maximum and minimum function values (1)

8/11/2019 Maximum and Minimum Function Values (1)

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Maximum and Minimum

Function Values


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Definition of a Relative Maximum ValueThe function f has a relative maximum value at the number c ifthere exists an open interval containing c, on which f is defined such

that f(c) f(x) for all x in this interval.

a c bx

a c bx

Definition of a Relative Minimum ValueThe function f has a relative minimum value at the number c ifthere exists an open interval containing c, on which f is defined suchthat f(c) f(x) for all x in this interval.

a c b x a c b x


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Theorem: If f(x) exists for all values of x in the open interval (a, b), and if fhas a relative extremum at c, where a < c < b, and if f (c) exists, then

f(c) = 0.

Illustration:Consider the function

f(x) = 2x 2 + 4x 5.The function is a parabolaopening downward with vertexat (1, -3). The function ismaximum there. Let c = 1.Then, from the abovetheorem, f (1) = 0.

Relative extremum:If a function f has either a relative maximum or a relative

minimum value at c, then f has a relative extremum at c.


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Critical Number

Definition of a Critical NumberIf c is a number in the domain of the function f, and if either f (c) = 0 orf (c) does not exist, then c is a critical number of f.

The critical number of

f(x) = 2x2 + 4x 5 is 1 sincef (1) = 0.

The critical number off(x) = x2/3 is 0 since f (0) does

not exist.


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Examples: Find the critical numbers. (Exercises on p.218)

1. f(x) = x 4 + 4x 3 2x 2 12x.

Solution: Since, f(x) is a polynomial function and f (x) is also a polynomial, wefind c only when f (c) = 0.

So, we find c for which f (c) = 0. f (x) = 4x 3 + 12x 2 4x 12 = 0,

x3 + 3x 2 x 3 = 0.

Solving for x by factoring by grouping or by synthetic division,(x + 3)(x 1)(x + 1) = 0,x = -3, 1, and -1. (critical numbers)

2. f(t) = (t 2 4) 2/3

Solution: The derivative is f (t) = 1/3

2 2 1/32 4t

t 4 2t 03 3(t 4)

Solving for t by cross multiplication,4t = 0. So, t = 0 (C.N. where f (c) = 0).

Next, we find c where f (c) does not exist. Since f (t) does not exist when t = 2and -2, and 2 and -2 are in the domain of f,

then 2 and -2 are critical numbers. So, the C.Ns are 0, 2, and -2


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3. f(x) =2x 2x 5

x 1

Solution: We find the derivative,

f (x) =2 2

2 2(x 1)(2x 2) (x 2x 5)(1) x 2x 7

(x 1) (x 1)

We find c where f (c) = 0. So, x 2 2x - 7 = 0.So,

x = (critical numbers)1 2Note that, f (x) does not exist when x = 1, but 1 is not in the domain of f.Therefore, 1 is not a critical number.

Solution: We find c for which f (c) = 0. So, f (x) = 2cos 2x 2sin 2x = 0.

tan 2x = 1,2x = /4

x = . In general, x = , where n is an integer.8

n8 2

4. f(x) = sin 2x + cos 2x


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Absolute Maximum and Absolute Minimum Values on an Interval(Refer to your book on pages 213-214)

Extreme Value Theorem: If a function f is continuous on the closed interval [a, b],then f has an absolute maximum value and an absolute minimum value on [a, b].

Absolute extremum either absolute maximum or absolute minimum.

The absolute extremum can be determined by the following procedure:1. Find the critical number of the function on (a, b).

2. Find the values of f(a), f(c), and f(b). (You may construct a table)3. The largest of the values in step 2 is the absolute maximum value and the

smallest is the absolute minimum value.

x f(x)

a f(a) - abs. max. valuec f(c) - abs. min. value

b f(b)

a c b


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Examples: (Exercises on p. 219)1. f(x) = 4 3x ; (-1, 2]

Solution:

First, find the critical number. We get f (x) = -3. So, no critical number.Next, we find the function value f(2).

x f(x)

2 -2 - abs. min.

Sketch:

We see that -2 is theabsolute minimum value.

There is no absolutemaximum since the intervalis open at -1 and there isno critical number.


