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MATH 137
Calculus I:
Section 003
Instructor:
Name: Sean Speziale
Email: [email protected]
Office: M3 2128
Advising Hours: Monday, Wednesday at 12:00-13:30, MC 4023
Office hours: starts week 3, details TBA ;wip
9/9/13
Preliminaries
IntegersZ = …,−2,−1,0,1,2,…
p
Rational numbersQ = q,p ∈ Z,q ∈ Z
Irrational numbers: cannot be expressed as ratio, non-terminating and non-repeating decimalπe,e,√s2,ln(2).
Real number (denotedR): containZ,Q , and irrationals; any number with decimal expansion as opposed to
complex/imaginary numbers.
Complex number (denotedC ): contains real and imaginary parts.
Intervals/Sets
[a,b]- inclusive (closed) intervaa tob.
(a,b]- left exclusive (left open) interato b, not includan.
[a,b)- right exclusive (right open) interaatob, not includib.
x ∈ (a,b) -x is an element (a,b)and {x ∈ R a < x < b} .
Infinity can be used as an interval endpoint, but is always excluded as it is a l[4,∞) n.t a number:
Inequalities
▯ - if and only if; first implies second, and second implies first.
Solve6 < 1 − 3x ≤ 10 : 6 < 1 − 3x ≤ 10
Isolate x: 5 < −3x ≤ 9
−5 > 3x ≥ −9
−5/3 > x ≥ −3
Solve 1 < 2 :
x
1
We know that x< 2 . 1
Ifx > 0 ,1 < 2x , sox > 2.
Ifx < 0 ,1 > 2x , sox < 1.
1 2 1
Ifx > 0 and x > 2 , thex > 2 .
Ifx < 0 and x < 1 , thex < 0 .
2
Therefore,x ∈ (−∞,0)∪ (1/2,∞) .
Absolute Value
Distance between number and 0 on the real number line.
x, if x >= 0
x = {
−x, if x < 0
|x − a| is the distance betwexnand a on the real number line:
x − a, if x >= 0
|x = { a − x, if x < 0
Solve by considering both cases. Deal with each case separately.
Solve|2x − 3 < 4 :
Open the inequality: − 4 < 2x − 3 < 4
|f < g ▯ −g < f < g
−1 < 2x < 7 ▯ − 1 < x < 7
2 2
Simplify|x + y ≤ 1 :
y ≤ 1 − x |
1 − x if x ≥ 0
y ≤ { 1 + x if x < 0
11/9/13
Functions
A function is a rule that assigns a single output to an input.
x (independent variable) -> f -> f(x) or y (dependent variable)
A function maps values in the set known as a domain into values in the set known as a range.
Domain: set of allowable values for the independent variable.
Common exclusions from domains:
Dividing by 0.
Logarithm of non-positive number.
Even root of negative number. A function has an inverse if and only if every unique value of the independent variable in the domain has a unique value of
the dependent variable - i.e., "one-to-one".
Range: set of possible values for the dependent variable.
Finding the range can be significantly more difficult than finding the domain, which is simply a matter of looking at the
function.
f(x) = √x − 1−
Restriction: x > 1
D = [1,∞)
R = [0,∞)
1
f(x) =
1 − x2
Restriction: x ≠ 0
D = (∞,−1)∪ (−1,1)∪ (1,∞)
R = (−∞,0)∪ [1,∞)
A true function must have a unique output for each input. Graphically, the vertical line test passes for all points (vertical line
drawn on graph will only intersect it at most once if it is a graph of a true function).
Non-functions can be described using functions by combining multiple functions into one:
−2−−−−2
Unit circy = ± √ x + y
−−−−−− −−−−−−
This can be written as two funcy = −:√ x + y 2 and y = √ x + y 2.
Horizontal line test: a function is one-to-one if any horizontal line drawn on its graph intersects it at most once.
Function examples:
n
Polynomialx , n is a positive integer (domain real numbers)
Rationalf(x)/g(x) (domain real numbers such thg(x) ≠ 0 )
Trigonometric/Inverse trigonometric
Exponentialxn,x is a positive real number (domain real numbers)
Logarithm:log(x)(domain positive real numbers)
13/9/13
Transformations
Sketchy = x 2,y = x − 1 y = (x + 1) 2:
2
y = x 2is simply a parabola.
y = x − 1 is the same parabola moved down 1 unit.
y = (x + 1)2 is the same parabola moved left one unit.
Given functiof(x):
f(x)+ c is a vertical transcaunits upward.
f(x + c) is a horizontal translcunits to the left.
cf(x) is a vertical stretch by a fc.tor of
f(cx) is a horizontal compression by a facc.r of
−f(x) is a reflection along the x-axis.
f(−x) is a reflection along the y-axis.
|f(x)|is a reflection of all parts of the function below the x-axis along the x-axis.
f( x )is a reflection of all parts of the function to the right of the y-axis along the y-axis replacing what was on the left.
Compositions Given two functionsf(x) and g(x) :
f ∘ g (equivalent tf(g(x)) ) is a composition of and g.
The domain of f ∘ g is all values xfin the domain ofg for whichg(x) is in the domain of .
1 −− −−−
Given f(x) = x,g(x) = √ 1 − x , finf ∘ g:
f ∘ g is equivalent to 1 .
√1−x
Restriction on square root1 − x ≥ 0 , sx ≤ 1 .
Restriction on divisiog(x) ≠ 0 , sx ≠ 1 .
Domain of f ∘ g:(−∞,1]∪ ((−∞,1)∪ (1,∞)) is equal t(−∞,1) .
16/9/13
Symmetry
A function is even if(−x) = f(x) - the function is symmetric about the y-axis.
2
Examples include y = x | and y = x .
A function is odd if(−x) = −f(x) .
Examples include y = x and y = sin(x) .
One-to-one Functions
A function is known as one-to-one on an interval if it never takes the same value twice - it passes the horizontal line test.
Methematically, for any two numbers x1∈ I,x ∈2I f(x ) ≠1f(x ) 2 whenever x 1 x 2.
