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
4.1 – Extreme Values of Functions
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)
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 bd
𝑓 (𝑐 )𝑓 (𝑏)
𝑓 (𝑑)
Absolute maximum value: f(c)
Absolute minimum value: f(d)
c
𝑓 (𝑎)
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.
𝑓 (𝑐 ):𝐷𝑁𝐸
Absolute maximum value: none
Absolute minimum value: f(d)
a bd
𝑓 (𝑏)
𝑓 (𝑑)c
𝑓 (𝑎)
F is not continuous at c.
Theorem does not apply.
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.
𝑓 (𝑐 )
Absolute maximum value: f(c)
Absolute minimum value: f(d)
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)
4.1 – Extreme Values of Functions
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
4.1 – Extreme Values of Functions
-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.
4.1 – Extreme Values of Functions
-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
𝑠𝑙𝑜𝑝𝑒𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑎𝑏=𝑓 (𝑏 )− 𝑓 (𝑎)
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
.
𝑠𝑙𝑜𝑝𝑒𝑠𝑒𝑔𝑚𝑒𝑛𝑡 𝑎𝑏=𝑓 (𝑏 )− 𝑓 (𝑎)
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
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