Transcript f (x)

§4.1 Graphing Using the First and
Second Derivatives.
The student will learn about
relative extreme points and
critical numbers, and
the first derivative test,
and concavity,
inflections points,
and the second
derivative test.
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Introduction
We are going to use derivatives in two
ways. First, in this section, we will look at
how the derivative helps us to graph a
function.
Then, in the next section, we will look at
how the derivative and graph help us to
optimize a function. That means we will find
the maximums and/or minimums of a
function.
Could be pretty useful stuff!!
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Relative Extreme Points and
Critical Numbers
f has a relative maximum at c if f (c) ≥ f (x) for
all x near c.
f has a relative minimum at c if f (c) ≤ f (x) for
all x near c.
The relative maximums and relative minimums
are called collectively relative extreme points.
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Critical Values
Definition. The values of x in the domain of f
where f ’(x) is 0 or does not exist are called
the critical values of f.
f′ (x) = 0
or
f′ (x) is undefined
A critical value for x tells us that something
important is happening.
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First – A Basic Fact about
Derivatives
For the interval (a, b)
f ' (x)
f (x)
Graph of f
+
Increasing
Rising
-
Decreasing
Falling
0
Flat
Steady
REMEMBER! The derivative describes the slope
and when the slope is positive the curve is rising
and when the slope is negative the curve is falling.
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Graphing Strategies for y = f (x).
Step 1: Analyze f (x): [Graphing calculator in blue.]
• Find the domain of f. Remember the two step
process, assume the domain is all real numbers and
delete division by 0 and the even roots of negative
numbers. (Check graph and table.)
• Find the x intercepts. Remember that this occurs
at y = 0. (Zeros or roots.)
• Find the y intercept. Remember that this occurs
at x = 0. (Value with x = 0.)
• Find the vertical asymptotes. (Check graph and table.)
This is algebra!
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Graphing Strategies for y = f (x).
Step 2: Analyze f ′ (x): [Graphing calculator in blue.]
Today's Lesson
• Find any critical values for f ′ (x).
• Construct a sign chart for f ′ (x) to determine the
intervals where f (x) is increasing and decreasing.
(Use the graph of f ′ (x) OR f (x).)
• Find local maxima and minima. (Use the graph
of f (x).)
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Graphing Strategies for y = f (x).
OR USE YOUR GRAPHING CALCULATOR.
• Make sure you know what you are looking at.
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Example 1
Find the intervals where f (x) = x2 + 6x + 7 is
rising and falling.
The function will be rising when the derivative is
positive.
WHY?
f ’ (x) = 2x + 6 so then 2x + 6 > 0
2x + 6 > 0 when 2x > - 6
or x > - 3
So the graph is rising when x > - 3.
- 3 is a critical
value.
f (x) will be falling when f ’ (x) = 2x + 6 < 0
OR 2x + 6 < 0 when 2x < - 6 or x < - 3
So the graph is falling when x < - 3.
Continued
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Example 1 continued
f (x) = x2 + 6x + 7
Sign charts of f ’ are helpful. f ’ (x) = 2x + 6
(- , -3)
f ’ (x)
f (x)
(-3, )
- - - - - - 0 + + + + + +
Decreasing
-3
- 3 is a critical
Increasing
value.
f (x) = x2 + 6x + 7
-3
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Example 2
f (x) = 1 + x3
f ’(x) = 3x2 Critical value at x = 0.
(- , 0)
f ’ (x)
f (x)
(0, )
+ + + + + 0 + + + + + +
Increasing
0
Increasing
f (x) = 1 + x3
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0
Example 3
f (x) = (1 – x)1/3
f ’(x) =
1
2
3 1  x 
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Critical value at x = 1.
f ’ (x)
(- , 1)
(1, )
- - - - - - ND - - - - - -
f (x)
Decreasing
1
Decreasing
f (x) = (1 – x)1/3
1
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Example 4
f (x) = 1/(1 – x).
f ’(x) = 1/(1 – x)2 Critical value at x = 1.
f ’ (x)
(- , 1)
(1, )
+ + + + + ND + + + + +
f (x)
Increasing
1
Increasing
f (x) = 1/(1 – x).
