Maximum_and_Minimum_Values.pdf

8/18/2019 Maximum_and_Minimum_Values.pdf

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Section 4.1 Maximum and Minimum Values 2010 Kiryl Tsishchanka

Maximum and Minimum Values

DEFINITION: A function f has an absolute maximum (or global maximum ) at c if f (c) ≥ f (x ) for all x in D, where D is the domain of f. The number f (c) is called themaximum value of f on D. Similarly, f has an absolute minimum (or global minimum )at c if f (c) ≤ f (x ) for all x in D and the number f (c) is called the minimum value of f onD. The maximum and minimum values of f are called the extreme values of f.

DEFINITION: A function f has a local maximum (or relative maximum ) at c if f (c) ≥f (x ) when x is near c. [This means that f (c) ≥ f (x ) for all x in some open interval containingc.] Similarly, f has a local minimum (or relative minimum ) at c if f (c) ≤ f (x ) when x isnear c.

EXAMPLES:

1. The function f (x ) = x 2 has the absolute (and local) minimum at x = 0 and has no absoluteor local maximum.

2. The function f (x ) = x2 , x

∈ (

−∞, 0)

∪

(0,

∞), has no absolute or local minimum and no

absolute or local maximum.

f (x ) = x 2 , x∈(−∞, 0)∪(0, ∞)3. The functions f (x ) = x, x 3 , x 5 , etc. have no absolute or local minima and no absolute orlocal maxima.

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4. The functions f (x ) = sin x, cosx have innitely many absolute and local minima andmaxima.

5. The functions f (x ) = sec x, cscx have innitely many local minima and maxima and noabsolute minima and maxima.

6. The functions f (x ) = tan x, cot x have no absolute or local minima and no absolute or localmaxima.

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7. The function f (x ) = x 4 + x 3 −11 x 2 −9 x + 18 = ( x −3)(x −1)(x +2)( x +3) has the absoluteminimum at x ≈2.2 and has no absolute maximum. It has two local minima at x ≈ −2.6 andx ≈2.2 and the local maximum at x ≈ −0.4.

8. The function f (x ) = x, 1 ≤ x ≤ 2, has the absolute minimum at x = 1 and absolutemaximum at x = 2 . It has no local maximum or minimum.9. The function f (x ) = x, 1 < x < 2, has no absolute or local minimum and no absolute orlocal maximum.

10. The function f (x ) = x has no absolute or local minimum and no absolute or local maximum.

1

2

−1

−2

1 2−1−2

1

2

−1

−2

1 2−1−2

1

2

−1

−2

1 2−1−2

11. The function f (x ) = 1 has the absolute (and local) minimum and absolute (and local)maximum at any point on the number line.

1

2

−1

−2

1 2−1−2

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THEOREM (The Extreme Value Theorem): If f is continuous on a closed interval [ a, b ], then f attains an absolute maximum value f (c) and an absolute minimum value f (d) at some numbersc and d in [a, b ].

THEOREM (Fermat’s Theorem): If f has a local maximum or minimum at c, and if f ′ (c)exists, then f ′ (c) = 0 .

REMARK 1: The converse of this theorem is not true. In other words, when f ′ (c) = 0 , f doesnot necessarily have a local maximum or minimum. For example, if f (x ) = x 3 , then f ′ (x ) = 3 x 2

equals 0 at x = 0 , but x = 0 is not a point of a local minimum or maximum.

REMARK 2: Sometimes f ′ (c) does not exist, but x = c is a point of a local maximum orminimum. For example, if f (x ) = |x|, then f

′ (0) does not exist. But f (x ) has its local (andabsolute) minimum at x = 0 .

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DEFINITION: A critical number of a function f is a number c in the domain of f such thateither f ′ (c) = 0 or f ′ (c) does not exist.

REMARK: From Fermat’s Theorem it follows that if f has a local maximum or minimum atc, then c is a critical number of f.

EXAMPLES:(a) If f (x ) = 2 x 2 + 5 x −1, then f

′ (x ) = 4 x + 5 . Hence the only critical number of f is x = −54

.

(b) If f (x ) = 3√ x 2 , then f ′ (x ) = 23

x − 1/ 3 = 23 3√ x . Hence the only critical number of f is x = 0 .

(c) If f (x ) = 1x

, then f ′ (x ) = − 1x 2

. Since x = 0 is not in the domain, f has no critical numbers.

