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Unformatted text preview: MATH 137 Exam Equation Sheet
PROPERTIES OF REAL NUMBERS
Properties of real numbers:
associativity, commutativity, existence of identity,
existence of inverses
Subsets of real numbers:
natural numbers (e.g. 1,2,3,…), integers (e.g. 2,1,0,1,2,…), rational numbers (i.e.
p for p, q
xx=
q
integers, q ≠ 0 ).
PROPERTIES OF FUNCTIONS
Domain:
The domain is the set of values of x for which f(x)
is defined
Range:
The range is the set of all possible values of f(x)
Even Function:
Even functions are functions that have the
property f (− x ) = f ( x ) for all x
Odd Function:
Odd functions are functions that have the
property f (− x ) = − f (x ) for all x
Invertible Function:
An invertible function is a function f is invertible if
there exists a function g such that y = f x if and () only if g ( y ) = x
Increasing Function:
An increasing function is a function is increasing
on an interval I if for all x, such that f (x1 ) < f ( x 2 )
whenever x1 < x2 on I
Decreasing Function:
A decreasing function is a function is decreasing
on an interval I if for all x, such that f ( x1 ) > f (x2 )
whenever x1 < x2 on I
MATHEMATICAL INDUCTION Induction step by step
Step 1—Base case. Show that the statement holds
in the simplest case (normally for n = 0 and/or for
n = 1).
Step 2—Induction Hypothesis. Assume the
statement holds for an arbitrary k, i.e. for n = k.
Step 3. Prove that it holds for n = k + 1 given the
above induction hypothesis.
Strong (or complete) induction:
Assume that the statement holds for n < k , and
then prove it true for n = k . Upper/lower bound:
A has an upper bound if there is a number a such
that for all x ∈ A , x ≤ a . A has a lower bound if
there is a number a such that for all x ∈ A , a ≤ x .
Bounded:
A is bounded iff A has both an upper bound and a
lower bound; iff there exists a number b such that
for all x ∈ A , x ≤ b .
Least Upper Bound:
b is the least upper bound of A iff b is an upper
bound of A and if for all other upper bounds a of
A, b ≤ a .
Greatest Lower Bound:
b is the greatest lower bound of A iff b is a lower
bound of A and if for all lower bounds a of A,
a ≤ b.
Completeness of real numbers:
If A is a non empty set of real numbers and A has
an upper bound then A has a least upper bound
If A is a non empty set of real numbers and A has
a lower bound then A has a greatest lower bound If you’re taking the limit as x → ∞ of a
quotient, and substitution yields an invalid
answer, then try first to divide both the
numerator and the denominator of the limit
expression by its highest power of x
If the limit expression contains absolute
values, then try breaking the limit up into
two onesided limits.
If the limit expression is a quotient, and if
factoring doesn’t work, then try l’Hopital’s
Rule:
f ( x)
f ′( x) • • • lim
x→a •
•
• f (a ) and f (b ) then there exists a c such that lim f ( x ) = lim h( x ) = L then lim g ( x ) exists and is
x→a x→ a x→a equal to L.
THE DERIVATIVE h→0 h Rules for differentiation:
for functions u,v and for constants c, n
•
dy
n
n −1 du
dx • log b b = 1 log b b r = r •
•
•
• g ′( x) a < c < b and f (c ) = k
Squeeze theorem:
If f ( x ) ≤ g (x ) ≤ h( x ) near a and f ′( x ) = lim log b 1 = 0 • x→a Differentiation by first principles
f ( x + h ) − f (x ) . log b m r = r log b m
1
log b = − log b m
m • = lim Intermediate value theorem:
If f ( x ) is continuous on [a, b] and if k is between LOGARITHMIC FUNCTIONS
Properties of logarithms:
•
log b (mn) = log b m + log b n
⎛ m⎞
•
log b ⎜ ⎟ = log b m − log b n
⎝ n⎠ g ( x) b logb m = m
log x
10
=x
e ln x = x
log a m
log b m =
log a b • LIMITS AND CONTINUITY (cu ) = cnu dx dy
(uv ) = u dv + v du
dx
dx
dx
du
dv
v
−u
dy ⎛ u ⎞
dx
dx
⎜ ⎟=
dx ⎝ v ⎠
v2 Chain rule:
If f ( x) = (a o b )( x) , then f ′( x) = a ′(b( x )) ⋅ b′( x)
Power rule:
d u(x )
du
c
= c u ( x ) ⋅ ln c ⋅
dx
dx
Higher order derivatives:
f ′′( x) = d ⎛ dy ⎞ .
