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Learning Outcomes: Pages:
1. Extreme values 1 – 3
2. Mean value theorem 3 – 5
3. Applied optimization 5 – 6
4. Newton’s method 6 – 7
5. Anti–derivatives 7 – 8
1. Extreme values
(Absolute) / (global) extrema:
o If 𝑓 is a function with a domain of D, then 𝑓 has an:
1. Absolute max. on D at c if 𝑓( 𝑥) ≤ 𝑓(c) for all 𝑥 in D
2. Absolute min. on D at c if 𝑓( 𝑥) ≥ 𝑓(c) for all 𝑥 in D
o There may be > 𝟏 pt. that correspond to that extrema (e.g. trigonometric
functions)
o Functions with the same eqn. may have diff. extrema depending on the domain
o If 𝑓 is continuous on a closed finite interval, [a, b], then 𝑓 has an absolute
max. value, M, and an absolute min. value, m, in [a, b] – that is, there are
𝑥1 and 𝑥2 in [a, b] such that 𝑓( 𝑥1) = m, 𝑓( 𝑥2) = M, and m ≤ 𝑓( 𝑥) ≤ M
for every other 𝑥 in [a, b]:
 E.g. for 𝑦 = 𝑥2
in the domain (0,2), there is neither an absolute max. nor
absolute min.
(Local) / (relative) extrema:
o If 𝑓 is a function with a domain of D, then 𝑓 has an:
1. Local max. on D at c if 𝑓( 𝑥) ≤ 𝑓(c) for all 𝑥 ∈ D lying in an open
interval containing c
2. Local min. value on D at c if 𝑓( 𝑥) ≥ 𝑓(c) for all x ∈ D lying in an
open interval containing c
o In this diagram, a and d are absolute extremum while b, c and e are local
extremum:

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o Note that 𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐞𝐱𝐭𝐫𝐞𝐦𝐮𝐦 ∈ 𝐥𝐨𝐜𝐚𝐥 𝐞𝐱𝐭𝐫𝐞𝐦𝐮𝐦
If 𝑓 has a critical pt. at an interior pt., c, of its domain, and if 𝑓′
is defined
at c, then:
𝑓′(c) ≡ 0 or 𝑓′(c) ≡ undefined − eqn. (1a)
o ♣! A stationary pt. is a critical pt. where the derivative is 0
o 𝑓 can have an extremum only at:
1. Critical pts. (interior pts. where 𝑓′
= 0 or where 𝑓′
is undefined)
2. Endpts. of the domain of 𝑓
o However, a critical pt. may not be a local extrema when it is inflection pt. – if a
local extrema is to be present, an additional requirement is that 𝑓′(c) for 𝑥 < c
and 𝑥 > c have opposite signs (according to the 1st–derivative test)
Functions are monotonic on an interval, I, if they are either ing or ing on I –
if 𝑓 is continuous on [a, b] and differentiable on (a, b), which may be finite or
infinite, then:
1. 𝑓 is ing on [a, b] if 𝑓′( 𝑥) > 0 at each pt. 𝑥 ∈ (a, b)
2. 𝑓 is ing on [a, b] if 𝑓′( 𝑥) < 0 at each pt. 𝑥 ∈ (a, b)
The 1st
derivative test for local extrema uses the idea that, at a local extrema,
the sign of 𝑓′( 𝑥) must  – if c is a critical pt. of a continuous function, 𝑓, and if
𝑓 is differentiable at every pt. in an interval containing c, except possibly at c
itself, then, moving across this interval from left to right:
1. If 𝑓′
s from –ve to +ve at c, 𝑓 has a local min. at c
2. If 𝑓′
s from +ve to –ve at c, 𝑓 has a local max. at c
3. If 𝑓′
does not  sign at c, then 𝑓 has no local extremum at c
The 2nd
derivative test for concavity describes how a graph bends – if 𝑦 = 𝑓( 𝑥)
is twice–differentiable on an interval, I, then the graph of 𝑦 = 𝑓( 𝑥) is:
1. Concave up over I if 𝑓′′
> 0 or if 𝑓′
is ing on I
x
y
b da c e