Graphically, when we take the inverse of a function, we reflect it along y = xin. So the horizontal line test is actually
the vertical line test of the inverse.
A function is monotonically increasing on an intervalI if for anx1∈ I,x ∈2I x < x1▯ 2 f(x 1 < f(x )1 .
Likewise, a function is monotonically decreasing on an intervaI if for anx ∈ I,x ∈ I ,
1 2
x1< x ▯2 f(x 1 > f(x )1 .
A function that is monotonically increasing or monotonically decreasing on an interval is always one-to-one oon that interval.
Show that f(x) = x 2is not one-to-one overR but is ove[0,∞) :
Counterexample: Since, f(−1) = f(1) , the function is not one-to-one.
We can prove the latter statement by proving that the function is increasing over the interval.
Let x ∈ [0,∞) and x ∈ [0,∞) represent two arbitrary values such thx < x .
1 2 1 2
Then f(x2)− f(x )1= x − 2 = (x1− x )(2 + x 1 1 2 .
Both terms are always positive, sf(x )− f(x ) > 0 and f(x ) > f(x ) .
2 1 2 1
Therefore, the function is monotonically increasing.
Inverse
A function has an inverse if and only if it is one-to-one. By definition:
Letf(x) be a one-to-one function with domainA and range B .
The inverse function of(x) , denotedf −1(x) , is defined fy−1 (y) = x ▯ f(x) = y .
The range of the inverse is the domain of the function, and the domain of the inverse is the range of the function.
An inverse of a function "undoes" the operation applied by the function. In other words, the cancellation equations
−1 −1
∀x ∈ A, (f(x)) = x ∀x ∈ B,f( (x)) = x ∀x ∈ A,f −1(f(x)) = x and ∀x ∈ B,f(f −1 (x)) = x hold for all functions with inverses.
−1 −1
Note thatf (x) is not the same f(x) - the former is the inverse and the latter is the reciprocal.
The inverse is found by swapping the variables and isolating variables.
If a function is not one-to-onR , we may restrict the domain to a region where it is, and define the inverse there. This is
useful with things like trigonometric functions, for example.
−−−−−−−
Find the inversef(x) = √ 10 − 3x :
10
D = (−∞, 3 ]
R = [0,−−−−−−−−−−−
−1
x = √ 10 − 3f (x)
2 −1
x = 10 − 3f (x)
x − 10 = −3f −1 (x)
2
10 − x −1
3 = f (x)
D = [0,∞)
10
R = (−∞, 3 ]
Graphically, if the (a,b) is on the graph f(x), the(b,a) is on the graph ff1(x) . Using this information, we can
sketch the inverse of a graph by reflecting it aloy = xe.line
18/9/13
The Constant e
e is a constant used in many places in mathematics. It is approximately equal to 2.718.
There are several ways to dee:ne
1 n
As a limie = x→∞ (1 + x ) .
As an infinite sere =:∑ ∞ 1.
n=0 n! x
Graphically with tangents: of all exponential equations oy = a ,a > 0 , only whera = e does the tangent
of the graph have a slope ofx = 0 .
a − 1
Also,a = e is the only value for wlimh = 1 .
h→0 h
Exponentials/Logarithms
The functiof(x) = e xis called the natural exponential functionf(x) = ln(x) is called the natural logarithm.
x
Sketchf(x) = 3 + 2e :
Start with graph f(x) = ex.
Reflect the graph along the Y axis.
Stretch graph vertically by a factor of 2.
Move the graph up by 3 units.
The functiof(x) = a x is either increasina > 1 or decreasing f0 < a < 1 , and therefore has an inverse. This
y
inverse is calloga(x).log ax) = y ▯ a = x .
x
The domain off(x) = a isR , and the range{y ∈ R x > 0} . Therefore, the domainf(x) = loga(x) is
{y ∈ R x > 0} and the range Rs.
ln(e) = log (x)
e
ln(e) = 1 ln(e) = 1
1
loga(x ) = rlog (a ) r . Solog(√x) = 2 log(x) .
2x+1
Find the inverse of(x) = e :
−1
x = e 2f (x)+1
−1
ln(x) = 2f (x)+ 1
1 1
f −1(x) = ln(x)−
2 2
20/9/13
Trignomonetric Functions
a
Angles are measured in radians. Radians are the ratio of the associated arc and the θ =iur:.
For a full circa = 2πr , sθ = 2πr = 2π .
r
Radians Degrees
π 180
π
2 90
π
3 60
π 45
4
π
6 30
These are based on the special triangles:
π π π
Right triangle with side lengt1,1, 2 and internal angles4 ,4 . 2.
Right triangle with side lengt1, 3,2 and internal anglesπ ,π .π .
6 3 2
2 2
Parametric equations of unit circx + y = 1 ):x = cosθ and y = sinθ together form a unit circle from polar
coordinates.
2 2
x + y = 1
sin θ + cos θ = 1
Properties
Bounds: −1 ≤ sinθ ≤ 1 ,−1 ≤ cosθ ≤ 1
Periodicitysin(θ + 2πk) = sinθ,k ∈ Z ,cos(θ + 2πk) = cosθ,k ∈ Z
Odd/even: sin(−θ) = −sin(θ) - odd functioncos(−θ) = cos(θ) - even function
Sign:cos is positive to the right of the ysiniis positive above the x-axis
Other trigoometric functions:
Tangent: tanθ = sinθ
c1sθ
Cosecant cscθ = sinθ
Secant: secθ = 1
cosθ
Cotangent: cotθ = cosθ
sinθ
Identities
2 2
θ + θ = 1 Square Identisin θ + cos θ = .
Addition formulsin(x + y) = sinxcosy+ cosxsiny ,cos(x + y) = cosxcosy− sinxsiny .
Find the subtraction formulas:
sin(x − y) = sinxcos(−y)+ cosxsin(−y)
= sinxcosy− cosxsiny
cos(x − y) = cosxcos(−y)− sinxsin(−y)
= cosxcosy+ sinxsiny
Inverse Trigonometric Functions
Trignometric functions are periodic. Therefore, they are not one-to-one on the real number line and cannot have an inverse
on this interval.
