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Unformatted text preview: Amputee;
V.A.: denominator = 0
”mx_,,:+ f (x) ==§ ltmxacHx) =% * VA at x = C Use your “brain“ to determine whether the limit is approaching inﬁnity or negative inﬁnity by
substituting a value close to c into the function. [fit is a positive value, the limit = no. [fit is a negative
value, the limit:  no. H.A.: degree of the numerator < degree of the denominator... y = O “Trix+1.“ f (x) = 0
degree of the numerator = degree of the denominator... y 2 leading coeﬁ‘s iimxnim f (x) = H. A.
degree of the numerator > degree of the denominator... long division limxniwf (x) = 00 or timxnimex) = ~00 Lmits:
“ A hole does not prevent a limit from existing.
* The limit of a constant is a constant. * If the limit is approaching no or  can, think about HA. or divide every term by the variable with'the
largest degree in the denominator. Then, remember that: llmxnimi = U. "‘ Ifthe answer is of the form it'mxnc % = 02’ we have an indeterminate form. Try one of the following
techniques: .
a Factor the numerator d: denominator, simplify, and then substitute the value ofc into the new
expression. I Rationalize the numerator or the denominator by multiplying by the conjugate.
0 Simplify the complex fractiou by getting a common denominator. 1—cosx
— 0 *Special Limits: zimxnus‘“‘”3= 1 timing x — ax " PiecewiseDeﬁned Functions I To ﬁnd the limit of the function as it approaches a value where the function does not Split:
Evaluate the limit of the ﬁltration for that speciﬁc piece of the function. a To ﬁnd the limit of the function as it approaches a value where the function does split: Evaluate a
left—hand limit using the piece of function on the left side & evaluate a righth limit using the
piece of ﬁmction on the right side. If the left—hand limit = the righthand limit, then an overall
limit exists. Intermediate Value Theorem:
A function that is continuous on a closed interval [o, b] takes on every value between ﬁn) and ﬁb). in
other words, if yo is between ﬂu) andﬁb), then y0 = c for some 6 in [a, b]. Continnigg:
i) ﬁe) must exist the point must exist “ Endpoints only need onesided limits
2) limxncf (3:) must exist to prove that the function is continuous 3) iimxnchx) = ﬁe) at that point. Avegge Rate o! Change vs. Instantaneous Rate of Change:
Average Rate of Change = slope of the secant line roHcaJ = nae—rm m .=
ba h
Instantaneous Rate of Change = slope of the tangent line (derivative)
m = timnng m * Algebraically: original denominator of it must cancel. n
Dem‘ ative:
" Rate of change (instantaneous) = slope of the function = slope of the tangent line
* The differentiable function must be continuous. But, a continuous ﬁmction doesn’t have to be
diﬁ‘erentiahle. Think about the graph of the absolute value of x.
* The derivative of a constant is zero.
" Ifa function has a derivative it is said to be differentiable. Qerivetive of an Inverse Function;
If f & g are inverses of each other, the derivative of f can be found as follows: f1") = 3:010,” "' The derivative of a function = 1f(derivative of the inverse)
" Twist: What value goes into the inverse? Not the value given for the function! [Inglisit Differentiation:
When you can’t isolate y in terms of x, it's time to take the derivative implicitly. l) Diﬁ‘erentiate both sides of the equation with respect to 3:. Any time you take a derivative of y label it
ajufdx. 2) Collect the terms with obi/ab: on one side of the equation. 3) FactOr out dyfdx. 4) Solve for ajzla’x by dividing. Normal Line:
A normal line is simply the line perpendicular to the tangent line at the same point. Therefore the slopes
are negative reciprocals of each other. Linearization:
Finding the equation for the tangent line. Diﬁ‘erentgg' :
dy = f'Cxo) dx Take the derivative, ajJ/dr, & multiply dx to the other side of the equation
a)»: approximate value of y between the tangent line & value of the function ctr: the change in the values of the x's
ftx) = dy + f (xu) Critigg ﬁghts:
Points on the graph of a function Where the derivative is zero or is undeﬁned. f (x— =:+:i "‘ Derivative = 0 when g(x) = 0. (Numerator = 0) * Derivative is undeﬁned when hot) = 0. (Denominator = 0) positive negative “cup” up “cup” down
slope slope original‘s f’>0 f’<0 f’ is f’ is
ﬁrst increasing decreasing
derivative f” > 0 ftr< 0 fit: 0
changes
 to + or
' + to —
No
endpoints Particle Motion:
* Position of the particle at time I, usually denoted x(t) or 50‘), is its location on the number line if it
moves in one dimension, or its height if it moves in two dimensions. "‘ Velocity PU) = s'(t) of the particle is the rate of change of position with respect to time, or the
derivative of position. I v(:) = 0 Particle at rest I v0) > 0 Particle moves to right (or up)  110‘) < 0 Particle mow/es to left (or down) 0 Sign of 12(1) changes Particle changes direction
a v(t) Speed is the absolute value of velocity * Acceleration (10') = v‘(t) =' s”(t) of the particle is the rate of change of velocity with respect to time,
or the derivative of velocity. "' When the velocity and acceleration of the particle have the same signs, the particle’s speed is
increasing. * When the velocity and acceleration of the particle have the opposite signs, the particle’s speed is
decreasing (or slowing down). Eewton’s Method:
x“; = x“ — ﬁx“) Calculator: Enter the function into 3/1 and derivative into y; f'[xn)
Homescreen: Enter initial guess [etc>1 as “it“
Enter: x— ylfyz [sto>] x
Keep on hitting enter until the values satisfy the desire of
accuracy (4 places) Mean Value Theorem:
If a function' l8 continuous on the closed interval [a, b] 8: differentiable on the open interval (a. b), then f (c): fajita) "' Instantaneous rate of change — Average rate of change
* Slope of the tangent line = Slope of the secant line Roﬂe’s Theorem:
If a function is continuous on the closed interval [0, b], differentiable on the open interval (a. b) and has the same yvalue at the endpoints a and b, then there must be at least one value of 2:, call it c,
between a and b where the function has a horizontal tangent (max or rain point). Riemann Sums: RRAM, MRAM, LRAM:
The left—hand, right—hand, and midpoint sums are all examples of a more general. approach to ﬁnding
areas called Riemann sums. In a Riemann sum, a function & an interval are given, the interval is
partitioned, and the height of each rectangle can be the value of the function at any point in the
subinterval. In fact, even the subintervals do not necessarily need to be the same length. All you need to
remember is that a Riemann sum uses rectangles to ﬁnd the area between a curve and the x—axis Area Under a Curve
"‘ An integral will calculate the area under the curve to the x—axis.
