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Chapter 4: APPLICATIONS of DIFFERENTIATION
Duong T. PHAM - EEIT2014
Fundamental Engineering Mathematics for EEIT2014
Vietnamese German UniversityBinh Duong Campus
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Outline
1 Maximum and minimum values
2 Derivative and shape of a graph
3 Indeterminate forms and LHopital Rule
4 Optimization problems
5
Newtons method
6 Antiderivatives
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Applications in Optimization
Important optimization problems require differential calculus, e.g.
What is the shape of a can that minimizes manufacturing costs?
What is the maximum acceleration of a space shuttle?
What is the interest rate that the banks earn most in the market?
what is the shape of a racing car or a aircraft to minimize drag?
etc.
= finding the minimum and maximum of a function.
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UMaximum and minimum of a function
Def: Let f :D R be a function and let c D.f has an absolute maximum(global maximum) at c if
f(x) f(c) x D.
Then, f(c)is called the maximum value of f.f has an absolute minimum (global minimum) at c if
f(x) f(c) x D.
Then, f(c)is called the minimum value of f.
The maximum and minimum values of fare called the extremevaluesof f.
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UMaximum and minimum of a function
x
y
y = f(x)
a b c d e
f(a)
f(d)
c
f(c)
c1 c2
f(a)is the absolute minimum of f;
f(d)is the absolute maximum of f
f(c)
f(x)
x
(c1, c2)=
f(c)is a local minimum of f
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ULocal maximum and minimum
Def: Let f :D R be a function and let c D.f has an local maximum (relative maximum) at c if there exists aninterval (c1, c2) D such that c (c1, c2) and
f(x)
f(c)
x
(c1, c
2).
f has an local minimum (relative minimum) at c if if there existsan interval (c1, c2) D such that c (c1, c2) and
f(x)
f(c)
x
(c1, c2).
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ULocal maximum and local minimum
x
y
y = f(x)
a b c d e
f(a)
f(d)
c
f(c)
c1 c2
f(c)is a local minimum
f(b)is the local maximum of f;
f(d)is the local maximum of f (also an absolute maximum)
f(e)is a local minimum of f
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UThe Extreme value Theorem
Theorem: Let f :[a, b] R be a continuous function on[a, b]. Then fattains an absolute maximum value f(c)and an absolute minimum valuef(d)at some numbers c and d in[a, b].
Remark: The theorem is not true if we replace the closed interval [a, b]by the open interval (a, b) or other interval (a, b] or [a, b).
Ex: The function f(x) = 1x
does not attain an absolute maximum value
on (0, 5].
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UThe Extreme value Theorem
y
x
y= 1
x
1
2
34
5
6
7
8
9
10
11
12
13
14
15
16
17
1819
1 2 3 4
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UThe Fermats Theorem
x
y
y = f(x)
(c, f(c)
c d
(d, f(d)
fattains local maximum at cand local minimum at d = What is specialabout f at x=c and x=d?
The Fermats Theorem: Iffattains local maximum or local minimum at candif f(c)exists, then f(c) = 0
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UProof of Fermats Theorem
The Fermats Theorem: Iffattains local maximum or local minimum at candif f
(c)exists, then f
(c) = 0
Proof: Suppose fattains local maximum at c.
=x (c1, c2) : f(x) f(c)
= f(x)
f(c)
x c 0 c1
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The Fermats Theorem
Remark: Local maximum and local minimum at c + existence off (c)=
f (c) = 0. But(
=)is NOT true.
Ex: Let f(x) =x3. We have f (x) = 3x2 and f (0) = 0 but f does NOTattain local max. or local min. at 0.
x
y
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Local minimum and local maximum
Ex: Function f(x) = |x| attains local minimum at x= 0 but it is NOTdifferentiable at x= 0.
x
y
0
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Critical numbers of a function
Corollary: If fhas a local minimum or local maximum at c then c is a
critical number of f
Proof: Fermats Theorem:
f (c) = 0
f
(c) does NOT exist = c is a critical number of f
fhas local min. or local max. at c = There are 2 cases:
f (c) does not exist=
c is a critical number
f (c) existsFermats Th.