1

2

−1−2−3−4

1−1−2−31

2

1 2 3−1−2−31

2

3

−1−2−3

1 2 3−1−2−3

EXAMPLES:(a) Find the critical numbers of f (x ) = 2 x 3 −9x 2 + 12 x −5.

(b) Find the critical numbers of f (x ) = 2 x + 3 3√ x 2 .

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EXAMPLES:

(a) Find the critical numbers of f (x ) = 2 x 3 −9x 2 + 12 x −5.Solution: We have

f ′ (x ) = 6 x 2 −18x + 12= 6( x 2 −3x + 2)= 6( x −1)(x −2)

thus f ′ (x ) = 0 at x = 1 and x = 2 . Since f ′ (x ) exists everywhere, x = 1 and x = 2 are the onlycritical numbers.

1

2

3

−1−2

1 2 3

(b) Find the critical numbers of f (x ) = 2 x + 3 3√ x 2 .Solution: We have

f ′

(x ) = (2 x + 3x 2/ 3)′

= 2 x ′ + 3( x 2/ 3) ′

= 2 ·1 + 3 · 23

x 2/ 3− 1

= 2 + 2 x − 1/ 3 =21

+ 2

3√ x = 2 3√ x

3√ x + 2

3√ x = 2 3√ x + 2

3√ x = 2( 3√ x + 1)

3√ xor

= 2 + 2 x − 1/ 3 = 2x − 1/ 3 ·x 1/ 3 + 2 x− 1/ 3 ·1 = 2x

− 1/ 3(x 1/ 3 + 1) = 2( 3

√ x + 1)3√ xIn short,

f ′ (x ) = 2 + 2 x − 1/ 3 = 2 x − 1/ 3(x 1/ 3 + 1) = 2( 3√ x + 1)

3√ xIt follows that f ′ (x ) equals 0 when 3√ x + 1 = 0 , which is at x = −1; f

′ (x ) does not exist when3√ x = 0 , which is at x = 0 . Thus the critical numbers are −1 and 0.

1

2

−11−1−2−3−4−5

EXAMPLE: Find the critical numbers of f (x ) = x 1/ 3(2 −x ).

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EXAMPLE: Find the critical numbers of f (x ) = x 1/ 3(2 −x ).Solution: We have

f ′ (x ) = [x 1/ 3(2 −x )]′

= ( x 1/ 3) ′ (2

−x ) + x1/ 3(2

−x ) ′

= 13

x − 2/ 3(2 −x ) + x1/ 3 ·(−1)

= 2−x3x 2/ 3 −x

1/ 3

=2 −x3x 2/ 3 −

x1/ 3 ·3x 2/ 33x 2/ 3

= 2−x3x 2/ 3 −

3x3x 2/ 3

= 2−x −3x

3x 2/ 3 =

2−4x3x 2/ 3

= 2(1 −2x )

3x 2/ 3

In short,f ′ (x ) = ( x 1/ 3) ′ (2 −x ) + x1/ 3(2 −x )

′ = 2−x3x 2/ 3 −x

1/ 3 = 2(1 −2x )

3x 2/ 3

Here is an other way to get the same result:

f ′ (x ) = [x 1/ 3(2 −x )]′

= (2 x 1/ 3 −x 4/ 3)′

= 2( x 1/ 3) ′ −(x 4/ 3)′

= 2 · 13

x 1/ 3− 1 − 43

x 4/ 3− 1

= 23

x − 2/ 3 − 43

x 1/ 3 = 2

3x 2/ 3 − 4x 1/ 3

3 =

23x 2/ 3 −

4x 1/ 3 ·x 2/ 33 ·x 2/ 3

= 23x 2/ 3 −

4x3x 2/ 3

= 2−4x

3x 2/ 3

= 2(1 −2x )

3x 2/ 3

or

= 23x− 2/ 3 − 43x 1/ 3 = 23x

− 2/ 3 ·1 − 23x− 2/ 3 ·2x = 23x

− 2/ 3(1 −2x ) = 2(1 −2x

)3x 2/ 3

In short,

f ′ (x ) = (2 x 1/ 3 −x 4/ 3)′ =

23

x − 2/ 3 − 43

x 1/ 3 = 23

x − 2/ 3(1 −2x ) = 2(1 −2x )

3x 2/ 3 .

It follows that f ′ (x ) equals 0 when 1 −2x = 0 , which is at x = 12

; f ′ (x ) does not exist when

x 2/ 3 = 0 , which is at x = 0 . Thus the critical numbers are 12

and 0.

