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5.2 Relative Extrema
• Find Relative Extrema of a function using the first derivative
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
Wesley
If a function f has a relative extreme value f (c) on an open interval; then c is a critical value. So,
f (c) = 0 or f (c) does not exist.
Relative extrema will only occur at points where the derivative is = 0 or where it is undefined.
Relative Extrema
Relative Minimum
Let I be the domain of f :f (c) is a relative minimum (bottom of a valley) if
there exists within I an open interval I1 containing c such that f (c) ≤ f (x) for all x in I1;
f has a relative minimum at c if f (x) < 0 on(a, c) and f (x) > 0 on (c, b). That is, f is decreasing to the left of c and increasing to the right of c.
If the graph is continuous (no break) at the point where the function changes from decreasing to increasing, that point is called a relative minimum point
Relative Maximum
Let I be the domain of f :F (c) is a relative maximum (top of a hill) if
there exists within I an open interval I2 containing c such that f (c) ≥ f (x) for all x in I2.
f has a relative maximum at c if f (x) > 0 on(a, c) and f (x) < 0 on (c, b). That is, f is increasing to the left of c and decreasing to the right of c.
If the graph is continuous (no break) at the point where the function changes from increasing to decreasing, that point is called a relative maximum point.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 7
Using First Derivatives to Find Maximum and Minimum Values
Graph over the interval (a, b)
f (c) Sign of f (x) for x in (a, c)
Sign of f (x) for x in (c, b)
Increasing or decreasing
Relative minimum
Relative maximum
–
+
+
–
Decreasing on (a, c];
increasing on [c, b)
Increasing on (a, c];
decreasing on [c, b)
c ba
c ba
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 8
Using First Derivatives to Find Maximum and Minimum Values
Graph over the interval (a, b)
f (c) Sign of f (x) for x in (a, c)
Sign of f (x) for x in (c, b)
Increasing or decreasing
No relative maxima
or minima
No relative maxima
or minima
–
+
–
+
Decreasing on (a, b)
Increasing on (a, b)
c ba
c ba
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 9
Using First Derivatives to Find Maximum and Minimum Values
Example 1: For the function f given by
find the relative extrema.
f (x) 2x3 3x2 12x 12.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 10
Using First Derivatives to Find Maximum and Minimum Values
Example 1 (continued): Find Derivative And set it = 0
These two critical values partition the number line into 3 intervals: A (– ∞, –1), B (–1, 2), and C (2, ∞).
CB A
2-1
6x2 6x 12 0
x2 x 2 0
(x 2)(x 1) 0
x 2 or x 1
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 11
Example 1 (continued):3rd analyze the sign of f (x) in each interval.
Test Value x = –2 x = 0 x = 4
Sign off (x)
+ – +
Resultf is increasing on (–∞, –1]
f is decreasing on [–1, 2]
f is increasing on [2, ∞)
Using First Derivatives to Find Maximum and Minimum Values
xInterval
CB A
2-1
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 12
Using First Derivatives to Find Maximum and Minimum Values
Example 1 (concluded):Therefore, by the First-Derivative Test,
f has a relative maximum at x = –1 given by
The relative maximum value is 19. It occurs where x = -1.
And f has a relative minimum at x = 2 given by
The relative minimum value is -8. It occurs where x is 2.
f ( 1) 2( 1)3 3( 1)2 12( 1)12 19
f (2) 2(2)3 3(2)2 12(2)12 8
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 13
Using First Derivatives to Find Maximum and Minimum Values
Example 1 (continued):
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 14
Using First Derivatives to Find Maximum and Minimum Values
Example 3: Find the relative extrema for the Function f (x) given by
Then sketch the graph.
1st find f (x).
f (x) (x 2)2 3 1
f (x) 2
3x 2
1 3
f (x) 2
3 x 23
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 15
Using First Derivatives to Find Maximum and Minimum Values
Example 3 (continued): 2nd find where f (x) does not exist or where f (x) = 0.
Note that f (x) does not exist where the denominator equals 0. Since the denominator equals 0 when x = 2, x = 2 is a critical value.
f (x) = 0 where the numerator equals 0. Since 2 ≠ 0, f (x) = 0 has no solution.
Thus, x = 2 is the only critical value.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 16
Using First Derivatives to Find Maximum and Minimum Values
Example 3 (continued): 3rd x = 2 partitions the number line into 2 intervals:
A (– ∞, 2) and B (2, ∞). So, analyze the signs of f (x) in both intervals.
Test Value x = 0 x = 3
Sign of f (x) – +
Result f is decreasing
on (– ∞, 2]f is increasing on
[2, ∞)
xInterval
B A
2
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 17
Using First Derivatives to Find Maximum and Minimum Values
Example 3 (continued):Therefore, by the First-Derivative Test,
f has a relative minimum at x = 2 given by
The relative minimum value is 1. It occurs at x = 2.
f (2) (2 2)2 3 1 1
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 18
Using First Derivatives to Find Maximum and Minimum Values
Example 3 (concluded):We use the information obtained to sketch the graph below, plotting other function values as needed.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 19
More Examples
3
2
.
( ) 1
( )4
4 21( )
2
Find the relativeextrema
f x x
xf x
x
x xf x
x
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-
WesleySlide 2.1- 20
C(q) above is the cost function. p(q) is the price function.
Finda)The number of units that will produce a maximum profit.b)The maximum profit.c)The price that will produce a maximum profit.
C( ) 25q 5000
p(q) 90 0.02q
q
a) 1625 b) $47812.50 c) $57.50