= f (c) = 0= c is a critical number
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Finding absolute minimum and absolute maximum
The closed interval method: Let f : [a, b] R be a continuous function. Tofind the absolute max. and absolute min. of f, we follows the steps:
1 Find the values of fat critical numbers of f in (a, b);
2 Find the values f(a)and f(b);
3 The largest number in steps 1 and 2 is the absolute maximum value of f,and the smallest number in steps 1 and 2 is the absolute minimum of f.
Ex: Find abs. max. and abs. min. of f(x) = 2x3 9x2 + 12x+ 2 in[0, 3]Ans: f(x) = 6x2 18x+ 12 = 6(x2 3x+ 2)
1 f(x) = 0 x= 1 or x= 2, and f(1) = 7, f(2) = 62 f(0) = 2and f(3) = 113 Comparing the 4 values in Steps 1 and 2, we conclude
max[0,3]
f =f(3) = 11 and min[0,3]
f =f(0) = 2
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Th l d i l h d
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The closed interval method
x
yy = 2x3 9x2 + 12x+ 5
0
2
7
1
6
2
11
3
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Exercises
4.1:
310, 1520, 2728, 2936, 4755
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R ll Th
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Rolles Theorem
Rolles Theorem: Let f : [a, b] R be a function satisfying1 f is continuous in [a, b]
2 f is differentiable in (a, b)
3 f(a) =f(b). Then there exists a c (a, b)such that f(c) = 0
Proof:
The case: f(x) =cx [a, b]. then clearly f(c) = 0 c (a, b).The case: x (a, b) s.t. f(x)> f(a) =f(b).Since f is continuous in [a, b]
ExtremeValueTh.= c (a, b) such that
f(c) = max[a,b]
f
= f has a local max. at c + f(c) exists (f is differentiable in (a, b))Fermats Th.
= f(c) = 0The case: x (a, b) s.t. f(x)< f(a) =f(b). (Similar as the secondcase)
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R ll Th
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Rolles Theorem
Ex: Prove that equation x3 + x2 + 3x+ 1 = 0 has exactly one real root.
Ans: Denote f(x) =x3 +x2 + 3x+ 1.
f is a polynomial= fis differentiable (and thus continuous) in (,).f(1) = 2 and f(0) = 1. Since f is continuous in [1, 0] and sincef(
1)< 0 < f(0), the Intermediate value Theorem=
c
(
1, 0) s.t.
f(c) = 0= the equation has one root c (1, 0)Suppose that d=cis another root of the equation.
Ifc 0 for all x R= ContradictionIfc>d, similar argument= Contradiction
The equation has exactly one root c (1, 0).
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R ll Th
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Rolle s Theorem
x
y
y = x3 +x2 + 3x+ 1
0
1
c1
2
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Recall the Rolles Theorem
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Recall the Rolle s Theorem
x
y
(a, f(a)) (b, f(b))L
t(c, f(c))
(a, f(a))
(b, f(b))
(c, f(c))t
The slope of secant connecting (a, f(a)) and (b, f(b)) is f(b)
f(a)
b a = 0Rolles Theorem:c (a, b) s.t. f(c) = 0
= f(c) = f(b) f(a)b
a
t//L
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The Mean Value Theorem
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The Mean Value Theorem
The Mean Value Theorem: Let f : [a, b] R be a function satisfying1 f is continuous in [a, b]
2 f is differentiable in (a, b)
Then there exists a c (a, b)such that
f(c) = f(b) f(a)
b aroof: Denote h(x) =f(x) f(b)f(a)
ba (x a) f(a). Then:h is continuous in [a, b] and h is differentiable in (a, b)
h(a)= f(a)
f(b)f(a)
b
a
(a
a)
f(a)= 0
h(b)= f(a) f(b)f(a)ba (b a) f(a)= 0
= h(a) =h(b)Rolles Theorem: there is a c (a, b)such that h(c) = 0=
f(c) = f(b) =f(a)
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The Mean Value Theorem
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The Mean Value Theorem
Theorem: Letf : (a, b) R be a function which is differentiable in (a, b)and f
(x) = 0 for all x (a, b). Then f is a constant function.Ans: Let x1, x2 (a, b]. Then f is continuous in [x1, x2] and differentiablein (x1, x2). By the Mean Value Theorem, there is a c (x1, x2) such that
f (c) = f
(x
2) f(x
1)x2 x1 .