However, they are one-to-one if we restrict the domain. Conven[−ona, ]for sine a[0,π]for cosine.
2 2
−1
The inverse of the sine function x = arcsinθ, ox = sin θ - "the angle whose sxnin the restricted
domain". Its domai[−1,1]and its range[−s2, 2.
The inverse of the cosine function arccosθ, ocos−1θ - "the angle whose cosine is x in the restricted domain". Its
domain i[−1,1] and its rang[0,π].
The inverse of the tangent function arctanθt, otan−1θ - "the angle whose tangent is x in the restricted
domain". Its domaiR and its rang(−isπ, ) .
2 2
Evaluatcos(arctan√ 3):
√3 opposite
=
1 adjacent
opposite = √3
adjacent = 1
−−−−−−−−−−−−−−−−−−
hypotenuse = √ opposite + adjacent2
adjacent 1
cos(arctan√3) = = −−−−−−−−−−−−−−−−−−
hypotenuse √ opposite + adjacent2
1 1
cos(arctan√3) = −−−−− = 2
√3 + 1
Simplifsin(arctanx)
x opposite
1 = adjacent
opposite = x
adjacent = 1
−−−−−−−−−−−−−−−−−−
hypotenuse = √ opposite + adjacent2
sin(arctanx) = opposite = x
hypotenuse −−−−−−−−2−−−−−−−−− 2
√ opposite + adjacent
x
sin(arctanx) = −−−−−−
√ x + 1
( ( π))
Evaluatarccos sin − 3 :
pi/6
|\
| \ 2
sqrt(3) | \
|___\
pi/2 1 pi/3
√
= = − −π opposite √ 3
sin ) = = −
3 hypotenuse 2
opposite = √3
hypotenuse = 2
−−−−−−−−−−2−−−−−−−−− 2 −−−−−
adjacent =√ hypotenuse − opposite = √4 − 3 = 1
π π 5π
arccos sin − 3 )) = π − 6 = 6
23/9/13
Hyperbolic Functions
The hyperbolic functions are expressible as combinations of exponential functions.
e −ex
sinhx = 2 ("sinch ex") - hyperbolic sine function. DomR.n and range is
e +ex
coshx = x2−x("cosh ex") - hyperbolic cosine functRoand range [1,∞].
tanhx = e +ex= coshx hyperbolic tangent function.RDand range (−1,1).
There are also hyperbolic secant, cosecant, and cotangent functions, but they are rarely used.
e−−ex 1 x
Hyperbolic sin is an odd sinh(−x) = 2 = −sinhx . Ax → ∞ ,sinhx approache2e
asymptoticallyx → −∞ ,sinhxapproache− 1e−xasymptotically.
ex+e
Hyperbolic cosine is an even cosh(−x) = 2 = coshx . Ax → ∞ ,sinhxapproaches2ex
1 −x
asymptoticallyx → −∞ ,sinhxapproaches2e −xsxmptotically.
Hyperbolic tangent is an odd tanh(−x) = e−+ex = −tanhx . Ax → ∞ ,tanhx approaches 1
asymptoticallyx → −∞ ,tanhx approaches -1 asymptotically.
2 2
If we x = coshtand y = sinh, thex − y = 1 . This is a hyperbola. The graph appears as an hourglass shape.
25/9/13
Inverse Hyperbolic Functions
Hyperbolic sine is one-to-Rn. over
To find the inverse functions, yewherex = sinhy = e −e:
2
y −y
x = sinhy = e − e
2
2x = e − e−y
y −y
0 = e − 2x − e
0 = e e − e 2x − e ey
y 2 y
(e ) − 2xe − 1 = 0 −−−−−−−−−−−−−−−
2
y 2x ±√ (−2x) − 4 ⋅1 ⋅−1 −−2−−−
e = = x ±√ x + 1
−−−−−− −−−−− 2
Since x −√ x + 1 < 0, e = x − √ x + 1 is extraneous.
y −2−−−
e = x + √ x + 1
√ −−2−−−
y = ln(x + x + 1)
With this technique, we discover the inverses of the hyperbolic functions are:
−1 −−−−−
x = ln(x +√ ) sinh−1x = ln(x + √ x + 1)
−1 √ −−2−−−
cosh−1x = ln1x +1+xx − 1),x ≥ 1
tanh x = 2ln(1−x),−1 < x < 1
Findtanh x1 :
e − e −y
x = tanhy = y −y
e + e
xe + xe −y = e − e −y
xe + x = e 2y− 1
2y
(x − 1)e = x + 1
x + 1
e = ± √
x − 1
x + 1
Since − √ < 0, the negative solution is extraneous.
x − 1
y −x + 1
e = √
x − 1
x + 1 2
y = ln( )
x − 1
1 x + 1
y = ln
2 x − 1
Limits
Limits are a core concept in calculus, because they appear in the definitions of derivatives and definite integrals, the two main
branches of calculus.
A limit does not depend on what the value of the function is at the point it is approaching. Consider the following function:
g(x) = { 3x − 1 if x ≠ 1
0 if x = 1
When we say the limitf(x)isL asx → a , we meanf(x) gets closer and closLrasoxgets closer and closea.to
The precise definition of a limit:
Letfbe a function defined on an open interval that contains a, except possiblaitself.
Then we say the limif(x) asx approachesa iL , if for every nuϵ > 0 there exists a numδ > 0
such tha0 < x − a < δ ▯ |f(x)− L < ϵ . We write thilimf(x) = L , of(x) → L as x → a.
x→a
In other word(∀ϵ > 0,∃δ > 0,0 < x − a < δ ▯ f(x)− L < ϵ) ▯ (lix→a f(x) = L) .
ϵis Epsilon (Greek) and is associated with error.
δ is Delta (Greek) and is associated with difference.
The key point is that the limit exists if the dif(x)candeLw(the errϵ) can be made as small as needed by
making the distance betwxeand a(the differenδ) sufficiently small. Hox ≠ a,.
We can also write the implic0 < x − a < δ ▯ |f(x)− L < ϵ as
a − δ < x < a + δ ▯ L − ϵ < f(x) < L + ϵ .