1

−1−2−3

1 2 3 4−1−2

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THE CLOSED INTERVAL METHOD: To nd the absolute maximum and minimum valuesof a continuous function f on a closed interval [a, b ]:

1. Find the values of f at the critical numbers of f in (a, b ).

2. Find the values of f at the endpoints of the interval.

3. The largest of the values from Step 1 and 2 is the absolute maximum value; the smallestvalue of these values is the absolute minimum value.

EXAMPLE: Find the absolute maximum and minimum values of f (x ) = 2 x 3 −15x 2 + 36 x onthe interval [1 , 5] and determine where these values occur.Solution:Step 1: Since

f ′ (x ) = 6 x 2 −30x + 36 = 6( x 2 −5x + 6) = 6( x −2)(x −3)there are two critical numbers x = 2 and x = 3 .

Step 2: We now evaluate f at these critical numbers and at the endpoints x = 1 and x = 5 .We have

f (1) = 23 , f (2) = 28 , f (3) = 27 , f (5) = 55

Step 3: The largest value is 55 and the smallest value is 23 . Therefore the absolute maximumof f on [1, 5] is 55, occurring at x = 5 and the absolute minimum of f on [1, 5] is 23, occurringat x = 1 .

EXAMPLES:

(a) Find the absolute maximum and minimum values of f (x ) = 2 x 3

−15x 2 + 24 x + 2 on [0, 2]

and determine where these values occur.

(b) Find the absolute maximum and minimum values of f (x ) = 6 x 4/ 3 −3x 1/ 3 on the interval[−1, 1] and determine where these values occur.

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EXAMPLES:

(a) Find the absolute maximum and minimum values of f (x ) = 2 x 3 −15x 2 + 24 x + 2 on [0, 2]and determine where these values occur.Solution:Step 1: Since

f ′ (x ) = 6 x 2 −30x + 24 = 6( x 2 −5x + 4) = 6( x −4)(x −1)there are two critical numbers x = 1 and x = 4 .

Step 2: Since x = 4 is not from [0, 2], we evaluate f only at x = 1 and at the endpoints x = 0and x = 2 . We have

f (0) = 2 , f (1) = 13 , f (2) = 6

Step 3: The largest value is 13 and the smallest value is 2 . Therefore the absolute maximumof f on [0, 2] is 13, occurring at x = 1 and the absolute minimum of f on [0, 2] is 2, occurringat x = 0 .

(b) Find the absolute maximum and minimum values of f (x ) = 6 x 4/ 3 −3x 1/ 3 on the interval[−1, 1] and determine where these values occur.Solution:Step 1: Since

f ′ (x ) = (6 x 4/ 3 −3x 1/ 3)′ = 6( x 4/ 3) ′ −3(x 1/ 3)

′ = 6 · 43

x 4/ 3− 1 −3 · 13

x 1/ 3− 1

= 8 x1/ 3

−x− 2/ 3

= 8 x − 2/ 3 ·x −1 ·x− 2/ 3

= x − 2/ 3(8x −1)=

8x −1x 2/ 3

there are two critical numbers x = 0 and x = 18

.

Step 2: We now evaluate f at these critical numbers and at the endpoints x =

−1 and x = 1 .

We havef (−1) = 9 , f (0) = 0 , f

18

= −98

, f (1) = 3

Step 3: The largest value is 9 and the smallest value is −98

. Therefore the absolute maximum

of f on [−1, 1] is 9, occurring at x = −1 and the absolute minimum of f on [−1, 1] is −98

,

occurring at x = 18

.

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Appendix

EXAMPLE: Find the absolute maximum and minimum values of f (x ) = 2 x + 3 on [−1, 4] anddetermine where these values occur.Solution:Step 1: Since

f ′ (x ) = (2 x + 3) ′ = (2 x ) ′ −3′ = 2( x ) ′ −3

′ = 2 ·1 −0 = 2there are no critical numbers.

Step 2: Since there are no critical numbers, we evaluate f at the endpoints x = −1 and x = 4only. We havef (−1) = 2(−1) + 3 = −2 + 3 = 1

f (4) = 2(4) + 3 = 8 + 3 = 11Step 3: The largest value is 11 and the smallest value is 1 . Therefore the absolute maximum of f on [−1, 4] is 11, occurring at x = 4 and the absolute minimum of f on [−1, 4] is 1, occurringat x = −1.EXAMPLE: Find the absolute maximum and minimum values of f (x ) = 2 x 2 −8x +1 on [−1, 3]and determine where these values occur.Solution:Step 1: Since

f ′ (x ) = (2 x 2 −8x + 1)′ = (2 x 2) ′ −(8x )

′ + 1 ′ = 2( x 2) ′ −8(x )′ + 1 ′ = 2 ·2x −8 ·1 + 0 = 4 x −8

the number at which f ′ is zero is 2. Therefore x = 2 is the critical number.