Since f (x) = 0 for all x (x1, x2), we havef(x2)
f(x1)
x2 x1 = 0.It means that f(x1) =f(x2) and this is true for any x1, x2 (a, b). Hence,f is a constant function in (a, b).
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Exercises
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Exercises
4.2:
15, 7, 1114, 151720, 2231, 36
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Derivative and shape of a graph
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Derivative and shape of a graph
Increasing and decreasing Test: Let f : (a, b) R be a differentiablefunction.
1 If f (x)>0,x (a, b), then f is increasing in (a, b).2 If f (x)
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Increasing and decreasing Test
Ex: Determine when the functionf(x) = 3x4
4x3
12x2 + 5is increas-
ing and decreasing.
Ans: f (x) = 12x3 12x2 24x= 12x(x 2)(x+ 1)= f (x) = 0 x= 1 x= 0 x= 2
x 1 0 2 x 2 | | 0 +x | 0 + | +
x+ 1 0 + | + | +f (x) 0 + 0 0 +f(x) 0 5 27
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Increasing and decreasing Test
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Increasing and decreasing Test
x
y
12
y = 3x4
4x3
12x2
+ 5
0
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The first derivative Test
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The first derivative Test
The First derivative Test: Suppose that c is a critical number of a continuousfunction f.
(a) If f changes from positive to negative at , then has a local maximum at c.
(b) Iff changes from negative to positive at c, then has a local minimum at c.
(c) If f does not change sign at c(for example, if f is positive on both sidesofcor negative on both sides), then fhas no local maximum or minimum
at c.
y
xc
f < 0 f > 0
local minimum
y
xc
f > 0 f < 0
local maximum
y
xc
f > 0
f > 0
no local min. or max.
y
xc
f < 0
f < 0
no local min. or max.
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The first derivative Test
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The first derivative Test
Ex: Find the local minimum and maximum values of the function f(x) =3x4
4x3
12x2 + 5
Ans: f (x) = 12x3 12x2 24x= 12x(x 2)(x+ 1)= f (x) = 0 x= 1 x= 0 x= 2
x 1 0 2 x 2 | | 0 +x | 0 + | +
x+ 1 0 + | + | +f (x)
0 + 0
0 +
f(x) 0 5 27
fattains local minimum at1 and 2; and attains local maximum at0.
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The first derivative Test
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The first derivative Test
x
y
12
y = 3x
4
4x
3
12x
2
+ 5
0
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Concave functions
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Def: Let f :I R.If the graph of f lies above all of its tangent lines, then f is said to
be upward concave on IIf the graph of f lies below all of its tangent lines, then f is said tobe downward concave on I
y
x
upwad concave
y
x
downward concave
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Concavity Test
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y
Concavity Test: Let f :I R. Then(a) If f(x)> 0 for all x
I, then the graph of f is concave upward on I.
(b) If f(x)< 0 for all x I, then the graph of f is concave downward on I.
Discussion:
f(x)> 0 for all x I = f(x) is increasing on If(x
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The second derivative Test
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The second derivative Test: Let f be a function such that f is continuousnear c. Then
(a) If f(c) = 0and f(c)> 0, then fhas a local minimum at c,(b) If f(c) = 0and f(c)< 0, then fhas a local maximum at c,
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The second derivative Test
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Ex: Discuss the concavity, inflection points, local maxima and local minima ofthe curve y=x4 4x3.
Ans: Denote f(x) =x4 4x3. Thenf(x) = 4x3 12x2 = 4x2(x 3)f(x) = 12x2 24x= 12x(x 2)
f(x
) = 0x
= 0 x
= 3 and f(x
) = 0x
= 0 x
= 2
x 0 2 3x2 + 0 + | + | +
x 3 | | 0 +x
2
| 0 +
| +
f(x) 0 | 0 +f(x) + 0 0 + | +f(x) 0 16 27
up.con. infl.p. down.con. infl.p. up.con. local min. up.con.
