, infinity^0
L'Hopital's rule
take the derivative of the functions int he numerator and the denominator if it results in intermediate form
instantaneous rate of change
limit of the average rate of change as h->0
instantaneous rate of change vs average rate of change
instantaneous change is defined as the derivative of the function evaluated at a single point. The average change is the same as always (pick two points, and divide the difference in the function values by the difference in the values).
to determine increasing and decreasing functions
derivatives with discontinuities
solve for critical values and use number line, if the derivative is negative then f(x) is decreasing, and vice versa
derivatives with discontinuities
when the derivative is not defined
To determine minimums, maximums, and points of inflection
cusps
find derivative/second derivative, find critical points, if second derivative is greater than 0 then c yields a local minimum and vice versa
cusps
don't have minimums or maximums, but can change in concavity
optimization problems
find equations, find equation you need to minimize, substitute original equation into equation for minimization, take the derivative, set equal to 0 and solve
relating a function and its derivatives graphically
refer to page 174
to construct graphs of f, f', f''
use number lines and know when they change signs, minimums, maximums, points of inflections
motion rules
if velocity is greater than 0 then the particles is moving to the right and vv, if acceleration if greater than 0 then velocity is increasing and vv, if acceleration and velocity are both positive then they are accelerating and if they have different signs then they are decelerating, if the position of the particle is continuous then the particle reverses its direction whenever velocity is 0 and the acceleration is different from 0
local linear approximation
local linear approximation
if f'(a) exists, then the local linear approximation of f(x) at a is f(a) + f'(a)(x-a)
related rates
related rates
refer to page 185
anti derivative formulas
refer to page 211-212
differential equations: motion problems
refer to page 223
Fundamental Theorem of Calculus
if f is continuous on the closed interval [a,b] and F' = f, then according to the FTC, F(b)-F(a)
properties of definite integrals
refer to page 242
definite integrals for F'(x)
use FTC and if it is a constant then you can ignore, just multiply by derivative
definition of definite integral as the limit of Riemann sum
refer to page 246
Riemann sum
start by taking (b-a)/n in interval to get width (where n is the number of sub intervals needed)

left: use left most x point and substitute into f(x), then multiply by width calculated initially, then do this process for the number of sub intervals needed while adding the width each time you move on to a new point

right: use right most x point out of first interval and substitute into f(x), then multiply by width calculated initially, then do this process for the number of sub intervals needed while adding the width each time you move on to a new point

midpoint: use left and right most x point on the first interval and divide by 2, then substitute into f(x), then multiply by width calculated initially, then do this process for the number of sub intervals needed while adding the width each time you move on to a new point
trapezoidal rule
T(n)= [(y0+y1)/2] h1 + [(y1+y2)/2] h2 + [(y2+y3)/2] h3 + [(y3+y4)/2] h4 +..........
average value
(intergal)/b-a
to find are between curves
find intersection points, then put into integral with top curve minus bottom curve, or right curve minus left curve
to find volume between curves
same process as area
finding volume with disks
same process but use (pi)r^2(dx)
finding volume using washers
same process but use (pi)(R^2-r^2)dx
finding volume with shells
same process but use 2(pi)rh(dx) *you can use instead of washer method if you can still go through 2 curves
to find distance traveled using intergals
find out time it is going through and input equation into derivative
order of derivatives for motion
x(t)
v(t)
a(t)
to determine slope fields with f'(x)
take antiderivative of f'(x) and match with graph
to solve for C
set the same variables to their respective sides, take the antiderivative, solve for C with given variable values
Exponential Growth and Decay
refer to page 363
for volume of cross sections
semircircle: pi/2
equilateral triangle: root 3/4
and then do top minus bottom squared
e^0=1
sin(0)=0
cos(0)=1
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