Step 2: We evaluate f at the critical number x = 2 and at the endpoints x = −1, 3. We havef (2) = 2 ·22 −8 ·2 + 1 = 2 ·4 −8 ·2 + 1 = 8 −16 + 1 = −7f (−1) = 2(−1)2 −8(−1) + 1 = 2 + 8 + 1 = 11f (3) = 2 ·32 −8 ·3 + 1 = 2 ·9 −8 ·3 + 1 = 18 −24 + 1 = −5

Step 3: The largest value is 11 and the smallest value is −7. Therefore the absolute maximumof f on [−1, 3] is 11, occurring at x = −1 and the absolute minimum of f on [−1, 3] is −7,occurring at x = 2 .EXAMPLE: Find the absolute maximum and minimum values of f (x ) = 5 −x 3 on [−2, 1] anddetermine where these values occur.

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EXAMPLE: Find the absolute maximum and minimum values of f (x ) = 5 −x 3 on [−2, 1] anddetermine where these values occur.Solution:Step 1: Since

f ′ (x ) = (5 −x 3)′ = 5 ′ −(x 3)

′ = 0 −3x 2 = −3x 2

the critical number is x = 0 .Step 2: We evaluate f at the critical number x = 0 and at the endpoints x = −2, 1. We have

f (0) = 5 −03 = 5 −0 = 5f (−2) = 5 −(−2)3 = 5 −(−8) = 5 + 8 = 13f (1) = 5 −13 = 5 −1 = 4

Step 3: The largest value is 13 and the smallest value is 4 . Therefore the absolute maximum of f on [

−2, 1] is 13, occurring at x =

−2 and the absolute minimum of f on [

−2, 1] is 4, occurring

at x = 1 .

EXAMPLE: Find the absolute maximum and minimum values of f (x ) = x 3 +3 x 2 −1 on [−3, 2]and determine where these values occur.Solution:Step 1: Since

f ′ (x ) = ( x 3 + 3 x 2 −1)′ = ( x 3) ′ + (3 x 2) ′ −1

′ = ( x 3) ′ + 3( x 2) ′ −1′ = 3 x 2 + 3 ·2x −0 = 3x 2 + 6 x

= 3 x (x + 2)

the critical numbers are x = 0 and x = −2.Step 2: We evaluate f at the critical numbers x = 0 , −2 and at the endpoints x = −3, 2. Wehave

f (0) = 0 3 + 3 ·02 −1 = 0 + 0 −1 = −1f (−2) = ( −2)3 + 3 ·(−2)2 −1 = −8 + 12 −1 = 3f (−3) = ( −3)3 + 3 ·(−3)2 −1 = −27 + 27 −1 = −1f (2) = 2 3 + 3 ·22 −1 = 8 + 12 −1 = 19

Step 3: The largest value is 19 and the smallest value is −1. Therefore the absolute maximumof f on [−3, 2] is 19, occurring at x = 2 and the absolute minimum of f on [−3, 2] is −1,occurring at x = 0 and x = −3.

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EXAMPLE: Find the absolute maximum and minimum values of f (x ) = 3 x 4 + 4 x 3 −36x 2 on[−2, 3] and determine where these values occur.Solution:Step 1: Since

f ′ (x ) = (3 x 4 + 4 x 3 −36x 2)′ = (3 x 4) ′ + (4 x 3) ′ −(36x 2)

′ = 3( x 4) ′ + 4( x 3) ′ −36(x 2)′

= 3 ·4x 3 + 4 ·3x 2 −36 ·2x= 12x 3 + 12 x 2 −72x= 12x (x 2 + x −6)= 12x (x −2)(x + 3)

the critical numbers are x = 0 , x = 2 , and x = −3.Step 2: Since −3 ∈ [−2, 3], we evaluate f only at the critical numbers x = 0 , 2 and at theendpoints x =

−2, 3. We have

f (0) = 3 ·04 + 4 ·03 −36 ·02 = 0 + 0 −0 = 0f (2) = 3 ·24 + 4 ·23 −36 ·22 = 48 + 32 −144 = −64f (−2) = 3 ·(−2)4 + 4 ·(−2)3 −36 ·(−2)2 = 48 −32 −144 = −128f (3) = 3 ·34 + 4 ·33 −36 ·32 = 243 + 108 −324 = 27

Step 3: The largest value is 27 and the smallest value is −128. Therefore the absolute maximumof f on [−2, 3] is 27, occurring at x = 3 and the absolute minimum of f on [−2, 3] is −128,occurring at x =−

2.