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The second derivative Test
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x
y
2
inflection point16
3
local min.27
0inflection point
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Exercises
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4.3:
12, 56, 912, 1921, 3340
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Indeterminate forms
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-VGUIndeterminate forms: consider limx
a
f(x)
g(x). Then
(a) If f(x) 0and g(x) 0when x a, then the limit is called anindeterminate form of type 00
(b) If f(x) (or) and g(x) (or) when x a, then the limitis called an indeterminate form of type
Ex: Evaluate limx1
x3 xx 1
Ans: limx1
x3 xx
1
= limx1
x(x2 1)x
1
= limx1
x(x 1)(x+ 1)x
1
= limx1
[x(x+ 1)] = 1(1 + 1) = 2.
Ex: limx1
ln x
x 1 = we need a tool to evaluate this limit.
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LHopitals Rule
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-VGULHopitals Rule: Let f and gbe differentiable functions on an interval (b, c),
except possibly at number x=a (b, c). Suppose further that g(x) = 0for allx (b, c)\{a}. Iflimxa
f(x) = limxa
g(x) = 0
or limxa
f(x) = and limxa
g(x) = ,then
limxa
f(x)
g(x) = lim
xaf(x)
g(x)
provided the limit on the right hand side exists (or isor)
Ex: limx1
ln x
x 1LHopital.
= limx1
(ln x)
(x 1) = limx1
1
x1
= 1.
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LHopitals Rule
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Ex: limx
ex
x2LHopital.
= limx
(ex)
(x2)= lim
xex
2x
LHopital.= lim
x(ex)
(2x)= lim
xex
2 = .
Ex: limx0
sin x
x
LHopital.= lim
x0(sin x)
(x)= lim
x0cos x
1 = 1.
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Indeterminate products
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Intermediate products: The limit limxa[f(x)g(x)] in which f(x) 0 andg(x) (or) as x a is called an indeterminate form of type 0
Ex: Evaluate limx0+
xln x
Ans: We have
limx0+
xln x= limx0+
ln x1x
LHopital.= lim
x0+(ln x)
1x
= lim
x0+
1x
1x2
= limx0+
(x) = 0.
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Indeterminate powers
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D.T.
PHAM
-VGUThe limit limxa[f(x)]g(x) when f and g satisfy
(a) limxa
f(x) = limxa
g(x) = 0 type 00
(b) limxa
f(x) = and limxa
g(x) = 0 type0
(c) limx
af(x) = 1 and lim
x
ag(x) = type 1 are indeterminate powers.
Remark: To evaluate the above limits we can either
1 write y= [f(x)]g(x) and thus ln y=g(x) ln f(x), and then evaluate limxa
ln y
first; after that, we deduce limxa
y.
2 We can write [f(x)]g(x) =eg(x) ln f(x).
3 By these ways, we transfer it to the problem of finding limit of the type0 .
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Indeterminate powers
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D.T.
PHAM
-VGUEx: Evaluate lim
x0+
(1 + 4 sin 4x)cot x
Ans: Denote y= (1 + 4 sin 4x)cot x = ln y= cot xln(1 + 4 sin 4x)Then
limx0
+ln y= lim
x0
+cot x ln(1 + 4 sin 4x) = lim
x0
+
ln(1 + 4 sin 4x)
tan xLHopital.
= limx0+
16cos4x1+4 sin 4x
1cos2 x
= limx0+
16cos4x cos2 x1 + 4 sin 4x
= 16
Hence, limx0+
y= limx0+
eln y =elimx0+ ln y =e16.
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Exercises
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4.4:
Each student should do a third of exercises from 564.
6970,
7576
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Optimization problems
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D.T.