“' If the area is under the xaxis, the integral will be equal to a negative value. Area Between Two Curves'
* Vertical rectangles: Top — Bottom
‘ Horizontal rectangles: Right  Left Tranezoidgl Rule:
b.—
Trap = 2—:0’0 + Zyl + 23/2 + + Zyrt—l + yn) Antiderivatives;
"' The process of ﬁnding an antiderivative is called antidifferentiation or integration. Fundamental Tlieoreg 9f Calculus, Part 1 & 2:
For) = la fCrJdt E3 = j—xfjrcodt = foe) ——————» d1 g = ifrﬂﬂtkt = f(v(x))1J’(x) ———* l = if: sec (0dr = 3::2 sec (x3)
f:f(t)dt = raw) — Ha) I: mm = F(a) — Fan = —[F(b) — Fcan
l:f(t)dt = o [nte all:
WI Total Area A1153 1) Find out if the function goes below the xaxis.
“m N l: 2) Take the absolute value of the areas... no negative values III f (x) Idx Mil‘1 h 1' mm  a
"' Position Shift orlPiepl_ee_emenﬁ= integral: L) f (x) (1.1: (Keep negative values) * Maximum Value, minimum value or yvalue: Displacement +lnitial value
(15 there an initial value given?) Average Value: 1 b
all f 0‘de
Volumes: Disk Method: 7r f:[f(x)]2dx "' Revolve around xaxis integrate w.r.L x’s
Washer Method: 11' I: [(R (26))2 — (r(x))2]dx * Revolve around yaxis integrate w.r.t. y’s.
Shell Method: 21: f:(mdius)(helght)dx * Height: deals with the functions: (top # bottom) or (right — left) * Radius: deals with axis of rotation & at variable (if revolved about y‘s) or ands of rotation 3: 3; variable (if revolved about :65) Slicing: 1) If crosssection is perpendicular to x—axis integrate w.r.t. x’s or if it is perpendicular to yaxis integrate w.r.t_ y‘s.
2) What is the shape of the cross—section? Find the area formula.
3) Incorporate the functions into the area formula.
4) Integrate L'Hugital’s Rule: to: )_
“m" C ctr—5 a (c) * Apply only when taking the limit of a quotient. If the expression is an indeterminate form. but not in
the than of a quotient, it must be rewritten in quotient form. * Remember to take the derivative of the numerator & the derivative of the denominator. Do not use the
Quotient Rule! "' Be sure to recheck each time L’ Hopital’s Rule is applied to see if the result is still indeterminate. it
may be applied more than once, but only if the limit results in an indeterminate form each time. Euler’s Method:
YﬂH = 3’11. + thxUJ’D) «+44% = UK y= In x
Horizontal Asymptote at y= 0 Horizontal Asymptotc at y= 0 Vertical Asymptotc at x: 0
Vertical Asymptote at x = 0 217 gr y=sinx y=cosx y=tanx
Period: 21: Period: 21: Period: 1:
Asymptotes at x =2): Exthngorean Identities: 5!:an + coszx = 1 1 + tartan: 2: seczx 1+cot2x = csczx PowerReduction Identities: 1—mst 1400st
sin2x= 2 caszx= z 'Mgmmmgg ysmx Quad:I.W y=cos“x Quad:I.I]
=csc“ (1m =sac"'[1!x) y=mii" x Quad:I.W MM; £(sinu) =cosuE £02") =e"%—‘: % (cosu) = ﬁlm?  720:“) = nubtag é (tuna) = seczu% far—(mu) = 5% 3: (com) = «sang iﬂognu) = 7:33;“ oruae the Change of Base formula
% (5.2qu = swarmE 109,11 i—E a an
E (cscu) — —cscucatu3 d  1 du
39*" “lbw; d .1 _ 1 all
63011“ n) _ 11111 a: a __ _ 1 «In
ECSEC Lu) _‘ uvu‘—ldx mg“ F9 mulu: f (sinquu = wcasu + C I(casu)du = sinu + C
f(sec2u)du = mm: + C Hcsc‘uwu = Hcot'u + C
fisecutanuMu == secu + C
j (csmcotu)du = —csm + C f(tanu)du = —!nlcosul + C =1nsecul + C
Kannada; = tnlsinul + C = £nlcscul + C I(secu)du = lulsecu + canal + C
f(cscu)du = —£nlcscu + catnl + C  Iﬁsinzuku = 1—?!" du PowerReducing ldcnﬁty
'I(tﬂ$zu)du = f “in" du PowerRedwing Idmlity
'J' tanzudu = J' (52:21; — 1) tin Pythngmmn Identity
I cotzudu = ﬂcsczu — 1) tin Pythagorean Identity
“my, u)du = Igdu Change :3me Formula ...
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 Fall '11
 KUSTIN
 Calculus

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