4.1 – Extreme Values of Functions Extreme Values of a function are created when the function changes from increasing to decreasing or from decreasing to

4.1 – Extreme Values of FunctionsExtreme Values of a function are created when the function changes from increasing to decreasing or from decreasing to increasing

Extreme value

decreasingincreasingincreasing

Extreme value

decreasing

decdec

inc

Extreme value

Extreme value

inc dec

inc

dec

Extreme value

Extreme value

Extreme value

4.1 – Extreme Values of Functions

Absolute Minimum – the smallest function value in the domain

Absolute Maximum – the largest function value in the domain

Local Minimum – the smallest function value in an open interval in the domain

Local Maximum – the largest function value in an open interval in the domain

Classifications of Extreme Values

Absolute Minimum Absolute Minimum

Absolute Maximum

Absolute Maximum

Local Minimum

Local Minimum Local MinimumLocal Minimum

Local Minimum

Local Maximum

Local Maximum Local Maximum Local Maximum

Local Maximum


Absolute Minimum – occurs at a point c if for x all values in the domain.

Absolute Maximum – occurs at a point c if for all x values in the domain.

Local Minimum – occurs at a point c in an open interval, , in the domain if for all x values in the open interval.

Local Maximum – occurs at a point c in an open interval, , in the domain if for all x values in the open interval.

Absolute Minimum at cc

Absolute Maximum at cc

Definitions:

Local Minimum at c

ca b

Local Maximum at cca b

4.1 – Extreme Values of FunctionsThe Extreme Value Theorem (Max-Min Existence Theorem)

If a function is continuous on a closed interval, [a, b], then the function will contain both an absolute maximum value and an absolute minimum value.

a bc

𝑓 (𝑎)

𝑓 (𝑏)

𝑓 (𝑐 )

Absolute maximum value: f(a)

Absolute minimum value: f(c)



a bd

𝑓 (𝑐 )𝑓 (𝑏)

𝑓 (𝑑)

Absolute maximum value: f(c)

Absolute minimum value: f(d)

c

𝑓 (𝑎)



𝑓 (𝑐 ):𝐷𝑁𝐸

Absolute maximum value: none


a bd

𝑓 (𝑏)

𝑓 (𝑑)c

𝑓 (𝑎)

F is not continuous at c.

Theorem does not apply.



𝑓 (𝑐 )

Absolute maximum value: f(c)


F is not continuous at c.

Theorem does not apply.

a bd

𝑓 (𝑏)

𝑓 (𝑑)c

𝑓 (𝑎)

4.1 – Extreme Values of FunctionsThe First Derivative Theorem for Local Extreme Values

If a function has a local maximum or minimum value at a point (c) in the domain and the derivative is defined at that point, then .

Slope of the tangent line at c is zero.

c

𝑓 (𝑐 )=0𝑓 (𝑐 )>0 𝑓 (𝑐 )<0

c

𝑓 (𝑐 )>0𝑓 (𝑐 )<0

4.1 – Extreme Values of FunctionsCritical Points

If a function has an extreme value, then the value of the domain at which it occurs is defined as a critical point.

Three Types of Critical Points(1 ) 𝐸𝑛𝑑𝑝𝑜𝑖𝑛𝑡𝑠𝑜𝑓 𝑎𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙

(1)

(2 )𝑆𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑃𝑜𝑖𝑛𝑡𝑠 : 𝑓 (𝑐 )=0(3 )𝑆𝑖𝑛𝑔𝑢𝑙𝑎𝑟 𝑃𝑜𝑖𝑛𝑡𝑠 : 𝑓 (𝑐 )𝑑𝑜𝑒𝑠𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡

(1)(2) (2) (2) (2)(3)


a b c d

a 27

b 0

c 0

d -5

a -30

b 5

c 0

d -7

a -22

b 0

c 0

d -9

Which table best describes the graph?

Table A Table B Table C


-1 4

Graph the function. State the location(s) of any absolute extreme values, if applicable. Does the Extreme Value Theorem apply?

Absolute maximum at x = 4

No absolute minimum

𝑓 (𝑥 )={ 1𝑥 𝑖𝑓 −1≤ 𝑥<0

√𝑥 𝑖𝑓 0≤ 𝑥≤ 4

The Extreme Value Theorem does not apply

The function is not continuous at x = 0.


-2 -1

Graph the function. Calculate any absolute extreme values, if applicable. Plot them on the graph and state the coordinates.

Critical points

𝑓 (𝑥 )=−𝑥− 1𝑓 (𝑥 )=− 1𝑥−2≤ 𝑥≤−1

Absolute minimum

𝑓 (𝑥 )=𝑥−2=1

𝑥2

𝑓 (𝑥 )≠0𝑓 (𝑥 ) 𝑖𝑠𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑𝑎𝑡 𝑥=0

𝑥=−2 ,−1

𝑥=0 𝑖𝑠𝑛𝑜𝑡 𝑎𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙𝑝𝑜𝑖𝑛𝑡 ;

𝑓 (−2)=12

𝑓 (−1)=1 Absolute maximum

𝑛𝑜𝑡 𝑖𝑛[−2,−1]

(−2 ,12)

(−1 ,1)

4.1 – Extreme Values of FunctionsCalculate any absolute extreme values. State their identities and coordinates.