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2. Concave down over I if 𝑓′′
< 0 or if 𝑓′
is ing on I
♣! An inflection point is a pt. where the graph has a tangent line (includes a
vertical tangent too) and where concavity s:
o At an inflection pt., (c, 𝑓(c)),:
𝑓′′(c) ≡ 0 or 𝑓′′(c) ≡ undefined − eqn. (1b)
o At a vertical tangent, both 𝑓′(c) and 𝑓′′(c) are undefined
o However, 𝑓′′(c) = 0 does not imply that an inflection pt. is present since an
extrema is also possible
The 2nd
derivative test for local extrema states that if 𝑓′′
is continuous on an
open interval containing 𝑥 = c and if 𝑓′(c) = 0, then:
1. 𝑓 has a local max. at 𝑥 = c if 𝑓′′(c) < 0 (𝑓 is concave down)
2. 𝑓 has a local min. at 𝑥 = c if 𝑓′′(c) > 0 (𝑓 is concave up)
3. The test is inconclusive if 𝑓′′(c) = 0
Procedures for graphing 𝑦 = 𝑓(𝑥):
1. Identify the domain of 𝑓 and symmetries
2. Calc. the derivatives, 𝑦′
and 𝑦′′
3. Find the x–intercepts and determine the regions where 𝑓 > 0 and where 𝑓 < 0
4. Find the critical pts. and determine the monotonicity in different regions:
Interval
Sign of 𝑓′
Monotonicity of 𝑓
5. Find the inflection pts. and determine the concavity in different regions:
Interval
Sign of 𝑓′′
Concavity of 𝑓
6. Determine the local extrema and absolute extrema
7. Plot key pts. such as intercepts, critical pts. and inflection pts., together with
any asymptotes
2. Mean value theorem
Rolle’s theorem states that if 𝑦 = 𝑓( 𝑥) is continuous in the closed interval,
[a, b], and differentiable at every pt. of its interior, (a, b), and if 𝑓(a) = 𝑓(b),
then there is ±𝟏 number, c, in (a, b) at which 𝑓′(c) = 0:

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o If a differentiable function crosses a horizontal line at 2 different pts., there is
±𝟏 pt. btw. them where the tangent to the graph is horizontal
o Rolle’s theorem may be combined with the intermediate value theorem to
show when there will be only 1 real solution of an eqn.1
:
1. If the limits on both sides equal –ve and +ve infinity and if the function
is continuous, then 𝑓(𝑥) hast ± 𝟏 root
2. If gradient is never 0 and if the function is continuous, then 𝑓(𝑥) has at
most 𝟏 root
Mean–value theorem states that if 𝑦 = 𝑓( 𝑥) is continuous on a closed interval,
[a, b], and differentiable on the interval’s interior, (a, b), then there is ±𝟏 pt., c,
on [a, b], at which
𝑓(b)−𝑓(a)
b−a
= 𝑓′(c):
o There is ±𝟏 pt. whose tangent line is parallel to the chord 𝐀𝐁:
a
x
y
bc2c1
𝑓′(c3) = 0
c3
𝑓′(c1) = 0
𝑓′(c2) = 0
a
x
y
b
slope =
𝑓(b) − 𝑓(a)
b − a
c
𝑓′(c)