δ depends onϵ.
27/9/13
Geometry and intuition are useful tools, but do not prove anything.
Considerf(x) = sin(x) . The graph has higher and higher frequency oscillations as it approaches 0 from either direction, to a limit of infinity.
Consider the following function:
f(x) = { 3x − 1 if x ≠ 1
0 if x = 1
How close must x be to 1 to ensure that h(x) is within 0.1 of 2?
0.1 plays the role ofϵ.
We need to find the corresponding value of δ > 0 such that 0 < x − 1 < δ ▯ h(x)− 2 < 0.1 .
So |3x − 1 − 2 < 0.1 , or3 x − 1 < 0.1 .
So |x − 1 < 0.1 , or1 .
3 30
To prove that a limit exists, we need to use an arbitrarϵ > 0 :
We need to find the corresponding value of δ > 0 such that 0 < x − 1 < δ ▯ h(x)− 2 < ϵ .
So |3x − 1 − 2 < ϵ , or3 x − 1 < ϵ .
So |x − 1 < ϵ.
3 ϵ
So if we choose δ = 3 , then we know that 0 < x − 1 < δ ▯ |h(x)− 2 < ϵ .
Therefore, the limit exists, by definition.
Heaviside Function
1 if x ≥ 0
H(x) = { 0 if x < 0
Prove that the limit does not exist ax = 0 :
Proof by contradiction.
Suppose the limit does exist.
So ∀ϵ > 0,∃δ > 0,0 < x < δ ▯ | |H(x)− L < ϵ | .
Consider ϵ = 0.2 .
So 0 < x < δ ▯ L − 0.2 < H(x) < L + 0.2 .
So the distance between L − 0.2 and L + 0.2 is 0.4.
Let −δ < x <10 and 0 < x <2δ .
So it must be that and .
H(x )1= 0 H(x )2= 1
So the distance between the two is 1.
But the distance between the two must be less than 0.4.
Therefore, the limit cannot exist foϵ ≤ 1.
2
30/9/13
Limit Laws/Theorems
For complicated functions it is often impractical to use the definition to prove limits. Instead, we use limit laws, which can be
proved from the definition.
Limit sum law
Iflim x→a f(x) = L and lim x→a g(x) = M , thenlim x→a (f(x)+ g(x)) = L + M .
Proof:
Iflim x→a (f(x)+ g(x)) = L + M , then
∀ϵ > 0,∃δ > 0,0 < x − a < δ ▯ | |(f(x)+ g(x))− (L + M) < ϵ | .
Clearly,0 < x − a < δ ▯ (f(x)+ g(x))− (L + M) < ϵ | is equivalent to
0 < x − a < δ ▯ (f(x)− L)+ (g(x)− M) < ϵ | .
By the triangle inequality,(f(x)− L)+ (g(x)− M) ≤ f(x)− L + g(x)− M | | .
ϵ
We want to make each term less than 2 , so later we can add them together to get ϵ. Sinceϵ is arbitrary,> 0,∃δ > 0 0 , x − a < δ ▯ |f(x)− L 0,∃δ 2 0 0,< x − a < δ ▯ 2 |g(x)− M 0
Clearly, this is equivalent to . Since the left side limit does not equal the right side limit, the limit
−1 if x < 0
does not exist.
Approaching from either side gives different results. This leads to the idea of a one sided limit.
lim x→a− f(x) is the left side limit - approacaifrom below/left.
+
lim x→a f(x) is the right side limit - approacaifrom above/right.
limf(x) = L ▯ lim−f(x) = lim f(+) = L
x→a x→a x→a
The limit exists if and only if the limits on either side exist.
2/10/13
Vertical Asymptotes
At vertical asymptotes, the limit approaches positive or negative infinity.
Consider f(x) = 12 nearx = 0 . Asx → 0 ,f(x) grows without bound.
x
We say that f(x) → ∞ asx → a , orlimx→a f(x) = ∞ , if for eveM > 0 there existsδ > 0 such thatf(x) > M
whenever 0 < x − a < δ .
In other words, the value f(x) can become arbitrarily large for some value xfclose enough to a.
Likewise, with a downwards asymptote likef(x) = − 1 near x = 0 , the same behavior applies, excepf(x) → −∞ as
x2
x → a , olim x→a f(x) = −∞ .
12 6
Given f(x) = e x , determine how close must x be to 0 so thaf(x) > 10 :
In other words, finδ > 0 such thatf(x) > 10 6 when 0 < x < δ .
6 1 1 2 1
Clearly,f(x) = 10 is equivalent tx2 = 6ln(10) , andx2 = 6ln(10) . Sox = 6 ln10
So |x < 1 .
√6 ln10
In rational functions, vertical asymptotes usually occur when the denominator becomes 0.
x +1
Find the left and right hand limitf(x) = 3x−2x2 at each asymptote:
x +1 3
Clearly,f(x) = x(−2x+3). So the asymptotes are ax = 0 and x = 2 .
− 2
As x → 0 ,x + 1 > 0 − ,x < 0 , and−2x + 3 > 0 .
So f(x) < 0 forx = 0 .
So f(x) → ;wip: use the sign analysis with infinismals
Continuity
We say f(x) is continuous aa iflim f(x) = f(a) .
x→a
If this is true foa ∈ I , withI being an interval, we sf(x) is continuous onI.
This implies the following conditions must be met:
f(a) must be defined -a must be in the domain off(x) .
The left and right limits are equal. f(a) is equal to the limit at that location
The most common discontinuities are:
1
Infinite discontinuity - vertical asx.ptote:
Removable discontinuity (hole) - where the limit exists but the function is not defined.
Jump discontinuity - where the limit doesn't exist because the left and right limits1are different.
Wild oscillations discontinuity - where the function oscillates too much f(x) = sinlxmit:
For removable discontinuities, we can remove the discontinuity by writing a related function filling in the undefined point:
1
f(x) = xsin
x
xsin 1 if x ≠ 0
becomes f(x) = { x
0 if x = 0
Iff(x) andg(x) are continuous at, thef(x)± g(x) ,f(x)g(x) , and(x),g(x) ≠ 0 are also continuousaa.
g(x)
2
Determine wheref(x) = ln(x−1)+ 16−x is continuous:
x −8
ln(x − 1) is continuous forx ∈ (1,∞) .
