Take it to the Extrema

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Take it to the Extrema

• Maximum and Minimum Value Problems
• By Rakesh Biswas

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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.

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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
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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
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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.

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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.

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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.

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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
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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.

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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)
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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.

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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.

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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.
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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

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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)

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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.

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Sketch of Original Function

• f(x)

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References

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

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The End