Transcript Concavity and Second Derivative Test Lesson 4.4 Concavity • Concave UP • Concave DOWN • Inflection point: Where concavity changes 

Concavity and Second
Derivative Test
Lesson 4.4
Concavity
• Concave UP
• Concave DOWN
• Inflection point:
Where concavity
changes

Inflection Point
• Consider the slope as curve changes
through concave up to concave down
At inflection point
slope reaches
maximum
positive value
Slope starts
negative

Slope
Becomes
Slopebecomes
becomes
zero
less negative (horizontal)
positive, then
more positive
After inflection
point, slope
becomes less
positive
Graph of the slope
Inflection Point
• What could you say about the slope function
when the original function has an inflection
point
• Slope function has a maximum (or minimum
• Thus second derivative = 0
Graph of the slope
Second Derivative
• This is really the rate of change of the slope
• When the original function has a relative
minimum
 Slope is increasing (left to right) and goes
through zero
 Second derivative is positive
 Original function is concave up
Second Derivative
• When the original function has a relative
maximum
 The slope is decreasing (left to right) and goes
through zero
 The second derivative is negative
 The original function is
concave down
View Geogebra
Demo
Second Derivative
• If the second derivative f ’’(x) = 0
 The slope is neither increasing nor decreasing
• If f ’’(x) = 0 at the same place f ’(x) = 0
 The 2nd derivative test fails
 You cannot tell what the function is doing
f ''( x)  12x
2
f ( x)  x
Not an inflection
point
4
Example
• Consider
f ( x)  x  3x  4
3
• Determine f ‘(x) and f ’’(x) and when they are
zero
f '( x)  3x  3  0 when x  1
2
f ''( x)  6 x  0 when x  0
Example
f ( x)  x  3x  4
3
f ’(x) = 0, f’’(x) > 0,
this is concave up, a
relative minimum
f ‘(x)
f(x)

f ‘(x) = 0, f ‘’(x) < 0
this is concave
down, a maximum
f ‘’(x)
f ‘’(x) = 0 this is an
inflection point
Example
• Try
f ( x)  x  1
2
• f ’(x) = ?
• f ’’(x) = ?
• Where are relative max, min, inflection point?
Algorithm for Curve Sketching
• Determine critical points
 Places where f ‘(x) = 0
• Plot these points on f(x)
• Use second derivative f’’(x) = 0
 Determine concavity, inflection points
• Use x = 0 (y intercept)
• Find f(x) = 0 (x intercepts)
• Sketch
Assignment
• Lesson 4.4
• Page 235
• Exercises 5, 12, 19, 26, 33,
40, 47, 54, 61, 68, 73, 83, 91,
95, 96