√ 16 − x2 is continuous forx ∈ (−4,4) .
3
The denominator must not be 0,x − 8 ≠ 0 , ox ∈ (−∞,2)∪ (2,∞) .
The function is continuous x ∈ (1,∞)∩ (−4,4)∩ ((−∞,2)∪ (2,∞)) , ox ∈ (1,2)∪ (2,4) .
Continuity theorem
Iff(x) is continuousx = b , anlimx→a g(x) = b , thelimx→a f(g(x)) = f(b).
In other words, the limit can be moved within compositions.
√−− −−−2
Evaluatelimx→0.5arcsin( 1 − x ) :
−−−−−− −− −−−−
lim arcsin( 1 − x ) = arcsin( lim √ 1 − x )
x→0.5 x→0.5
−−−−−−−−− 2
= arcsin(√ x→0.5 − x )
−−−−−−−−− 2
= arcsin(√ x→0.5 − x )
√ 3
= arcsin( )
2
= π
3
Intermediate value theorem (IVT)
Suppose f(x) is continuous ov[a,b]and f(a) < n < f(b). Then there exists a numc ∈ (a,b)such thaf(c) = n .
In other words, for a function continuous on an interval, any value of the function in a range has a cox.esponding value of
This is used most often to find the roots of complicated equatn = 0-.when
This is an existance theorem - it tells us that a certain number with a certain property exists, but not its value.
Prove thalnx = 3 − 2x has at least one real root.
Constructf(x) = lnx − (3 − 2x) .
We want to prove there exixsuch thatf(x) = 0.
Clearly, this is continuoux > 0r.
Constructa = 1,b = 2 (we could find these values using a graph).
Clearlyf(x) is continuous over the interval.
Clearlyf(1) = −1 andf(2) = ln2 − 1 .
Clearlyf(1) < 0 andf(2) > 0 .
So by the IVT, there must c ∈ [a,b]such thaf(c) = 0.
So there must be a real root. This is a trancendental function and cannot be solved analytically.
;wip: prove this theorem
Bisection method
We can use the bisection to get a better approximation. Currently, we know th[1,2]. is in
We can divide[a,b]into two equally sized intervals. Using IVT, we can determine that the root must be in one of the two
intervals.
So we have made the possible interval smaller, and therefore gotten a better estimate.
This process can be repeated as many times as needed to get as good an estimate as needed.
Limits at Infinity
limx→∞ f(x) = L means ∀ϵ > 0,∃n > 0,x > n ▯ |f(x)− L < ϵ .
So oncex exceeds some valuen,f(x) must stay withiϵ ofn. Heren has a role similarδ.o
limx→−∞ f(x) = L means ∀ϵ > 0,∃n < 0,x < n ▯ f(x)− L < ϵ .
Limit laws also applyx → ±∞ , including the squeeze theorem.
1
Prove limx→∞ k= 0,k > 0 :
x
Letϵ be an arbitrary positive real number.
Constructn = k .
√ϵ
Assume x > n . Sox > 1.
√ϵ
So 1 < √ ϵand 1 − 0 < ϵ .
x xk 1
Thereforelim x→∞ k= 0,k > 0 .
x
r 1
Infinite limit theoremr ∈ Q,r > 0 , whenx is definelimx→∞ xr = 0 . In other words, wreis rational and
definedlim 1r= 0 .
x→∞ x
2
Evaluatelimx→∞ x2+2 :
3x −4x
2
2 x +2
lim x + 2 = lim x2
x→∞ 3x − 4x x→∞ 3x −4x
x2
x2 2
x2+ x2
= lim 2
x→∞ 3x2 − 42
x x
1 + 2
= lim x
x→∞ 3 − 4
x
1 + 0
=
3 − 0
1
=
3
−−−2− −−
Evaluatelimx→∞ ( 9x + x − 3x) :
√ −−−−−− −
= ( − 3x) + 3x −−−−−−− −−−−−− √ 9x + x + 3x
lim( 9x + x − 3x) = lim( 9x + x − 3x)
x→∞ x→∞ √ 9x + x + 3x
2 2
= lim 9x + x − 9x
x→∞ √−9x + x + 3x
= lim 1
x→∞ √9x +x
x + 3
1
= lim −−−−
x→∞ √ 9x +x+ 3
x2
1
= lim −−−−−
x→∞ x (9x )
√ x2 + 3
1
= x→∞ √9 + 3
1
= 6
Evalualim −2 :
x→−∞ x− x −2x
Note that ∀x > 2,x > 2x
lim −2 = −2 = −2 = 0
x→−∞ x −√ x − 2x −∞ − √ ∞ − ∞− −∞
Evalualimx→∞ e−xsinx:
−1 ≤ sinx ≤ 1
1
e−x=
ex
lim ex= 0
x→∞
By the squeeze theorem: lim ⋅−1 ≤ lim e−xsinx ≤ lim ex⋅1
x→∞ x→∞ x→∞
0 ≤ lim exsinx ≤ 0
x→∞
lim exsinx = 0
x→∞
7/10/13
General techniquϵδfproofs:
1. Leϵbe an arbitrary positive real number.
2. Construct a parδas some functio. of
3. Show thδ > , so it's still in the domain.
4. Assum0 < x − a < δ.
5. Provf(x)− L < ϵ.
6. By the definition oflix→af(x) = L.
A horizontal asymptote exy = Lif and onlimx→∞ f(x) = Lorlimx→−∞f(x) = L.
1
Considef(x) = e. Ax → 0 −,f(x) → 0. Butx → 0+ f(x) → ∞ . This is a one-sided asymptote - a vertical
asymptote from the right.
