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