Critical points𝑓 (−2 )=−0.5

𝑓 (𝑥 )= 𝑥+1𝑥2+2𝑥+2

Absolute minimum

𝑓 (𝑥 )=(𝑥¿¿ 2+2𝑥+2) (1 )−(𝑥+1)(2𝑥+2)

(𝑥2+2 𝑥+2)2¿

𝑓 (𝑥 )=0

𝐼𝑠 𝑓 (𝑥 )𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑?𝑥=−2 ,0

𝑛𝑜𝑟𝑒𝑎𝑙𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠

Absolute maximum

(−2 ,−0.5)

𝑓 (𝑥 )= −𝑥2−2 𝑥(𝑥2+2 𝑥+2)2

¿−𝑥 (𝑥+2)

(𝑥2+2𝑥+2)2

𝑥2+2𝑥+2=0

𝑥=−2±√22−4 (1)(2)

2(1)

𝑓 (0 )=0.5(0 ,0.5)

4.2 – The Mean Value TheoremRolle’s Theorem

A function is given that is continuous on every point of a closed interval,[a, b], and it is differentiable on every point of the open interval (a, b). If , then there exists at least one value in the open interval,(a, b), where .

𝑠𝑙𝑜𝑝𝑒𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑎𝑏=𝑓 (𝑏 )− 𝑓 (𝑎)

𝑏−𝑎=

0𝑏−𝑎

=0

a b

𝑓 (𝑐 )=0

c

𝑓 (𝑎 )= 𝑓 (𝑏)

𝑠𝑙𝑜𝑝𝑒𝑡𝑎𝑛𝑔𝑒𝑛𝑡@𝑐=0

𝑓 (𝑐 )=0

4.2 – The Mean Value TheoremRolle’s Theorem

A function is given that is continuous on every point of a closed interval,[a, b], and it is differentiable on every point of the open interval (a, b). If , then there exists at least one value in the open interval,(a, b), where .

da bc

𝑓 (𝑎 )= 𝑓 (𝑏)

𝑓 (𝑐 )=0

𝑓 (𝑑 )=0


𝑏−𝑎=

0𝑏−𝑎

=0

𝑠𝑙𝑜𝑝𝑒𝑡𝑎𝑛𝑔𝑒𝑛𝑡@𝑐=0

𝑓 (𝑐 )=0

𝑠𝑙𝑜𝑝𝑒𝑡𝑎𝑛𝑔𝑒𝑛𝑡@𝑑=0

𝑓 (𝑑 )=0

4.2 – The Mean Value TheoremThe Mean Value Theorem

A function is given that is continuous on every point of a closed interval,[a, b], and it is differentiable on every point of the open interval (a, b). If , then there exists at least one value (c) in the open interval,(a, b), where

.


𝑏−𝑎

a b

𝑓 (𝑐)=

𝑓 (𝑏 )− 𝑓(𝑎)

𝑏−𝑎

c

𝑓 (𝑎 )𝑠𝑙𝑜𝑝𝑒𝑡𝑎𝑛𝑔𝑒𝑛𝑡@𝑐= 𝑓 (𝑐 )

𝑓 (𝑐 )=𝑓 (𝑏 )− 𝑓 (𝑎)

𝑏−𝑎

𝑓 (𝑏)

4.2 – The Mean Value TheoremThe Mean Value Theorem

A function is given that is continuous on every point of a closed interval,[a, b], and it is differentiable on every point of the open interval (a, b). If , then there exists at least one value (c) in the open interval,(a, b), where

.

da bc

𝑓 (𝑎) 𝑠𝑙𝑜𝑝𝑒𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑎𝑏=𝑓 (𝑏 )− 𝑓 (𝑎)

𝑏−𝑎

𝑠𝑙𝑜𝑝𝑒𝑡𝑎𝑛𝑔𝑒𝑛𝑡@𝑐= 𝑓 (𝑐 )

𝑓 (𝑐 )=𝑓 (𝑏 )− 𝑓 (𝑎)

𝑏−𝑎

𝑠𝑙𝑜𝑝𝑒𝑡𝑎𝑛𝑔𝑒𝑛𝑡@𝑑= 𝑓 (𝑑)

𝑓 (𝑑 )=𝑓 (𝑏)− 𝑓 (𝑎)

𝑏−𝑎

𝑓 (𝑏)

4.2 – The Mean Value TheoremFind the values of x that satisfy the Mean Value Theorem:

√22√𝑥−1=2

√2√𝑥−1=1

√𝑥−1= 1

√2

𝑥−1=12

𝑥=32

Documents

4.1 – Extreme Values of Functions Extreme Values of a function are created when the function changes from increasing to decreasing or from decreasing to