AP Calculus AB
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07/20/22 Perkins AP Calculus AB Day 1 Section 3.1
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AP Calculus AB. Day 1 Section 3.1. Extreme Value Theorem (EVT). If f is continuous for all x in [ a,b ] then f must have both an absolute maximum and an absolute minimum value in the interval [ a,b ]. Examples:. Absolute Maximum. Abs Max. Abs Max. Abs Min. Abs Min. - PowerPoint PPT Presentation
Transcript of AP Calculus AB
04/19/23 Perkins
AP Calculus AB
Day 1Section 3.1
Extreme Value Theorem (EVT)
4
2
-2
-4
-5 5
If f is continuous for all x in [a,b] then f must have both an absolute maximum and an absolute minimum value in the interval [a,b].
a b
Absolute Maximum
4
2
-2
-4
-5 5
a b
4
2
-2
-4
-5 5
a b
Examples:
Absolute Minimum
Abs Max
Abs Min
Abs Max
Abs Min
Extrema – highest and lowest function values3 types
Absolute (See previous slide)
Relative -- occur only at critical numbers
(cannot be at endpoints of an interval)
‘Local’ -- compared to the points we can see
(can be at endpoints and critical numbers)
4
2
-2
-4
-5 5
a b
A/R/L4
2
-2
-4
-5 5
a b
4
2
-2
-4
-5 5
a b
A/R/L
A/L
A/R/L
A/L
Local
R/L
R/L
LocalLocal
A/L
when the derivative is either zero or undefined
Horizontal tangents
Vertical tangents
Extrema will only occur at endpoints or critical numbers!
Find the extrema of on [-1,3]. 232 3f x x x
13' 2 2f x x
f(x) is continuous on [-1,3]. Extrema occur at either endpoints or critical numbers.
13
22
x
13
13
2 2x
x
Horizontal tangents
Vertical tangents
130 2 2x 1
30 x131 x
1 x0 x
1f
3f 0f
1f
2
32 1 3 1 5
0.240
1
2
-2
-4
-6
Minimum
Maximum
Perkins
AP Calculus AB
Day 1Section 3.1
Extreme Value Theorem (EVT)
4
2
-2
-4
-5 5
If f is continuous for all x in [a,b] then f must have both an absolute maximum and an absolute minimum value in the interval [a,b].
a b
4
2
-2
-4
-5 5
a b
4
2
-2
-4
-5 5
a b
Examples:
Extrema3 types
Absolute
Relative
‘Local’
4
2
-2
-4
-5 5
a b
4
2
-2
-4
-5 5
a b
4
2
-2
-4
-5 5
a b
Find the extrema of on [-1,3]. 232 3f x x x
2
-2
-4
-6