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2. f(x) = x 2 2x + 4; (- , + )

Solution:Observe that f is defined on the open interval (- , + ). So the only function value

that can be obtained is from the critical number c.

We find the critical number,f (x) = 2x 2 = 0. So, x = 1 is the critical number.

If x = 1, y = 1 2 -2(1) + 4 = 3. Since f is a parabola opening upward then 3 must bethe absolute minimum value. Also, we see that the vertex is at the point (1, 3).

x f(x)

1 3 - abs. min.


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3. G(x) = -3sin x ;3

0,4

Solution:Finding the critical number,

G (x) = -3cos x = 0,x = /2, C.N.

Since G is always differentiable, there is no critical number c where G (c) dne.

x G(x)

0 0 - abs. max. value

/2 -3 - abs. min. value


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4. ;[3, 5]2

if x 5f(x) x 5

2 if x 5

Solution:No critical number for which f (x) = 0. But 5 is a critical number since f (5) dne. Since f(x) = 2 when x = 5, then 2 mustbe an absolute extremum.

x f(x)

3 -15 2 - abs max value


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Applications Involving an Absolute Extremum on a Closed Interval

Examples: (pp. 226 227)

1. Find the number in the interval [-1, 1] such that the difference of the numberand its square is a) a maximum, and b) a minimum.

Solution:Let x be the number. We want to maximize or minimize x minus its

square. Let D be the difference of x and its square.

D = x x2.So, D(x) = 1 2x = 0.Thus, x = (critical number)

x D

-1 -2 abs. min value

1/4 abs. max value

1 0


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2. Suppose that the weight is to be held10 ft below a horizontal line ABby a wire in a shape of a Y. If the points A and B are 8 ft apart. What isthe shortest length of the wire that can be used?

Let x be the length of the line segmentCD.

We would like to minimize the total lengthL of the wire.

So, L = AC BC CW

L = ; [0,10]2 216 16 10 x x x

B A

C

W

8

10-x

x 4 B

C

D

x

22 16 10 x x

L (x) = 22

1 0

16

x

x

2

2 2

2

2 16

4 16

3 16

4 4 333

x x

x x

x

x or

Solving for the criticalnumber,

x L(x)

0 18

- abs.min

10

4 3

3

4 29 21.54 ft

10 4 3 16.93 ft


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3. An island is at point A, 4 km offshore from the nearest point B on a straightbeach. A woman on the island wishes to go to a point C, 6 km down the beachfrom B. She Can go by a rowboat at 5 km/hr to a point P between B and C andthen walk at 8 km/hr along a straight path from P to C. Find the route from A to C

that she can travel in the least time.Solution: Let x be the distance from B to P.We would like to find the distance AP + PC where the travel time isminimum.

B

A

P C

4 km

x6 - x

Let T be the total time travelled by the woman. So,T = T AP + T PC .

Since, time = distance/speed,

; [0, 6]2x 16 6 x

T5 8

Since, T is to be minimized,2x 1T '(x) 0

85 x 16

Solving for x,

km

28x 5 x 16 0,

20 20 39x

3939


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x T

0 31/20 hr

1.37 hr least time

6 1.44 hr

20 39

39

Thus, the distance that the woman can travel is

22 20 20x 16 6 x 16 6

39 39

7.92km


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4. Find the dimensions of the right-circular cylinder of greatest lateral surface areathat can be inscribed in a sphere with a radius of 6 in.

r

h6 in

Solution:Find r and h of the cylinder with greatest lateral

surface area.

Lateral surface area of the cylinder is A = 2 rh.

By Pythagorean theorem,12 2 = (2r) 2 + h 2.

So, 144 = 4r 2 + h 2.

Solving for h, 2 2h 144 4r 2 36 r

So, A = ; [0, 6]2

4 r 36 r Since A is to be max.,

A(r) = 22

2r (4 r) (4 ) 36 r 0

2 36 r

2 2r 36 r 0,

r 3 2 in

h

2r

12


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If r = , then h =3 2 i n 6 2 in

r h A

0 6 in 0

72 in 2

6 in 0 0

3 2 in 6 2 in

So, A = 22 rh 2 (3 2 )(6 2) 72in

Thus, the dimensions are

r = and h =3 2 i n 6 2 in


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Assign number 2:Page 227,

numbers 25, 27, and 31

maximum and minimum function values (1)

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