1. Extreme Values of Functions

2. What you’ll learn… . Absolute (global) extreme values
Local (relative) extreme values
Finding extreme values
Inflection point and concavity
Graph a function
Why?
Finding maximum and minimum values of functions, called optimization, is an important issue in real-world problems .

3. The Extreme Value Theorem
If f is continuous on a closed interval [a,b], then f has both a maximum value and a minimum value on the interval.
2 conditions for f: continuous & closed interval
If either condition does not exist the E.V.T. does not apply.

4. First Derivative Test for Local Extrema

5. First Derivative Test for Local Extrema
The following test applies to a continuous function f(x).
At a critical point c:
If f ‘ changes signs from positive to negative at c, then f has a local maximum value at c.
If f ‘ changes signs from negative to positive at c, then f has a local minimum value at c.
If f ‘ does not change signs at c, then f has no local extreme value at c.

6. First Derivative Test for Local Extrema
The following test applies to a continuous function f(x).
At a left endpoint a:
If f ‘ < 0 for x > a, then f has a local maximum value at a.
If f ‘ > 0 for x > a, then f has a local minimum value at a.
At a right endpoint b:
If f ‘ < 0 for x < b, then f has a local minimum value at b.
If f ‘ > 0 for x < b, then f has a local maximum value at b.

7. Example Using the First Derivative Test1.
Since f is differentiable for all real numbers, the only critical points are the zeros of f ‘.
l l
inc.
dec.
inc.
Local Maximum at x = – 3
Local Minimum at x = 3
The range of f (x) is
No absolute extrema.

9. Concavity : Compare f(x) and g(x)
Both are increasing functions but they don’t look quite the same.

10. Concavity
Concavity Test

12. Example Determining Concavity2.

13. Inflection Points
An inflection point is a point on the graph where the concavity changes from upward to downward or downward to upward.
This means that if f ’’(x) exists in a neighborhood of an inflection point, then it must change sign at that point.
Theorem 1. If y = f (x) is continuous on (a,b) and has an inflection point at x = c, then either f ’’(c) = 0 or f ’’(c) does not exist.

14. Point of Inflection .
Points of Inflection:

15. Example 2
Find the inflection point(s) of f(x) = x3 – 9x2 +24x -10
F’(x) = 3x2 – 18x + 24
F’’(x) = 6x – 18 = 0
6(x-3) = 0
x = 3
Note: It’s important to do this test because the second derivative must change sign in order for the graph to have an inflection point. x 2 3 4
F’’ -- +
Concave down
Concave up
Therefore 3 is the infection point of f(x)

16. Example 3: A special case
Find the inflection point(s) of f(x) = x4
F’(x) = 4x3
F’’(x) = 12x2 = 0
x = 0 x -1 1
F’’ + +
Concave up
Concave up
Therefore 0 is not the inflection point of f(x)
There is no inflection point for this graph

17. Example 4
Find the inflection point(s) of f(x) = ln(x2 - 2x + 5)
x = -1 and x =3 x -2 -1 3 4
F’’ - + -
Concave up
Concave down
Concave down
Therefore there are two inflection points at x= -1 and x=3

18. Second Derivative Test for Local Extrema
Example Using the Second Derivative Test 4.
Local Maximum at:
Local Minimum at:

19. Learning about Functions from Derivatives

21. Use the following function to find each of the following.
Local min. at x = 3
Identify where the extrema of f occur.
Find the intervals on which f is increasing and decreasing.
Find where the graph of f is concave up or down.
Sketch a possible graph for f.
l l
Dec
Dec
Inc
l l Up
Down Up

23. With today technology, graphing calculator and computer can produce graphs. However, important points on a plot may be difficult to identify.
Therefore, it’s useful to learn how to sketch a graph by hand.

24. Curve Stretching
Analyze f(x). Find the domain and intercepts. (Set x=0, solve for f(x); set f(x)=0, solve for x).
Analyze f’(x): Find critical values. Determine increasing and decreasing intervals as well as local maximum and/or minimum. (set f’(x)=0).
Analyze f’’(x): Find inflection point. Determine the intervals on which the graph is concave upward and concave downward. (set f’’(x)=0).
Plot additional points as needed and sketch the graph.

25. Example Sketch f(x) = x4 + 4x3 by hand Step 1: Domain: (-∞,∞)
X: intercept: x4 + 4 x3 = 0
x3 (x+4) = 0, so x=0 or x = -4
Y: intercept: f(0) = 0
Step 2: f’(x) = 4x3 + 12x2 = 0
4x2 (x+3) = 0 so x= 0 or x=-3 both critical v.
Test numbers on the left and on the right of 0 and -3, we see that -3 is a
local minimum. Also, f(x) is decreasing on (- ∞, -3) and increasing on
(-3, ∞).
Step 3: f’’(x) = 12x2 + 24x = 0
12x(x+2) = 0 so x = 0 or x = -2
Test numbers on the left and on the right of -2 and 0, we see that both of
them are inflection points. Also, the graph is concave upward on (- ∞, -2),
concave downward on (-2,0), and concave upward on (0, ∞)

26. Continue: Sketch f(x) = x4 + 4x3
Note -4
x-int -3
-27
min -2
-16
Inflection point
x-int, y-int
(- ∞, -3)
decreasing
(-3, ∞)
increasing
(- ∞, -2)
concave up
(-2,0)
concave down
(0, ∞)

27. Example Sketch f(x) = 3x2/3 - x by hand Step 1: Domain: (-∞,∞)
X: intercept: 3x2/3 - x = 0
x (3x-1/3 - 1) = 0, so x=0 or 3x-1/3 – 1 = 0
x-1/ = 1/3, (x-1/3)-3 = (1/3)-3 , so x = 27
Y: intercept: f(0) = 0
Step 2: f’(x) = 2x-1/3 -1 = 0
x-1/ = 1/2, (x-1/3)-3 = (1/2)-3 , so x= 8
Also, f’(x) is discontinuous at 0.
Test numbers on the left and on the right of 0 and 8, we see that 0 is a min
and 8 is a max. Also, f(x) is decreasing on (- ∞, 0) and (8, ∞) and increasing
on (0,8)
Step 3: f’’(x) = (-2/3) x-4/3 = 0
x-4/3 = 0; so x = 0
F’’ is also discontinuous at 0. Test numbers on the left and on the right of 0,
we see that there is no inflection point. The graph is concave downward on
(- ∞, 0) and on (0, ∞)

28. Continue: Sketch f(x) = 3x2/3 - x
Note
X-int, y-int, min 8 4
max 27
X-int
May need to add more points on the left of 0 to have a better graph
(- ∞, 0)
Decreasing
Concave down
(0,8)
increasing
(8, ∞)
decreasing
(0, ∞)

29. Quick Quiz

30. Quick Quiz