√x−x
Evalualimx→1 :
1−√x
2 2
− −
= ⋅ 2 2
lim √ x − x = lim √ x − x ⋅ 1 + √ x
x→1 1 − √ x x→1 1 − √ x 1 + √ x
2
(√x − x )(1 + √ x)
= lim
x→1 1 − x
√ x − x + x − x 2.5
= lim
x→1 1 − x
x(1 − x)+ √ x(1 − x )
= lim
x→1 1 − x
(1 − x)(x + √ x(1 + x))
= lim
x→1 1 − x
= lim(x + √ x(1 + x))
x→1
= 1 + √1(1 + 1)) = 3
Derivatives
The derivative of a funcf(x) at the poiat, denotef(a):
f(x)− f(a)
f (a) = lim
x→a x − a
This can also be written as follows:
f(x + h)− f(x)
f (x) = lim
h→0 h
The first form is better for determining differentiability at a point, while the second is better for finding the derivative.
If the limit exisa, the function is differentiaa.e at
Two interpretations:
f (x) is the slope of a tangef(x)fdrawn atx .
′
f (x) is the instantaneous rate of chanf(x).
The following are equivalent notatf (x),y , dy, df, d f(x) .
dx dx dx
1
xsin x ,x ≠ 0
Determine whetherf(x) = { 0,x = 0 :
1
′ xsin x − 0
f (0) = x→0
x − 0
= limsin 1
x→0 x
The limit does not exist
1
x sin x ,x ≠ 0
Determine whetherf(x) = { 0,x = 0 :
2 1
′ x sin x − 0
f (0) = lim
x→0 x − 0
1
= x→0xsin x
= 0(squeeze theorem)
Power law
d
dxx = nx n−1 .
Proof for positive intn:er
d n
Consider dxx .
Letn be a positive integer. (x+h − n
By definition, x = lim h→0 ) x .
dx h Using the binomial theorem:
(x + h) − x = (x + nx n−1h + n(n−1)xn−2h + …+ nxh n−1 + h )− x n.
(x+h) −x 2 n(n−1)
Divide all termh:by = nx n−1+ xn−2h + …+ nxh n−2 + hn−1.
hn(n−1) 2 n−2 n−1
Clearllimh→0(nx n−1+ 2 xn−2h + …+ nxh + h ) = nxx−1.
d n n−1
So dxx = nx .
Proof for all n(requires implicit differentiation, described later):
n ln x nlnx
Cledrln = ed nlnxe nlnx d
So dxx = dxe = e ⋅dx (nlnx) .
So d x = e nlnx⋅n⋅ 1= n xn= nx n−1.
dx x x
9/10/13
−−−−−
d √x + h − √ x
√ x = lim
dx h→0 −−−−− −−−−−
√x + h − √ x √x + h + √ x
= lim −−−−−
h→0 h √x + h + √ x
x + h − x
= lim
h→0 h(√x + h + √x)
= lim 1
h→0√x + h + √ x
1
= 2 x
√
Differentiability implies continuity
Iff(x)is differentiabx = a , then it is also continuous at that point. Likewise, if a functiox = a,ot continuous at
then it is also not differentiable - the contrapositive.
Proof:
f(x)is continuous x = aif and onlylimx→a f(x) = f(a), olimx→a(f(x)− f(a)) x−a = 0 .
f(x)−f(a) f(x)−f(a)
Clearly, this is equivlimx→ato x−a (x − a) = 0, olimx→a x−a limx→a(x − a) = 0.
Clearly, this is equivf (a)⋅0 = 0 .
′
Clearly, this is true wf (a)eis defined.
Sof(x) is continuousx = a only if (a)is defined - the function is diffex = aa.le at
Some continuous functions are not differentiable.
Considerf(x) = x . Clearf (0) = lim f(x)−f(0= lim ∣x∣= undefined . So the function is not differentiable
x→0 x−0 x20 x
atx = 0 due to a cusp. Others include piecewise funcxi.ns and
′ f(x)−f(0) 1
Considerf(x) = √ x. Clearf (0) = limx→0 x−0 = limx→0 2 = ∞ . So the function is not differentiable at
x3
3
x = 0 due to a vertical tangent. Others arcsinx andx .
Exponential functions
x
Considerf(x) = a .
x+h x
= − x+h x
′ a − a
f (x) = lh→0 h
x h
a (a − 1)
= lh→0
h
x a − 1
= a lim
h→0 h
h
So the derivative of any power function is just a power function multiplied by a constant of propolimoh→0ity−1 .
h
′ a −1
Consider f (0) = lim h→0 h . Clearly, the constant of propertionality is the slope of thx = 0g.nt at
e −1
We know that the slope of the tangentf(x) is 1 only whena = e. So we know thatlimh→0 h = 1 , and so
d e = e x.
dx
Derivative Rules
d
Ifcis a constantdx c = 0.
Sum/difference rule
Iff(x) and g(x) are differentiablex,td (f(x)± g(x)) is defined xtand d (f(x)± g(x)) = d f(x)± d g(x) .
dx dx dx dx
Constant multiple rule
Ifcis a constant anf(x) is differentiablex,td cf(x) is defined ax and d cf(x) = c d f(x) .
dx dx dx
Product rule
d d ′ ′
Iff(x) and g(x) are differentiablex, thendx f(x)g(x) is defined xtand dx f(x)g(x) = f (x)g(x)+ f(x)g (x) .
Proof:
d f(x + h)g(x + h)− f(x)g(x)
f(x)g(x) = lh→0
dx h
f(x + h)g(x + h)− f(x)g(x)+ (f(x)g(x + h)− f(x)g(x + h))
= lim
h→0 h
f(x + h)− f(x) g(x + h)− g(x)
= lim g(x + h) + f(x) )
h→0 h h
f(x + h)− f(x) g(x + h)− g(x)
= limg(x + h) + limf(x)
h→0 h h→0 h
= g(x)f (x)+ f(x)g (x)′
11/10/13
2 x
Differentiatx e :
d 2 x x d 2 2 d x x 2 x
(x e ) = e x + x e = 2xe + x e
dx dx dx
Reciprocal rule
g (x)
Ifg(x) is differentiablx , thend 1 is defined xtand d 1 = − .
