Download 3.2 - El Camino College

Sec 3.2 Extrema with 1st derivative Test
rev0216
Objective: Recognize the occurrence of relative extrema of functions.
Use the First Derivative Test to find the relatie extrema
Find absolute exrema of continuous functions on a closed interval
—————————————————————
First Derivative Test for relative Extrema on interval (a,b)
1. If f ′ x > 0 for all x < c and f ′ x < 0 for all x > c, then fc is the relative
maximum value of f. (cup down-∩)
f(x) is changing from ↑ to ⇂ function at c
2. If f ′ x < 0 for all x < c and f ′ x > 0 for all x > c, then fc is the relative minimum
value of f. (cup up-∪)
f(x) is changing from ⇂ to ↑ function at c
———————————————————–
.Example 1
Find the open intervals on which the function is increasing or decreasing
fx = 2x 3 − 3x 2 − 36x + 14
dfx
= 6x 2 − 6x − 36
dx
6x 2 − 6x − 36 = 0
6x + 2x − 3 = 0,
Solution is: −2, 3
the critical numbers
Make a table of the regions
−∞, −2
Test Intervals
Test number
′
Sign of f x = 6x + 2x − 3
Conclusion fx
−2, 3
3, ∞
-3
0
4
+
-
+
increasing decreasing increasing
Between the interval
there is a relative maximum (cup down-∩)
−∞, −2 and −2, 3
Between the interval
there is a relative maximum (cup up-∪)
−2, 3 and 3, ∞
and
f−2 = 58
f3 = −67
y
-4
50
-2
2
-50
4
x
-100
———————————————————Example 2 Finding Relative extrema
fx = x 4 − x 3
step 1:f ′ x = 4x 3 − 3x 2
step 2: Solve
f ′ x = 0 for critical numbers
1
x 2 4x − 3 = 0,
Solution is:
x = 0, 34
Define regions:
region1=−∞, 0
region2=0,
Region
1
2
3
Test no.
-1
1
2
1
1st factor x 2
+
+
+
2nd factor 4x − 3 -
-
+
product=f’(x)
-
+
f(x)
⇂
4
3
f = x − x 
y
⇂
↑
3
4

region3= 34 , ∞
 relative minimum at x= 34
2.0
1.5
1.0
0.5
-1
1
-0.5
2
x
Do NOW cp1
————————————————————————–
First Derivative Test for Absolute Extrema on interval [a,b]
Suppose that c is a critical number of a continuous function f defined on an interval.
Numbers 1 and 2 below are relative max. and min. in previous examples+
f()is evaluated
After finding the critical number by taking the 1st derivative.....
1. Evaluate the function f at each critical number c on a, b
2. Evaluate fa
and
fb the end points of [a,b]
3. Choose from among 1-2 to get absolute max (greatest value) and min (least
value)
——————————————————
Example 4: Find extrema on closed interval
Find the min and max values of fx = x 2 − 6x + 2
on 0, 5
—————————————————Find relative min and max
f ′ x = 2x − 6 = 0
x = 3 is critical no.
2
Test Intervals
0, 3
3, 5
Test number
1
4
-
+
′
Sign of f x = 2x − 3
Conclusion fx
decreasing increasing
x = 3 is relative minimum and f3 = − 7
On closed interval
x value
endpoint x=0
critical no.
endpoint x=5
fx
f0 = 2
f3 = − 7
f5 = − 3
Conclusion Maximum at x=2 Minimum at x=3 neith max or min
y
0
1
2
3
4
5
x
-5
———————————
Do NOW
checkpoint 4
————————————
Applications of the Extrema
—————————————————
Example 5-Finding Maximum Profit
Reference Sec 2.3 Examples 7 and 8
A fast food hamburger resturant has a
profit function
2
Px = 2. 44x − x
− 5000, 0 < x < 50, 000 hamburgers
20000
Find the sales level that yields maximum profit.
————————————————Solution:
1.Find critical numbers
dP = 2. 44 − 0. 000 1x = 0
dx
Solution is: x=24400. 0 units (hamburgers),
the sales level that yields maximum profit
2.Evaluate
24400 2
P24400 = 2. 44 ⋅ 24000 −
− 5000
20000
P24400 = $24768. , maximum profit
3
y 20000
10000
0
20000
40000
x
——————————————————Do NOW # 47 sec 3.2
Minimize cost: Cx = 3x + 20000
x
ans:
C82 = 20 086 = 489. 9
41
y 4000
2000
0
0
50
100
150
200
x
————————————————-
4