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Take it to the Extrema

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Take it to the Extrema Maximum and Minimum Value Problems By: Rakesh Biswas Decreasing and Increasing Functions In this study of graphs it is important that we take a ... – PowerPoint PPT presentation

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Title: Take it to the Extrema


1
Take it to the Extrema
  • Maximum and Minimum Value Problems
  • By Rakesh Biswas

2
Decreasing and Increasing Functions
  • In this study of graphs it is important that we
    take a very close look as the when function are
    increasing and decreasing.
  • Knowledge of when a function is increasing or
    decreasing will aid our understanding of what a
    graph may look like if we do not have a
    calculator.

3
Increasing Function
  • A function is said to be increasing if
  • f(x1)ltf(x2) if x1ltx2
  • This graph shows that if the above conditions are
    satisfied the function is increasing.

f(x)
f(x2)
f(x1)
x1
x2
4
Decreasing Function
  • A function is said to be decreasing if
  • f(x1)gtf(x2) if x1ltx2
  • This graph shows that if the above conditions are
    satisfied the function is decreasing.

f(x)
f(x1)
f(x2)
x1
x2
5
First Derivative Test
  • The First Derivative Test will tell you when the
    function is increasing, decreasing or constant.
  • First Derivative Tests can also be used to find
    the relative extremas.
  • It will also give you the critical points of the
    function.
  • Used in conjunction with the Second Derivative
    Test the graph of a function can be sketched.

6
Second Derivative Test
  • Second Derivative Tests can be used to find the
    concavity of a function.
  • It can also be used to find the inflection points
    of the function by setting the second derivative
    of the function of equal to zero and solving.
  • Once the concavity of the function is found it
    can be used along with the results of the First
    Derivative Test to sketch the function.

7
Using the First Derivative
  • f(x) x2-3x8
  • (a) Find when f is increasing and decreasing.
  • (b) Find where f has a relative minima and
    relative maxima.
  • Solution
  • f(x) 2x-3
  • 0 2x-3
  • X3/2
  • f(1) 2(1)-3 -1 (a) f(x) is
    increasing when xgt3/2
  • f(x) when xlt 3/2 is negative f(x) is
    decreasing when xlt3/2
  • f(2) 2(2)-31 (b) f(x) has a
    relative minimum at x3/2 because the
  • f(x) when xgt3/2 is positive the graph
    f(x)lt0 (negative) when xlt3/2 and

  • f(x)gt0 (positive) when xgt3/2. f(x) has no
    relative

  • extrema.

8
Using the Second Derivative
  • Find the intervals of concavity and the
    inflection points of g(x) x4-486x2
  • g(x) 4x3 972x
  • g(x) 12x2 -972
  • 0 12x2 -972
  • 0 12(x2 81)
  • X 9 and x-9
  • g(-10) 12(-10) 2 972 228
  • g(0) 12(0) 2 972 -972
  • g(10) 12(10) 2 972 228
  • f is concave up on and
  • f is concave down on (-9,9)
  • f has inflection points at x-9 and x9 because
    the concavity changes form concave up to concave
    down at x-9 and concave down to concave up at x9

f(x)
x9
x-9
9
Visualizing Graphs
  • The First and Second Derivative tests can be used
    to sketch the graphs of a function.
  • The First Derivative test tells us if the
    function is increasing or decreasing and the
    Second Derivative test tells us if the graph is
    concave up or down.
  • Both the First and Second Derivative tests can be
    used to find the inflection points of the of the
    function.

10
Graph Sketching
  • In each part sketch a continuous curve yf(x)
    with the stated properties.
  • f(2)4, f(2)0, f(x)lt0 for all x.
  • If the first derivative of a function is set
    equal to zero and the solved for x. The
    horizontal tangents can be acquired. In this
    problem when x2 the slope of the tangent is zero
    (f(2)0), and at x2 the value of y is 4.
  • The problem also states that the second
    derivative is less than zero for all values of x.
    Since f(x)lt0 graph of the function is concave
    down on the interval

(2,4)
11
More Graphs
  • In each part sketch a continuous curve yf(x)
    with the stated properties
  • f(2)4, f(2)0, f(x)gt0 for xlt2, f(x)lt0 for
    xgt2
  • This conditions state that at x2 there is a
    horizontal
  • tangent (f(2)0).
  • It is also stated that when xlt2 the function is
    concave up
  • f(x)gt0 and when xgt2 the function is concave
    down f(x)lt0.
  • There is also an inflection point at x2 because
    the concavity of the graph changes for concave up
    to concave down.

12
Finding Information form Derivative Graphs
  • The graph of the first derivative can be used to
    find crucial information about the function for
    the original graph.
  • For example, by using the graph of the first
    derivative function, we can find the relative
    extremas, points of inflection and concavity.

13
Using the First Derivative Graph
  • The figure below shows the graph of f, the
    derivative of f. The domain of f is -5,6
  • Find
  • relative extrema
  • concavity
  • all inflection points

f(x)
Note Problem is form second test.
14
Relative Extrema
  • The relative extrema occur when the graph goes
    form positive to negative or form negative to
    positive. In other words if the graph of the
    derivative function is below the x-axis the
    function has a negative slope in that interval.
    If the graph of the derivative function is above
    the x-axis the function has a positive slope in
    that interval. This means that when a graph goes
    for positive to negative, there is a relative
    maximum and when that graph goes form negative to
    positive there is a relative minimum. At x - 4
    the graph goes from negative to positive which
    means that there is a relative minimum there. At
    x0 there is a relative maximum because graph
    goes form positive to negative. And finally, at
    x5 there exists another relative minimum because
    the graph goes form negative to positive.
  • Relative minima at x-4 and x5
  • Relative maximum at x0

15
Concavity
  • (b) We can find the concavity of the function by
    finding the intervals on which the slope of the
    derivative function is positive or negative. If
    the slope is increasing, the function is concave
    up and if the slope is decreasing the function is
    concave down. Since the slope of the derivative
    function is increasing form
    and the function is concave up. Since
    the derivative function has a negative slope in
    the interval (-2,2) the function is concave down
    on this interval.
  • f is concave up on and
  • f is concave down on (-2,2)

16
Inflection Points
  • (c) The horizontal tangent(s) of the derivative
    function is the inflection point. Another way to
    find the inflection points is to find the
    concavity of the function. When the concavity of
    the function changes form positive to negative or
    vice versa, an inflection point exists at that
    point.
  • f has inflection points at x-2 and x-2 because
    there are horizontal tangents at these points.

17
Sketch of Original Function
  • f(x)

18
References
  • Slide 2 Definition 5.1.1
  • Slide 3 Definition 5.1.1
  • Slide 4 Problem is form test two

19
The End
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