dx g(x) dx g(x) g(x)
Proof:
1 1
−
= 1 − 1
d 1 g(x+h) g(x)
dx g(x) = h→0 h
g(x)− g(x + h)
= lim
h→0 h(g(x + h)g(x))
g(x + h)− g(x) 1
= lim− ⋅ lim
h→0 h h→0g(x + h)g(x)
1
= −g (x)⋅ 2
g(x)
Quotient rule
d f(x) d f(x) f (x)g(x)−f(x)g (x)
If(x) andg(x) are differentiaxlandg(x) ≠ 0, thedx g(x)s definedxaand dxg(x)= g(x)
Proof:
d f(x) d −1
dx g(x) = dx (f(x)g(x) )
′ −1 d −1
= f (x)g(x) + f(x) dx g(x)
= f (x)g(x)−1+ f(x) d g(x)−1
dx
−g (x)
= f (x)g(x)−1+ f(x)
g(x)2
′ ′
f (x)g(x)− f(x)g (x)
= 2
g(x)
The functif(x) = x 2is known as a serpentine curve. Find the equations of the tangents where t:ey have a slope of
1+x 8
2 2
f (x) = 1 = 1 + x − 2x
8 (1 + x )
1 1 − x2
= 2
8 (1 + x )
22 2
(1 + x ) = 8 − 8x
(1 + x ) − (8 − 8x ) = x + 10x − 7 = 0
−−−− −−−−−−−
2 −10 ± √100 + 4 ⋅1 ⋅7
x =
−−−−−−−−−−−
128
x = ± √ −5 ± √
4
−−−−−−−
x = ± √ 4√2 − 5
−−−−−−− −−−−−−− 1 −−−−−−−
t1(−√ 4√2 − 5) = f(−√ 4√2 − 5) = 8 √ 4√2 − 5 + b
−−−−−−−
−−−−−−− √ 4 2 − 5 1 −−−−−−−
t1(−√ 4√2 − 5) = − = √ 4√2 − 5 + b
−−−−−−− −−−−−−−−− 8
−−−−−−−
−−−−−−− 2√ 4 2 − 5 √ ( 2 − 1)(4 2− 5)
t1(− √ 4√2 − 5 ) = − − = b
8( 2 − 1) 8( 2 − 1)
;wip
Higher order derivatives
2
The second order derivatf(x)is d2f(x) = d ( d f(x) = f (x) .
dx dx dx
It can be interpreted as the rate of change of the rate of change.
In the same way, we can define arbitrarily high orders of derivatives, but this is rarely used.
d d
Consider dxas an operator.
d2 x
Find dx 1+x2:
d 2 x d d x
=
dx 1 + x 2 dx dx 1 + x2
2 2
= d 1 + x − 2x
dx (x + 1) 2
d 1 − x2
= 2
dx (x + 1)
2
−2x(x + 1) − (1 − x )(4x + 4x)
= 4
(x + 1)
2 2 2 3
−2x(x + 1) − (1 − x )(4x + 4x)
= 2 4
(x + 1)
2 2 2 2
= −2x(x + 1) − 4x(1 − x )(x + 1)
2 4
(x + 1)
−2x(x + 1)− 4x(1 − x ) 2
=
(x + 1) 3
16/10/13
Derivatives of Trigonometric and Hyperbolic Functions
Considerlimx→0 sinx= 1
x
;wip: big ugly geometric proof, maybe find something better here: http://www.proofwiki.org/wiki/Limit_of_Sine_of_X_over_X
sin(3x)
Evaluatelimx→0 x :
sin(3x) 3sin(3x)
lim = lim
x→0 x x→0 3x
sin(3x)
= 3 lim
x→0 3x
= 3
Sine rule
d sin(x) = cos(x).
dx
Proof:
sin(x + h)− sin(x)
= d sin(x + h)− sin(x)
sin(x) = lim
dx h→0 h
sin(x)cos(h)+ sin(h)cos(x)− sin(x)
= lh→0 h
= sin(x)lim cos(x)− 1 + cos(x)lim sin(h)
h→0 h h→0 h
cos(h)− 1
= sin(x)lim + cos(x)⋅1
h→0 h
cos(h)− 1 cos(h)− 1 cos(h)+ 1
lim = lim ⋅
h→0 h h→0 h cos(h)+ 1
cos (h)− 1
= lim
h→0h(cos(h)+ 1)
2
= lim 1 lim(1 − sin (h))− 1
h→0cos(h)+ 1 h→0 h
2
= 1 lim −sin (h)
2 h→0 h
1 sin(h)
= ⋅− limsin(h)⋅ lim
2 h→0 h→0 h
1
= ⋅−0 ⋅1 = 0
2
d sin(x) = sin(x)⋅0 + cos(x)⋅1
dx
= cos(x)
Cosine rule
d cos(x) = −sin(x).
dx
Proof:
d cos(x + h)− cos(x)
dx cos(x) =h→0m h
cos(x)cos(h)− sin(h)sin(x)− cos(x)
= h→0 h
= cos(x)lim cos(h)− 1 − sin(x)lim sin(h)
h→0 h h→0 h
cos(h)− 1
= cos(x)lim − sin(x)⋅1
h→0 h
cos(h)− 1
lim = 0, from the proof of the sine rule
h→0 h
d
dx sin(x) = cos(x)⋅0 − sin(x)⋅1
= −sin(x)
Tangent rule
d 2
dx tan(x) = sec(x)
Proof:
d d sin(x)
tan(x) =
dx dx cos(x)
cos(x) d sin(x)− sin(x)d cos(x)
= dx dx
cos(x)2
2 2
cos(x) + sin(x)
= 2
cos(x)
1
= 2
cos(x)
= sec(x)2 Hyperbolic sine rule
d
dx sinh(x) = cosh(x.
Proof:
d 1 d d
sinh(x) = ( e − e−)
dx 2 dx dx
1 d d
= ( e − e−)
2 dx dx
1 −1
= ( e − x )
2 e
e + e−x
= = cosh(x)
2
Hyperbolic cosine rule
d cosh(x) = sinh(x.
dx
Proof:
d 1 d x d −x
dx cosh(x) = 2 ( dx e + dx e )
1 d x d −x
= 2 ( dx e + dx e )
= 1 (e + −1 )
2 ex
x −x
= e − e = sinh(x)
2
Hyperbolic tangent rule
d 1
dx tanh(x) = cosh(x)sech(x).