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o If
𝑓(b)−𝑓(a)
b−a
is the average  in 𝑓 over [a, b] and 𝑓′(c) is the instantaneous ,
then, at some interior pt., the instantaneous  must equal the average  over
the entire interval
Mathematical corollaries (apply to both finite & infinite intervals):
o If 𝑓′( 𝑥) = 0 at each pt., 𝑥, of an open interval, (a, b), then 𝑓( 𝑥) = C for
all 𝑥 ∈ (a, b), where C is a constant
o If 𝑓′( 𝑥) = 𝑔′( 𝑥) at each pt., 𝑥, in an open interval, (a, b), then there
exists a constant, C, such that 𝑓( 𝑥) = 𝑔( 𝑥) + C for all 𝑥 ∈ (a, b)
3. Applied optimization
Strategy:
1. Determine what is given and what is the unknown quantity to be optimized
2. Draw a diagram and label any part that may be impt. to the problem
3. Introduce variables and list every relation in the picture and in the problem as
an eqn. or algebraic expression, and identify the unknown variable
4. Write an eqn. for the unknown quantity and, if possible, express the unknown
quantity as a function of a single variable or in 2 eqns. in 2 unknowns
5. Analyze the critical pts. and endpts. in the domain of the unknown for the
desired optimization using the techniques in section 1
♥! Be careful about the domain of variables since they are mostly natural
numbers only
Marginal cost & marginal revenue:
o If 𝑥 is the quantity of items, then we let:
1. r(𝑥) be the revenue from selling 𝑥 items
2. c(𝑥) be the cost of producing 𝑥 items
x
$
Break–even pt.
c(𝑥)
Max. profit
r(𝑥)
Min. profit

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3. p(𝑥) = r(𝑥) − c(𝑥) be the profit from producing & selling 𝑥 items
o Marginal revenue, marginal cost, and marginal profit refer to r′(𝑥), c′(𝑥) and
p′(𝑥)
o If r(𝑥) and c(𝑥) are differentiable in an interval of production possibilities, and
if p(𝑥) has a max. value, it must occur at a critical pt. or at the endpt.
o If it occurs at a critical pt., then, at the production lvl. that yields the
max. profit, marginal revenue equals marginal cost:
p′(𝑥∗) = r′(𝑥∗) − c′(𝑥∗) = 0 − eqn. (3a)
4. Newton’s method
Used to approximate the solution to the eqn., 𝑓(𝑥) = 0 using a tangent line in place of
the graph of 𝑦 = 𝑓(𝑥) near the pts. where 𝑓 is 0
The strategy is to produce a sequence of approximations that approach
the solution:
𝑥n+1 ≡ 𝑥n −
𝑓(𝑥n)
𝑓′(𝑥n)
, 𝑓′(𝑥n) ≠ 0 − eqn. (4a)
o We pick the 1st number, 𝑥0, of the sequence:
n 𝑥n 𝑓( 𝑥n) 𝑓′( 𝑥n) 𝑥n+1 = 𝑥n −
𝑓( 𝑥n)
𝑓′( 𝑥n)
o The method uses the tangent to the curve at (𝑥0, 𝑓(𝑥0)) to approximate the next
pt., 𝑥1, where the tangent meets the x–axis – the pt.–slope eqn. of the tangent to
the curve at (𝑥0, 𝑓(𝑥0)) is:
𝑦 = 𝑓(𝑥0) + 𝑓′(𝑥0)(𝑥1 − 𝑥0) − eqn. (4a(i))
o We now find where it crosses the x–axis by setting 𝑦 = 0:
0 = 𝑓(𝑥0) + 𝑓′(𝑥0)(𝑥1 − 𝑥0) → 𝑥1 = 𝑥0 −
𝑓(𝑥0)
𝑓′(𝑥0)
 𝑥1, is often a better approximation to the solution than 𝑥0 is
o The pt., 𝑥2, where the tangent to the curve at (𝑥1, 𝑓(𝑥1)), crosses the x–axis, is
the next approximation in the sequence:

