Section 4'3 Derivatives and the Shape of Curves

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Section 4.3Derivatives and the Shape of Curves

• Goals
• Apply the Mean Value Theorem to finding where
  functions are increasing and decreasing
• Discuss the
• first derivative test and
• second derivative test
• for local max/minima

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Mean Value Theorem

• This theorem is the key to connecting derivatives
  with the shape of curves

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Geometric Interpretation

• The next slide shows the points A(a, f(a)) and
  B(b, f(b)) on the graphs of two differentiable
  functions.
• The slope of the secant line AB is
• the same expression as in the Theorem.
• Also f??(c) is the slope of the tangent at (c,
  f(c)) .

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Geometric (contd)
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Geometric (contd)

• Thus Equation 1 of the Theorem says that there is
  at least one point P(c, f(c)) on the graph where
• the slope of the tangent line is the same as
• the slope of the secant line AB .
• In other words, there is a point P where the
  tangent line is parallel to the secant line AB .

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Example

• If an object moves in a straight line with
  position function s f(t) , then the average
  velocity between t a and t b is
• and the velocity at t c is f??(c) .
• Thus the Mean Value Theorem says that

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Example (contd)

• at some time t c between a and b the
• instantaneous velocity f??(c) is equal to the
• average velocity.
• For instance, if a car traveled 180 km in2 h,
  then the speedometer must have read 90 km/h at
  least once.

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Increasing/Decreasing Functions

• Earlier we observed from graphs that a function
  with a positive derivative is increasing.
• This fact can also be deduced from the Mean Value
  Theorem

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Example

• Find where the function
• f(x) 3x4 4x3 12x2 5
• is increasing and decreasing.
• Solution First we compute
• f??(x) 12x3 12x2 24x 12x(x 2)(x 1)
• To find where f??(x) gt 0 and where f??(x) lt 0 we
  arrange our work in a chart.

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Solution (contd)

• The chart shows the sign of
• each factor of f??(x) over
• each interval between the critical numbers 1,
  0, and 2
• The next slide shows the graph of f

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Solution (contd)
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First Derivative Test

• Recall that at a critical point a function can
  have a
• a local maximum,
• a local minimum, or
• neither.
• We can look at whether, and how, f?? changes sign
  at the critical point to decide which of the
  above possibilities is the case

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First Derivative Test (contd)

• The figures on the next two slides illustrate the
  First Derivative Test

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First Derivative Test (contd)
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First Derivative Test (contd)
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Concavity

• We recall the following definition of concavity
  from Chapter 2
• In the figure on the next slide, the
• slopes of the tangent lines increase from left to
  right on the interval (a, b) , and so
• f?? is increasing, and f is concave upward.

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Concavity (contd)

• Likewise, the
• slopes of the tangent lines decrease from left to
  right on (b, c) , and so
• f?? is decreasing, and f is concave downward

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Concavity (contd)

• A point where a curve changes its direction of
  concavity is called an inflection point.
• On the preceding graph, both P and Q are
  inflection points.
• The fact that f?? (f??)? leads to a test for
  concavity based on the second derivative

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Concavity (contd)

• Thus there is a point of inflection at any point
  where the second derivative changes sign.
• This leads to the following test for maximum and
  minimum values

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Concavity (contd)

• As the next slide illustrates, part (a) is true
  because f??(x) gt 0 near c , and so f is concave
  upward near c .
• This means the graph of f lies above its
  horizontal tangent at c , and so f has a local
  minimum at c

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Concavity (contd)
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Example

• Discuss y x4 4x3 with respect to
• concavity,
• points of inflection, and
• local maxima and minima.
• Solution If f(x) x4 4x3 , then
• f??(x) 4x3 12x2 4x2(x 3)
• and
• f??(x) 12x2 24x 12x(x 2) .

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Solution (contd)

• Setting f??(x) 0 gives x 0 and x 3 .
• To apply the Second Derivative Test we evaluate
  f?? at these critical numbers
• f??(0) 0 f??(3) 36 gt 0
• Thus
• f(3) 27 is a local minimum, whereas
• the Second Derivative Test gives no information
  about the critical number 0 .

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Solution (contd)

• However, since f??(x) lt 0 for
• x lt 0 and also for
• 0 lt x lt 3 ,
• the First Derivative Test says that f does not
  have a maximum or minimum at 0 .
• Next, since f??(x) 0 when x 0 or 2 , we
  complete the following chart

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Solution (contd)

• The points (0, 0) and (2, 16) are inflection
  points since the curve changes concavity at each
  of these points.
• On the next slide is a sketch of f

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Solution (contd)
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Example

• Sketch the graph of f(x) x2/3(6 x)1/3 .
• Solution Calculation gives
• Since f??(x) 0 when x 4 and f??(x) does not
  exist when x 0 or 6, the critical numbers are
  0, 4, and 6.

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Solution (contd)

• Here is a chart of the sign of f??
• By the First Derivative Test, f has
• a local minimum at x 0 with f(0) 0
• a local maximum at x 4 with f(4) 25/3
• neither at x 6 .

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Solution (contd)

• Studying f??(x) in a similar way shows
• f??(x) lt 0 for x gt 0 and for 0 lt x lt 6
• f??(x) gt 0 for x gt 6 .
• Therefore f is concave
• downward on (8, 0) and (0, 6) , and
• upward on (6, 8) , with one
• inflection point at (6, 0) .
• On the next slide is the graph of f

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Solution (contd)
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Review

• Mean Value Theorem
• Increasing/Decreasing Test
• First Derivative Test
• Concavity Test
• Second Derivative Test
• Homework
• 1,3,5,7,9,10,17-25 odd,29-31