Difference between revisions of "Math 22 Concavity and the Second-Derivative Test"

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|'''Step 4''': By the concavity test, <math>f(x)</math> is concave up in <math>(-\infty,0)\cup (1,\infty)</math> and <math>f(x)</math> is concave down in <math>(0,1)</math>
 
|'''Step 4''': By the concavity test, <math>f(x)</math> is concave up in <math>(-\infty,0)\cup (1,\infty)</math> and <math>f(x)</math> is concave down in <math>(0,1)</math>
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==Points of Inflection==
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  If the graph of a continuous function has a tangent line at a point
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  where its concavity changes from upward to downward (or downward to upward),
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  then the point is a point of inflection.
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  If <math>(c,f(c))</math> is a point of inflection of the graph of <math>f</math>,
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  then either <math>f''(c)=0</math> or <math>f''(c)</math> is undefined.
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In '''exercises 1''', at <math>x=0</math>, the concavity changes from concave down to concave up, so <math>(0,f(0))</math> is a point of inflection.
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Therefore, <math>(0,0)</math> is a point of inflection
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 +
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In '''exercises 2''', at <math>x=0</math> and <math>x=1</math>, the concavity changes from concave up to concave down and from concave down to concave up, respectively. So, <math>(0,f(0))</math> and <math>(1,f(1))</math> are points of inflection.
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Therefore, <math>(0,10)</math> and <math>(1,9)</math> are points of inflection.
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==The Second-Derivative Test==
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  Let <math>f'(c)=0</math>, and let <math>f''(x)</math> exist on an open interval containing <math>c</math>,
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  1. If <math>f''(c)>0</math>, then <math>f(c)</math> is relative minimum.
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  2. If <math>f''(c)<0</math>, then <math>f(c)</math> is relative maximum.
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  3. If <math>f''(c)=0</math>, then the test fails. Use the first derivative test.
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'''Exercises:''' Find all relative extrema of
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'''1)''' <math>f(x)=2x^3+3x^2-5</math>
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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
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!Solution: &nbsp;
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|-
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|Notice, <math>f'(x)=6x^2-6x=6x(x-1)=0</math>, then critical numbers are <math>x=0</math> and <math>x=1</math>
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|-
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|<math>f''(x)=12x-6</math>, then <math>f''(0)=-6<0</math> and <math>f''(1)=6>0</math>
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|-
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|By the second derivative test, <math>f(0)=-5</math> is relative maximum and <math>f(1)=0</math> is relative minimum
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|-
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|Therefore, relative maximum: <math>(0,-5)</math> and relative minimum : <math>(1,0)</math>
 
|}
 
|}
  

Latest revision as of 08:29, 31 July 2020

Formal Definition of Concavity

 Let  be differentiable on an open interval . The graph of  is
 1. Concave upward on  when  is increasing on the interval.
 2. Concave downward on  when  is decreasing on the interval.

Test for Concavity

 Let  be a function whose second derivative exists on an open interval 
 1. If  for all  in , then the graph of  is concave upward on .
 2. If  for all  in , then the graph of  is concave downward on .

Guidelines for Applying the Concavity Test

 1. Locate the -values at which  or  is undefined.
 2. Use these -values to determine the test intervals.
 3. Determine the sign of  at an arbitrary number in each test intervals
 4. Apply the concavity test


Exercises: Find the second derivative of and discuss the concavity of its graph.

1)

Solution:  
Step 1: , so
Step 2: So , so the test intervals are and
Step 3: Choose for the interval , and for the interval .
Then we have: and
Step 4: By the concavity test, is concave up in and is concave down in

2)

Solution:  
Step 1: , so
Step 2: So, and , so the test intervals are and
Step 3: Choose for the interval , for the interval and for the interval .
Then we have: , and
Step 4: By the concavity test, is concave up in and is concave down in

Points of Inflection

 If the graph of a continuous function has a tangent line at a point 
 where its concavity changes from upward to downward (or downward to upward), 
 then the point is a point of inflection.
 
 If  is a point of inflection of the graph of , 
 then either  or  is undefined.

In exercises 1, at , the concavity changes from concave down to concave up, so is a point of inflection.

Therefore, is a point of inflection


In exercises 2, at and , the concavity changes from concave up to concave down and from concave down to concave up, respectively. So, and are points of inflection.

Therefore, and are points of inflection.

The Second-Derivative Test

  Let , and let  exist on an open interval containing ,
 1. If , then  is relative minimum.
 2. If , then  is relative maximum.
 3. If , then the test fails. Use the first derivative test.

Exercises: Find all relative extrema of

1)

Solution:  
Notice, , then critical numbers are and
, then and
By the second derivative test, is relative maximum and is relative minimum
Therefore, relative maximum: and relative minimum :

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