⎜⎟
dx ⎝ dx ⎠ More free study sheet and practice tests at: INEQUALITIES AND ABSOLUTE VALUES
Absolute value rules:
•
a+b ≤ a + b
• a−b ≥ a − b • a ⋅b = a ⋅ b Formal Definition:
The function f(x) has limit L, as x approaches a,
denoted lim f (x ) = L if given any ε > 0 , there DERIVATIVES OF LOGARITHMIC AND
EXPONENTIAL FUNCTIONS x→a exists a δ > 0 such that f(x)  L < ε for all x
satisfying 0 < x – a< δ.
Properties of limits:
•
lim[ f ( x ) ± g ( x )] = lim f (x ) ± lim g ( x )
x→a • x→ a x→a x→a •
• x→a x→a lim[ f (x ) ⋅ g (x )] = lim f ( x ) ⋅ lim g (x )
lim f ( x ) ⎡ f (x )⎤ x →a
=
lim ⎢
x→a g (x ) ⎥
⎦ lim g ( x )
⎣
x→a
n [ x→a dx •
•
• lim[ f ( x )] = lim f ( x )
x →a For u, a function of x:
•
d
1 n Techniques for finding limits:
•
First try substitution, factoring (ln x ) = x d
1 du
(ln u) = ⋅
du
u dx
d
1
du
(log b u) =
⋅
(ln b)u dx
dx
dx
e = ex
dx • du
du
(e ) = e u
dx
dx • du
du
(a ) = a u (ln a )
dx
dx More free Study Sheets and Practice Tests at: w ww.prep101.com Need help for exams? Check out our classroom prep sessions  customized to your exact course  at www.prep101.com TECHNIQUES OF DIFFERENTIATION
Implicit differentiation:
•
to differentiate an implicit function
Step 1. Take the derivative of both sides with
respect to x. Use the Chain Rule on terms
involving y (and note that the derivative of y with
respect to x must be left as dy dx .)
Step 2. Collect all terms involving dy dx on one
side of the equation.
Step 3. Solve for dy dx .
Logarithmic differentiation:
•
to differentiate a function with complicated
exponent/a product of several functions, etc.
•
to differentiate functions with an x in both
the base and the exponent
Step 1. Take the ‘ln’ of both sides (to find an
expression of the form ln y = ln[ f ( x )] ).
Step 2. Simplify ln( f ( x )) by using the properties of
logarithms.
Step 3. Differentiate both sides with respect to x
(thus: 1 dy d
).
⋅
=
(ln f (x ))
y dx dx Step 4. Solve for dy dx .
Step 5. Express the answer in terms of x only
(substitution f(x) for y).
APPLICATIONS OF THE DERIVATIVE
L’Hopital’s rule:
If the limit of the quotient of differentiable functions f (x ) and g ( x ) are of types 0 or ∞ ,
0
∞
and if g ′(a ) is not 0, then
lim
x →a Vertical asymptote:
The line x = a is a vertical asymptote for the graph
of the function f(x) if and only if lim f ( x ) = ±∞
x→a + or lim f ( x) = ±∞ .
Horizontal asymptote:
The line x = b is a horizontal asymptote for the
graph of the function f(x) if and only if lim f ( x ) = b x → +∞ or lim f ( x) = b .
x → −∞ Mean value theorem:
If f is continuous on [a, b] and differentiable on (a,
b), there exists some x in (a, b) for which
f (b ) − f ( a ) .
b−a INTEGRATION • minimum at p. If f ′( p ) = 0 and f ′′( p ) < 0 , then f has a relative maximum at p. If f ′( p ) = 0 and
f ′′( p ) = 0 , then the test fails. Concavity:
If f ′′( p ) > 0 on an interval, then f(x) is concave up
on that interval; if f ′′( p ) < 0 , then f(x) is concave
down on that interval. A point of inflection occurs
when f ′′( p ) changes sign (and thus concavity). •
•
• n
∫ x ⋅ dx = ∫ dx
= ln x + C
x ∫ sin xdx = − cos x + C
∫ cos xdx = sin x + C
∫ tan xdx = ln sec x + C
b a ∫ kf ( x) ⋅ dx = k ∫ f ( x) ⋅ dx
b b b ∫ [ f (x ) ± g (x )]⋅ dx = ∫ f (x ) ⋅ dx ± ∫ g (x ) ⋅ dx
a • 0 if , then −a TECHNIQUES OF INTEGRATION
Integration by substitution
Step 1. Choose a substitution u such that the
resulting integral, written in terms of u and du,
will be easy to evaluate (easiest with practice).