EXAMPLE: Find the absolute maximum and minimum values of f (x ) = − 1x 2

on [0.4, 5] anddetermine where these values occur.

Solution:Step 1: We have

f ′ (x ) = − 1x 2

′

= − 1x 2

′

= − x− 2 ′ = −(−2)x

− 2− 1 = 2 x − 3 = 2x 3

Note that there are no numbers at which f ′ is zero. The number at which f ′ does not exist is

x = 0, but this number is not from the domain of f . Therefore f does not have critical numbers.Step 2: Since f does not have critical numbers, we evaluate it only at the endpoints x = 0 .4and x = 5 . We have

f (0.4) = − 1

(0.4)2 = −

10.16

= −6.25 f (5) = −152

= − 125

= −0.04Step 3: The largest value is −0.04 and the smallest value is −6.25. Therefore the absolutemaximum of f on [0.4, 5] is−0.04, occurring at x = 5 and the absolute minimum of f on [0.4, 5]is −6.25, occurring at x = 0 .4.

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EXAMPLE: Find the absolute maximum and minimum values of f (x ) = 20 −3x −12x

on [2, 4]and determine where these values occur.


f ′ (x ) = 20 −3x − 12x′

= 20 −3x −12x− 1

′

= 20 ′ −(3x )′

− 12x− 1

′

= 20 ′ −3 ·x′

−12 x− 1 ′

= 0 −3 ·1 −12(−1)x− 1− 1

= −3 + 12x− 2

Since

−3 + 12x− 2

= −3 + 12 · 1x 2 = −3 +

12x 2 = −

3x 2

x 2 + 12x 2 = −

3x 2 + 12x 2

= −3(x 2 −4)x 2

= −3(x 2 −22)x 2

= −3(x −2)(x + 2)x 2

we havef ′ (x ) = −3(x −2)(x + 2)x 2

The numbers at which f ′ is zero are x = 2 and x = −2. The number at which f ′ does not exist

is x = 0. But f is not dened at x = 0, therefore the critical numbers of f are x = −2 andx = 2 only.Step 2: Since −2 ∈ [2, 4], we evaluate f only at the critical number x = 2 (which is also oneof the endpoints) and at the second endpoint x = 4. We have

f (2) = 20 −3 ·2 − 12

2 = 20 −6 −6 = 8

f (4) = 20 −3 ·4 − 124 = 20 −12 −3 = 5Step 3: The largest value is 8 and the smallest value is 5 . Therefore the absolute maximum of f on [2, 4] is 8, occurring at x = 2 and the absolute minimum of f on [2, 4] is 5, occurring atx = 4 .

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EXAMPLE: Find the absolute maximum and minimum values of f (x ) = 2 3√ x on [−8, 1] anddetermine where these values occur.Solution:Step 1: We have

f ′ (x ) = 2x 1/ 3′

= 2 x 1/ 3′

= 2

· 1

3x 1/ 3− 1 =

2

3x − 2/ 3 =

2

3x 2/ 3Note that there are no numbers at which f ′ is zero. The number at which f ′ does not exist isx = 0, so the critical number is x = 0 .

Step 2: We evaluate f at the critical number x = 0 and at the endpoints x = −8 and x = 1 .We havef (0) = 2 3√ 0 = 2 ·0 = 0f (−8) = 2

3√ −8 = 2 ·(−2) = −4

f (1) = 2 3

√ 1 = 2 ·1 = 2Step 3: The largest value is 2 and the smallest value is −4. Therefore the absolute maximum of f on [−8, 1] is 2, occurring at x = 1 and the absolute minimum of f on [−8, 1] is−4, occurringat x = −8.EXAMPLE: Find the absolute maximum and minimum values of f (x ) = −5x 2/ 3 on [−1, 1] anddetermine where these values occur.Solution:Step 1: We have

f ′ (x ) = −5x 2/ 3′

= −5 x 2/ 3′

= −5 · 23x2/ 3− 1 = −103 x

− 1/ 3 = − 103x 1/ 3Note that there are no numbers at which f ′ is zero. The number at which f ′ does not exist isx = 0, so the critical number is x = 0 .