Proof:
d tanh(x) = d sinh(x)
dx dx cosh(x)
d d
cosh(x)dx sinh(x)− sinh(x)dxcosh(x)
= 2
cosh(x)
cosh(x) − sinh(x)2
=
cosh(x)2
= 1 = sech(x)
cosh(x)2
Alternatively, we can prove it from the definition and derivative rules.
18/10/13
n n
The triangle inequality statx + y ≤ x + y ,x ∈ R ,y ∈ R
Chain Rule
′ ′ ′ d dg dh
f (x) = g (h(x))h (x) =dxg ∘ h =dh dx.
Proof: ;wip
Simplify d cos(3x) :
dx
d d
cos(3x) = −sin(3x) 3x = −3sin(3x)
dx dx
LetT(x) be the temperature at a height above the surface of the earth.h(t) be the height of a skydiver above the earth
at timet. What does dT represent?
dt
Clearly,dT = dT dh .
dT dt dh dt
So dt is the rate of change in temperature with respect to height times rate of change in height with respect to
time.
So it is the rate of change in temperature with respect to height time velocity.
Note that dT is independent of the skydiver - it is simply the change in temperature in the atmosphere.
dh
Likewise, dh is independent of the atmospheric temperature.
dT dt
So dt represents the rate of change in temperature with respect to time felt by the skydiver.
Exponential Rule
We can use the chain rule to find a x:
dx
x
Clearly, d a = d eln(a )= d exlna.
dx dx x ddx xlna d
By the chain rule,dxa = dxlnae ⋅ dx xlna .
d xlna d lna x x d x x
Clearly,dxlna e ⋅ dxxlna = (e ) ⋅lna = a ⋅lna , sincedx e = e .
So d a = a lnx
dx
Power Rule
d n d n n−1
Ifg(x) is differentiablexa, thendx g(x) is defined xt and dx g(x) = n⋅g(x) ⋅g (x) .
Proof:
Let u = g(x),y = u n.
Note that in this casu,is called an intermediate variable, since it is dependentxobut treated as an
independent variable in dy.
du
Clearly, d g(x) = dy = dy du = nu n−1 u = ng(x) n−1 g (x) .
dx dx du dx
21/10/13
Implicit Differentiation
Useful when we do not have an explicit relation ly = f(x) . For example,x + y = 1 or y + 2x y + x = 1 4 .
Useful for finding the derivatives of inverse functions.
Contrasts with ordinary differentiation. This is differentiating for a general curve.
dy
Find dx for x + y = 1 :
d d
( 2 + 2) = 1 d d
(x + y ) = 1
dx dx
d d d
= x + y = 1
dx dx dx
d 2
= 2x + y = 0
dx
d 2
= 2x + y(x)
dx
d 2 d
= 2x + y ⋅ y
dy dx
d
= 2x + 2y y
dx
d
2x + 2y y = 0
dx
d −2x −x
dx y = 2y = y
3 3
Where does x + y = 3xy have a horizontal tangent?
d 3 d 3 d
x + y = 3xy
dx dx dx
= 3x + y 3y = 3y+ 3xy ′
2 2 ′
3x − 3y = (3x − 3y )y
2 2 ′
3x − 3y = (3x − 3y )y
′ 3x − 3y x − y
y = = = 0
3x − 3y 2 x − y 2
2 2 2 2
x − y = 0 ∧ x − y ≠ 0 ≡ y = x ∧ y ≠ √ x ≡ y = x ∧ x ≠ 0 ∧ y ≠ 0
3 2 3 2 3 6 3 3 3
x + (x ) = 3xx ≡ x + x = 3x ≡ x (x − 1) = 0
x = 0 (extraneous),x = 1
2
y = 1 = 1
There is a horizontal tangent at (1,1)
We can use implicit differentiation to find the derivative of inverse functions.
Considery = log a :
y
Clearlyx = a .
So d x = d a = d a ⋅ dy.
dx dx dy dx
So 1 = a lna ⋅ dy.
dy dx
Clearly, = y1 .
dx a lna
;wip: write this as a rule
Derivative of Inverse Trignonometric Functions
Arcsine rule
d 1
dxarcsinx = √ 1−x2.
Proof:
Lety = arcsinx . Thenx = siny .
dy
Implicitly differentia1 = cosy dx .
So dy = 1 .
dx cosy
We want to write this in termx.of
−−−−−− −−
Clearlycosy = ± √ 1 − sin y .
π π
We pick the positive answer sicosy ≥ 0 whenever − 2 ≤ y ≤ 2 , the rangearcsin .
dy 1
So dx = 2 .
√ 1−sin y
So dy = 1 .
dx √1−x2 Arccosine rule
d arccosx = − 1 2 .
dx √1−x
Proof:
Lety = arccosx . Thenx = cosy .
Implicitly differentiatin1 = −siny dy .
dx
So dy = − 1 .
dx siny
We want to write this is terms ofx.
−−− −−−−−2
Clearly,siny = ± √ 1 − cos y .
We pick the positive answer since siny ≥ 0 whenever 0 ≤ y ≤ π , the range ofarccos .
dy
So = − 1 .
dx √1−cos y
dy 1
So dx = − √1−x 2.
Arctangent rule
d 1
dxarctanx = 2.
1+x
Proof:
Lety = arctanx . Then x = tany .
dy
Implicitly differentiatin1 = sec y .
dy dx
So = − 2 .
dx sec y
We want to write this in terms ofx .
sin x cos x 1
Clearly,cos x + cos x = cos x.
2 2
Clearly,tan x + 1 = sec x .
So dy = − 1 .
dx tan x+1
So dy = − 1 .
dx x +1
23/10/13
Logarithmic Differentiation
Logarithmic differentiation is a technique useful for differentiating complicated products and quotients with powers, or
h(x)
functions of the form f(x) = g(x) .
The general process is as fo

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