Step 2. Differentiate u with respect to x to find
du = u ′(x )dx .
Step 3. Replace dx with du u ′(x ) .
Step 4. Integrate with respect to u.
Step 5. Substitute back to the initial variable x.
Integration by parts
Step 1. Write the given integral ∫ f ( x) g ( x) ⋅ dx . Step 2. Introduce the intermediary functions u(x)
and v(x):
⎧ u = f (x )
⎨
⎩dv = g ( x )dx
Step 3. Differentiate u and integrate dv to get:
⎧
⎪ du = f ′( x ) .
⎨v = g ( x )dx
⎪
∫
⎩
Step 4. Use the following formula:
u (x )dv = u (x )v ( x ) − v( x )du and solve. ∫ ∫ P(x )
Q( x ) Integration by Trigonometric substitution
•
To be used when the integrand contains the b a • −a Step 1. If the degree (P) ≥ degree (Q), perform
polynomial longdivision. If it is not, proceed to
step 2.
Step 2. Factor the denominator Q(x) into
irreducible polynomials: linear and irreducible
quadratic polynomials.
Step 3. Find the partial fraction decomposition.
Step 4. Integrate the result of step 3. x n +1
+C
n +1 Properties of integration: • w ww.prep101.com
•
f (− x ) = − f ( x ) f (x ) = ∫ positive at p, then f has a relative minimum at p. If
f ’ changes from positive to negative at p, then f
has a relative maximum at p.
Second derivative test:
If f ′( p ) = 0 and f ′′( p ) > 0 , then f has a relative a Partial fractions:
•
to integrate rational functions of the form Rules of integration:
•
dx = x + c • a ∫ f (x )dx = 2∫ f (x )dx ∫ f (x )dx = 0 Optimization problems
Step 1. Determine what we’re trying to
maximize/minimize and write an equation for
this.
Step 2. Write a second equation from additional
information given in the problem, isolate one of
the two variables, and substitute this into the first
equation.
Step 3. Take the derivative of this equation, set it
to zero and solve for the remaining variable.
Step 4. Plug the value of this variable into the
original equations to solve for the remaining
variables. First derivative test:
If f ′( p ) = 0 and f ’ changes from negative to Odd and even functions:
•
if f (− x ) = f ( x ) , then a Curve sketching
Step 1. Find the intercepts.
Step 2. Find all the asymptotes.
Step 3. Find critical points and the intervals of
increase/decrease.
Step 4. Find inflection points and the intervals in
which the function is concave up/down. f ( x)
f ′( x )
= lim
g ( x ) x → a g ′( x) Critical points:
The values of x ∈ domain of f(x) such that
f ′( x ) = 0 or f ′( x ) is not defined. b ∫ f (x )dx = F (b) − F (a )
a x →a − f ′( x) = The fundamental theorem of Calculus:
For F ′(x ) = f (x ) a a b c b a a expressions a 2 ± x 2 or x 2 ± a 2 .
o
For a 2 − x 2 set x = a sin(t)
o
for a 2 + x 2 set x = a tan(t)
o for x 2 − a 2 set x = a sec(t) c ∫ f ( x) ⋅ dx = ∫ f ( x) ⋅ dx + ∫ f ( x) ⋅ dx Our Course Booklets  free at prep sessions  are the “Perfect Study Guides.” Need help for exams? Check out our classroom prep sessions  customized to your exact course  at www.prep101.com Improper Integration
•
If the integral exists (and equals, or can be
written as, a number), then it’s convergent
•
If the integral doesn’t exist (is infinite), then
it’s divergent The moment about the xaxis is given by
b
n
1
1
2
2
M x = lim ∑ ρ ⋅ [ f ( xi )] ∆x = ρ ∫ [ f ( x) ] dx
n →∞
2
2
i =1
a
SEQUENCES AND SERIES
Definitions, Rules and Properties APPLICATIONS OF INTEGRATION Arithmetic:
•
constant growth a n +1 = a n + d where d is a Area:
The area between functions f(x) and g(x) (whose
intersection points are a and b, and where
b f (x ) > g ( x ) on (a,b)) is given by ∫ [ f ( x ) − g ( x)]dx Average Value
The average value of function f(x) on the interval
b
[a, b] (or (a, b)) is given by
.