PHAM
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Find max(xy)?2x+y= 50
Ans: We have A= xyand since 2x+y= 50, we replace y= 50 2x in A toobtain
A(x) =x(50 2x).Our task is now to find: max A(x), where 0 x 25. We have A = 50 4x,and
A= 0 50 4x= 0 x= 15.x 0 15 25A 50 + 0 50A 0 300 0
We can deduce from the table that
maxA= 300m2 when x= 15m
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Optimization problems
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20m
15m
15m
20m
Area = 300m2
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Applications in business and economics
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-VGUThe cost function C(x)= the cost of producing xunits of a certainproduct.
The marginal cost (=C(x)) is the change ofC(x)w.r.t. x
The demand function (or price function), denoted by p(x), is the priceper unit that the company can charge if it sells x units
If the company sells xunits and the price per unit is p(x), then the total
revenue is denoted by the revenue function,
R(x) =xp(x)
The derivative R is the marginal revenue function
Ifxunits are sold, the total profit
P(x) =R(x) C(x)and P is called the profit function.
P is called the marginal profit function
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Applications in business and economics
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Ex: A store sells 200 DVD burners a week at $350 each.
Survey: if each $10 rebate is offered, 20 more units will be sold every week.
Q.: Find demand and revenue functions. How large a rebate should be tomaximize its revenue?
Ans: Denote by xthe number of units sold every week= weekly increase insales is x
200.
If the price per unit decreases by $10, more 20 units are sold.
= the price per unit so that the weekly sale is xunits, is
p(x)= 350 x 20020
10 =450 x2.
The revenue function is R(x)= xp(x) =450xx2
2
Our task: Find the absolute maximum ofR(x) = 450xx2
2
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Applications in business and economics
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M-VGUOur task: Find the absolute maximum ofR(x) = 450x
x2
2R(x) = 450 x, and R(x) = 0 x= 450.consider the table
x 350 450
R(x) 100 + 0 R(x) 96250 101250
The revenue has an absolute maximum (101250) when the number ofweekly sold units is 450.
The price per unit is then p(450) = 450 4502 = 225The rebate should then be offered as350 225 = 125
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Exercises
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4.7:
26, 11, 1719, 35
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Newtons method
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Some examples:Solve 2x+ 5 = 0 x= 52Solve x2 5x+ 4 = 0.
= (5)2 4 1 4 = 9The solutions
x1 = 5
2 =
x2 = 5+
2
x1 = 5
9
2
x2 = 5+
9
2
x1 = 1
x2 = 4
Solve cos x x= 0?
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Newtons method
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x
y
y = f(x)
r
x1
f(x1)
t1
t1 : y = f (x1)(x x1) +f(x1)
x2
f(x2)
t2
t2 : y = f (x2)(x x2) +f(x2)
x3
Step 1: Choose some x1
x2 = x1 f(x1)
f
(x1)
x3 = x2 f(x2)
f (x2)
Step 2: xn = xn1 f(xn1)
f (xn1)
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Newtons method
E Fi d i d i l l h f h i
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D.T.
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Ex: Find, correct to six decimal places, the root of the equation cos x=x
Ans: The equation is equivalent tocos x
x= 0.
Denote f(x) = cos x x. Then f(x) = sin x 1Newtons method:
xn+1 =xn f(xn)f(xn)
=xn+cos xn xn
sin xn+ 1
Choose x1 = 0. Then
x2 = 0 +cos0 0sin 0 + 1
= 1; x3 = 1 +cos1 1sin 1 + 1
= 0.750363868
x4 = 0.737151911; x5 = 0.739446670
x6 = 0.739018516; x7 = 0.739097442
x8 = 0.739082860; x9 = 0.739085553
x10 = 0.739085055; x11 = 0.739085147
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Exercises
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PHAM-V
GU
4.8:
58, 11, 12, 1722
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Antiderivatives
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GUDef: A function F is called an antiderivative of f on an interval I ifF
(x) =f(x) for all x in I.
Ex: F(x) = x3
3 is an antiderivative of f(x) =x2 for any x R.
Indeed, F(x)= x33
=x2 =f(x)
Theorem: IfF is an antiderivative of f on an interval I, then the mostgeneral antiderivative of fon the interval I is
F(x) +C,
where C is an arbitrary constant.
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Exercises
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GU
4.9:
120, 2328, 48
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