Step 2: We evaluate f at the critical number x = 0 and at the endpoints x = −1 and x = 1 .We havef (0) = −5(0)2/ 3 = −5 ·0 = 0f (

−1) =

−5(

−1)2/ 3 =

−5 (

−1)2

1/ 3=

−5(1)1/ 3 =

−5

·1 =

−5

f (1) = −5(1)2/ 3 = −5 ·1 = −5Step 3: The largest value is 0 and the smallest value is −5. Therefore the absolute maximum of f on [−1, 1] is 0, occurring at x = 0 and the absolute minimum of f on [−1, 1] is−5, occurringat x = ±1.EXAMPLE: Find the absolute maximum and minimum values of f (x ) = x 4/ 3 on [−27, 27] anddetermine where these values occur.

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EXAMPLE: Find the absolute maximum and minimum values of f (x ) = x 4/ 3 on [−27, 27] anddetermine where these values occur.Solution:Step 1: We have

f ′ (x ) = x 4/ 3′

= 43

x 4/ 3− 1 = 43

x 1/ 3

Note that there are no numbers at which f ′ does not exist. The number at which f ′ is zero isx = 0, so the critical number is x = 0 .

Step 2: We evaluate f at the critical number x = 0 and at the endpoints x = −27 and x = 27.We havef (0) = 0 4/ 3 = 0

f (−27) = (−27)4/ 3 = (−27)1/ 34

= ( −3)4 = 81

f (27) = 27 4/ 3 = 271/ 34

= 3 4 = 81

Step 3: The largest value is 81 and the smallest value is 0 . Therefore the absolute maximumof f on [−27, 27] is 81, occurring at x = ±27 and the absolute minimum of f on [−27, 27] is 0,occurring at x = 0 .EXAMPLE: Find the absolute maximum and minimum values of f (x ) = √ 4 −x 2 on [−2, 1]and determine where these values occur.Solution:Step 1: We have

f ′

(x ) = (4 −x2

)1/ 2 ′

= 12(4 −x

2

)1/ 2− 1

·(4 −x2

)′

= 12(4 −x

2

)− 1/ 2

·(4′

−(x2

)′

)

= 12

(4 −x 2)− 1/ 2 ·(0 −2x ) =

12

(4 −x 2)− 1/ 2 ·(−2x ) = −x (4 −x 2)

− 1/ 2

= − x

(4 −x 2)1/ 2 = −

x√ 4 −x 2

Note that the number at which f ′ is zero is x = 0 and the numbers at which f ′ does not existare x = ±2. Therefore the critical numbers are x = 0 , −2, 2.Step 2: Since 2∈[−2, 1], we evaluate f only at the critical numbers x = 0 , −2 (which is alsoone of the endpoints) and at the second endpoint x = 1 . We have

f (0) = √ 4 −02 = √ 4 −0 = √ 4 = 2f (−2) = 4 −(−2)2 = √ 4 −4 = √ 0 = 0f (1) = √ 4 −12 = √ 4 −1 = √ 3

Step 3: The largest value is 2 and the smallest value is 0 . Therefore the absolute maximum of f on [−2, 1] is 2, occurring at x = 0 and the absolute minimum of f on [−2, 1] is 0, occurringat x = −2.

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EXAMPLE: Find the absolute maximum and minimum values of f (x ) = cos x on −π2

, 5π

2and determine where these values occur.


f ′ (x ) =

−sin x

Note that there are no numbers at which f ′ does not exist. The numbers in −π2

, 5π

2 at which

f ′ is zero on are x = 0 , π, 2π , so the critical number is x = 0 , π, 2π.

Step 2: We evaluate f at the critical numbers x = 0 , π, 2π and at the endpoints x = −π2

, 5π

2 .

We have

f (0) = cos 0 = 1

f (π ) = cos π = −1f (2π ) = cos(2 π ) = 1

f −π2

= cos −π2

= 0

f 5π2

= cos5π2

= 0

Step 3: The largest value is 1 and the smallest value is −1. Therefore the absolute maximumof f on −

π2

, 5π

2 is 1, occurring at x = 0 , 2π and the absolute minimum of f on −

π2

, 5π

2 is

−1, occurring at x = π.

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