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o This goes on till we are close enough to the root to stop
o As n s, 𝑥n get arbitrarily close to the desired root
In practice, Newton’s method often gives convergence with impressive speed, but this
is not guaranteed:
o Newton’s method does not always converge (e.g. when the root is at a vertical
tangent, successive approximations will oscillate btw. 𝑥 = ±a)
o There are rare situations in which it converges but there is no root there
o Also, when it converges, it may not be the root you have in mind, esp. when
there are 2 roots are close to one another
In general, Newton’s method is often used in conjunction with:
1. Intermediate–value theorem1
, which makes sure that there is ± 𝟏 zero
2. Mean–value theorem, which makes sure that there is at most 1 zero
5. Anti–derivatives
A function, F, is an anti–derivative of 𝑓 on an interval, I, if, for all 𝑥 in I,:
F′(𝑥) = 𝑓(𝑥) − eqn. (5a)
♣! Anti–differentiation is the process of recovering a function, F(𝑥), from its
derivative, 𝑓(𝑥)
If F is an anti–derivative of 𝑓 on an interval, I, then the general anti–derivative
of 𝑓 on I is a family of curves whose graphs are vertical translations of one
another and which have the form F( 𝑥) + C:
o We can select a particular anti–derivative by assigning a specific value to C
Anti–derivative formulas, where k is a non–zero constant and where, for
the trigonometric functions, 𝑥 is in radians:
x
y
Tangent line
(linearization of 𝑓 at 𝑥n)
(𝑥n, 𝑓(𝑥n))
𝑥n+1 = 𝑥n −
𝑓(𝑥n)
𝑓′(𝑥n)
Root sought

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Function, 𝒇( 𝒙) General anti–derivative, 𝐅( 𝒙) ≡ Function, 𝒇( 𝒙) General anti–derivative, 𝐅( 𝒙) ≡
k k𝑥 + C − eqn. (5b(i)) sec2
k𝑥
1
k
tan k𝑥 + C − eqn. (5b(v))
𝑥n
, n ≠ −1
1
n + 1
𝑥n+1
+ C − eqn. (5b(ii)) csc2
k𝑥 −
1
k
cot k𝑥 + C − eqn. (5b(vi))
sin k𝑥 −
1
k
cos k𝑥 + C − eqn. (5b(iii)) sec k𝑥 tan k𝑥
1
k
sec k𝑥 + C − eqn. (5b(vii))
cos k𝑥
1
k
sin k𝑥 + C − eqn. (5b(iv)) csc k𝑥 cot k𝑥 −
1
k
csc k𝑥 + C − eqn. (5b(viii))
Anti–derivative rules:
∫(𝑓(𝑥) ± 𝑔(𝑥)) d𝑥 ≡ ∫ 𝑓(𝑥) d𝑥 ± ∫ 𝑔(𝑥) d𝑥 − eqn. (5c(i))
∫ k ∙ 𝑓(𝑥) d𝑥 = k ∫ 𝑓(𝑥) d𝑥 − eqn. (5c(ii))
♣! Initial value problem is the combination of a differential eqn. and an initial
condition:
o Finding an anti–derivative for a function, 𝑓(𝑥), is the same as finding a
function, 𝑦(𝑥), that satisfies the differential eqn. below:
d𝑦
d𝑥
= 𝑓(𝑥) − eqn. (5d(i))
o To solve it, we need a function, 𝑦(𝑥), that satisfies the eqn. – this function is
found by specifying an initial condition:
𝑦(𝑥0) = 𝑦0 − eqn. (5d(ii))
o The most general anti–derivative gives the general solution, 𝑦 = F(𝑥) + C, of
the differential eqn.,
d𝑦
d𝑥
= 𝑓(𝑥)
o We solve the initial value problem by finding the particular solution that
satisfies the initial condition
o The collection of all anti–derivatives of 𝑓 is called the indefinite integral
of 𝑓 with respect to 𝑥, and is denoted by ∫ 𝑓( 𝑥) d𝑥:
 The function, 𝑓, is the integrand and 𝑥 is the variable of integration
Reference:
1. Thomas, G. B., Thomas’ Calculus with Second–Order Differential Equations, Pearson
Education, Boston (2011), pp. 184 – 245