1
Arc Length
The arc length of the function f ( x ) from
to x = b is given by b
2 Fibonacci:
•
defined recursively with a1 = 1 , a 2 = 1 ,
a n = a n −1 + a n − 2 for n ≥ 3 f= ∫ b−a∫
a f ( x)dx x=a 1 + ( f ′( x )) ⋅ dx a Surface Area
The surface area from a to b of the solid obtained
by revolving y = f (x ) around the xaxis is given
by b 2π ∫ f (x ) ⋅ 1 + ( f ′( x )) ⋅ dx
2 • then the volume is
V = ∫ A( x )dx
a If the cross section is perpendicular to the yaxis and its area is a function of y, say A( y ) ,
then the volume is
b V = ∫ A( y )dy Algebra of series:
N Volumes of solids by revolution:
•
The volume of the solid obtained by rotating
the graph of f x from x = a to x = b () about the yaxis is given by N ∑a ± ∑b
n =1
N n ∑a n n =1
N n n is convergent, then ∞ ∑b ∞ ∑a is divergent then n that ai ≥ ai +1 ∞ ∑ (− 1) n +1 i =1 an for all i and lim a = 0 , then
n
n →∞ converges. Limit comparison test:
a
if
such that
lim n = L
n →∞ b
n a ⋅ (r M −1 − r N )
=
1− r
SEQUENCES AND SERIES
Convergence Tests a The volume of the solid obtained by rotating
the region bounded between curves f ( x ) and g (x ) between x = a and x = b (assume
g (x ) > f ( x ) for a < x < b ) about the line y = c π ∫ (c − g ( x )) − (c − f ( x )) dx
2 2 a Centroid:
The moment about the yaxis is given by
n b i =1 . then and
n k ∞ ∑b and L≠∞, converge or diverge
n k together.
Ratio test:
•
Given a n > 0 and Pseries:
•
A Pseries • then
• converges and if L > 1 , n If L = 1 , then the test fails. n →∞ If L < 1 , then ∞ a n = f (n ) where f is continuous, n Root test:
•
Given a n > 0 and lim(a )1 n = L .
n
∞ ∑a
i =1 then
If converges and if L > 1 ,
n ∞ ∑ a diverges.
L = 1 , then the test fails.
i =1 • function ∞ ∑a .
a n+1
=L
an diverges. ∞ ∑a
i =1 1 is convergent if p > 1 , and is
∑ np
n =1
divergent if p ≤ 1 . i =1 If L < 1 , then i =1 • Integral test:
•
let ∞
be a series such that there exists a
∑ ai lim
n →∞ n Absolute and conditional convergence:
•
A series of the form
a converges ∑ a M y = lim ∑ ρ xi f ( xi )∆x = ρ ∫ xf ( x)dx ∞ ∑a L≠0 n =1 N ⋅ (N + 1)
2
n =1
N
N ⋅ ( N + 1) ⋅ ( 2 N + 1)
2
∑n =
6
n =1
n=M is is divergent. n ∑n = ∑a ⋅r n Alternating series test:
•
for {a } a sequence of positive numbers such
n n =1
c −1 N n −1 ∞ ∑a
i =1 i =1 N n =1 2π ∫ x ⋅ f ( x ) ⋅ dx n →∞ ∞ ∑b convergent. If = ∑ an − ∑ an b [ If i =1 = ∑ (a n ± bn ) Summation formulas: N a b i =1 for all n. {a n }⋅ {bn } = {a n ⋅ bn } n=c b is given by such that 0 < a n ≤ bn n w ww.prep101.com
i =1 Algebra of sequences:
{a n } ± {b } = {a n ± bn } c ⋅ {a n } = {c ⋅ a n } ∞ ∑b and n • Volume:
•
If the cross section is perpendicular to the xaxis and its area is a function of x, say A(x ) , • ∞ ∑a for i =1 a • ∫ f (x ) ⋅ dx Comparison test: 1 n converges, then the series ∞ converges and if it diverges, then the series
diverges. Geometric:
•
proportional growth g = rg where r is a
n +1
n
constant, arbitrary terms are given by
g = r n −1 ⋅ g . a If 1 • constant
. • n absolutely if the series of absolute values (of
the form
an ) is convergent. positive and decreasing. ∑ • If it converges absolutely, then the original
series
a will also converge. ∑ n Our Course Booklets  free at prep sessions  are the “Perfect Study Guides.” ...
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