1
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First Derivative Test
Let c be a critical value of f (i.e. f (c) is defined and
either f ’ (c) = 0 or f ’ (c) is not defined). Construct a
sign chart for f ’ (x) close to and on either sign of c.
f (x) left of c
f (x) right of c
f (c)
Decreasing
Increasing
Local minimum at c
y’ is negative
Increasing
y’ is positive
Decreasing
y’ is positive
y’ is negative
Decreasing
Decreasing
y’ is negative
Increasing
y’ is positive
Local maximum at c
Not an extremum
y’ is negative
Increasing
y’ is positive
Not an extremum
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First Derivative Test
Graphing Calculators
Local extrema are easy to recognize on a
graphing calculator.
• Graph the derivative and use built-in root
approximations routines to find the critical values.
Use the zeros subroutine.
• Graph the function and use built-in routines that
approximate local maxima and minima. Use the
MAX or min subroutine.
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* * Example * *
Example. f (x) = x3 – 12x + 2.
f ‘ (x) = 3x2 – 12. Graph
this derivative and look
for critical values
f (x) = x3 – 12x + 2.
Graph this and look for
maxima and minima.
-10 ≤ x ≤ 10 and -15 ≤ y ≤ 20
-10 ≤ x ≤ 10 and -15 ≤ y ≤ 20
f ‘ (x) + + + + + 0 - - - 0 + + + + +
increases decrease
f (x)
increases f (x)
increases decrs increases
Critical values at –2 and 2
Maximum at - 2 and
minimum at 2.
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Summary.
• We can use the first derivative test to
determine where relative maxima or minima
occur.
• Use the first derivative to determine where
the critical values are located.
• Then examine the function on either side of
these critical values to find if there is a max or
min at the critical value.
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More
Let’s continue this study of using
derivatives to tell us about the original
function by examining what the second
derivative tells us.
Concavity.
The term concave upward is used to describe a
portion of a graph that opens upward. Concave
downward is used to describe a portion of a
graph that opens downward.
Concave down
Concave up
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More
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Summary of Facts
f is increasing
f ‘ (x) > 0
f is decreasing
f ‘ (x) < 0
f is constant
f ‘ (x) = 0
f is concave up
f ‘ (x) increasing
f “ (x) > 0
f concave down
f ‘ (x) decreasing
f “ (x) < 0
Inflection point
f ‘ (x) is constant
f “ (x) = 0
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Example 2
Find the inflection points of
f (x) = x3 + 3x2 - 9x + 5
f ‘ (x) = 3x2 + 6x – 9 = 3 ( x 2 + 2x – 3) = 3 (x + 3)(x – 1)
with critical values at x = - 3 and x = 1
f “ (x) = 6x + 6 with a critical value at x = - 1.
Sign charts for f ‘ (x) and f “ (x) may be useful.
Next slide please.
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What It All Means
f ‘ (x) = 3x2 + 6x – 9 = 3 (x + 3)(x – 1) with critical values at x = - 3 and x = 1
f “ (x) = 6x + 6 with a critical value at x = - 1.
f “ (x) - - - - - - -
0 + + + + + + + +
f ’ (x) + + + 0 - - - - - -3
f (x) increasing
f (x)
f (x)
-1
1
decreasing
Maximum
0 + + + + +
increasing
minimum
concave down - inflect - concave up
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What It All Means
f (x) increasing
f (x)
f (x)
-3
-1
decreasing
Maximum
1
increasing
minimum
concave down - inflect - concave up
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Note: It is difficult to find the
inflection point on a calculator.
You can graph y’ = 3x 2 + 6x - 9
and find where the max or min
are located.
You can graph y” = 6x + 6
and find the x-intercepts.
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Second Derivative Test
Let c be a critical value for f ' (x),
f ' (c)
f '' (c)
Graph of f is
f (c)
0
+
Concave up
Local
minimum
0
-
Concave down
Local
maximum
0
0
?
Test fails
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Summary.
• We can use the first derivative test to
determine where relative maxima or minima
occur.
• Use the first derivative to determine where
the critical values are located.
• Then examine the function on either side of
these critical values to find if there is a max or
min at the critical value.
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Summary.
• We can use the second derivative to
determine when a function is concave up or
concave down.
• When the second derivative is zero we get an
inflection point in f (x).
• The second derivative test may be used to
determine if a point is a local maxima or
minima.
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ASSIGNMENT
§4.1; Page 76; 1 - 9, odd.
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