Difference between revisions of "Math 22 Extrema and First Derivative Test"

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!Solution:  
 
!Solution:  
 
|-
 
|-
|Step 1: <math>f'(x)=2x+2=2(x+1)=0</math>,  
+
|'''Step 1''': <math>f'(x)=2x+2=2(x+1)=0</math>,  
 
|-
 
|-
|Step 2: Critical number is <math>x=-1</math>, so the test intervals are <math>(-\infty,-1)</math> and <math>(-1,\infty)</math>
+
|'''Step 2''': Critical number is <math>x=-1</math>, so the test intervals are <math>(-\infty,-1)</math> and <math>(-1,\infty)</math>
 
|-
 
|-
|Step 3: Choose <math>x=-2</math> for the interval <math>(-\infty,-1)</math>, and <math>x=0</math> for the interval <math>(-1,\infty)</math>.
+
|'''Step 3''': Choose <math>x=-2</math> for the interval <math>(-\infty,-1)</math>, and <math>x=0</math> for the interval <math>(-1,\infty)</math>.
 
|-
 
|-
 
|Then we have: <math>f'(-2)=-2<0</math> and <math>f'(0)=2>0</math>
 
|Then we have: <math>f'(-2)=-2<0</math> and <math>f'(0)=2>0</math>
 
|-
 
|-
|Step 4: <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
+
|'''Step 4''': By the first derivative test, <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
 
|-
 
|-
 
|Therefore, Relative minimum: <math>(1,3)</math>
 
|Therefore, Relative minimum: <math>(1,3)</math>
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|}
 
|}
  
'''2)''' <math>f(x)=6x^{\frac{2}{3}}+4x^2+3</math>
+
'''2)''' <math>f(x)=5x^3-10x^2+3</math>
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Solution: &nbsp;
 
!Solution: &nbsp;
 
|-
 
|-
|Step 1: <math>f'(x)=6\frac{2}{3}x^{(\frac{2}{3}-1)}+8x=4x^{\frac{-1}{3}}+8x=4x(x^{\frac{-4}{3}}+2)</math>,  
+
|'''Step 1''': <math>f'(x)=15x^2-20x=5x(3x-4)=0</math>,  
 
|-
 
|-
|Step 2: Critical number is <math>x=0</math>, so the test intervals are <math>(-\infty,-1)</math> and <math>(-1,\infty)</math>
+
|'''Step 2''': Critical number is <math>x=0</math> and <math>x=\frac{4}{3}</math>, so the test intervals are <math>(-\infty,0),(0,\frac{4}{3})</math> and <math>(\frac{4}{3},\infty)</math>
 
|-
 
|-
|Step 3: Choose <math>x=-2</math> for the interval <math>(-\infty,-1)</math>, and <math>x=0</math> for the interval <math>(-1,\infty)</math>.
+
|'''Step 3''': Choose <math>x=-1</math> for the interval <math>(-\infty,0)</math>, <math>x=1</math> for the interval <math>(0,\frac{4}{3})</math> and <math>x=2</math> for the interval <math>(\frac{4}{3},\infty)</math>.
 
|-
 
|-
|Then we have: <math>f'(-2)=-2<0</math> and <math>f'(0)=2>0</math>
+
|Then we have: <math>f'(-1)=35>0</math>, <math>f'(1)=-5<0</math> and <math>f'(2)=20>0</math>
 +
|-
 +
|'''Step 4''': By the first derivative test, <math>f'(x)</math> is positive to the left of <math>x=0</math> and negative to the right of <math>x=0</math>, then <math>f(0)=3</math> is a relative maximum,
 +
|-
 +
| and <math>f'(x)</math> is negative to the left of <math>x=\frac{4}{3}</math> and positive to the right of <math>x=\frac{4}{3}</math>, then <math>f(\frac{4}{3})=\frac{-79}{27}</math> is a relative minimum.
 
|-
 
|-
|Step 4: <math>f'(x)</math> is negative to the left of <math>x=-1</math> and positive to the right of <math>x=-1</math>, then <math>f(1)=3</math> is a relative minimum
+
|Therefore, Relative minimum: <math>(\frac{4}{3},\frac{-79}{27})</math> and Relative maximum: <math>(0,3)</math>
 
|}
 
|}
  
 
==Absolute Extrema==
 
==Absolute Extrema==
The page is under Construction
+
  Let <math>f</math> be defined on an interval <math>I</math> containing <math>c</math>.
 +
  1. <math>f(c)</math> is an absolute minimum of <math>f</math> on <math>I</math> when <math>f(c)\le f(x)</math> for every <math>x</math> in <math>I</math>
 +
  2. <math>f(c)</math> is an absolute maximum of <math>f</math> on <math>I</math> when <math>f(c)\ge f(x)</math> for every <math>x</math> in <math>I</math>
 +
 
 +
==Extreme Value Theorem==
 +
 
 +
  If <math>f</math> is continuous on a closed interval <math>[a,b]</math>, then <math>f</math> has both a minimum value and a maximum value on <math>[a,b]</math>.
 +
 
 +
==Guidelines for Finding Extrema on a Closed Interval==
 +
  To find the extrema of a continuous function <math>f</math> on a closed interval <math>[a,b]</math>, use the following steps.
 +
  1. Find all critical numbers of <math>f</math>
 +
  2. Evaluate <math>f</math> at each of its critical number
 +
  3. Evaluate <math>f</math> at each end point <math>a</math> and <math>b</math>
 +
  4. The least of these values is the absolute minimum, and the greatest is the maximum.
 +
 
 +
'''Exercises:''' Find all absolute extrema of the function below
 +
 
 +
'''1)''' <math>f(x)=5-2x^2</math> on <math>[-3,1]</math>
 +
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Solution: &nbsp;
 +
|-
 +
|'''Step 1''': <math>f'(x)=-4x=0</math>, So, critical number is <math>x=0</math>
 +
|-
 +
|'''Step 2''': <math>f(0)=5</math>
 +
|-
 +
|'''Step 3''': <math>f(-3)=-13</math> and <math>f(1)=3</math>
 +
|-
 +
|'''Step 4''': Absolute Maximum is <math>5</math> at <math>x=0</math>
 +
|-
 +
|and absolute minimum is <math>-13</math> at <math>x=-3</math>
 +
|}
  
 
[[Math_22| '''Return to Topics Page''']]
 
[[Math_22| '''Return to Topics Page''']]
  
 
'''This page were made by [[Contributors|Tri Phan]]'''
 
'''This page were made by [[Contributors|Tri Phan]]'''

Latest revision as of 09:17, 30 July 2020

Relative Extrema

 Let  be a function defined at .
 1.  is a relative maximum of  when there exists an interval  containing  such that  for all  in .
 2.  is a relative minimum of  when there exists an interval  containing  such that  for all  in .

If has a relative minimum or relative maximum at , then is a critical number of . That is, either or is undefined.

Relative extrema must occur at critical numbers as shown in picture below.

Relative extrema.png

The First-Derivative Test

 Let  be continuous on the interval  in which  is the only critical number, then
 
 On the interval , if  is negative to the left of  and positive to the right of , then  is a relative minimum.
 
 On the interval , if  is positive to the left of  and negative to the right of , then  is a relative maximum.

Guidelines for Finding Relative Extrema

 1. Find the derivative of 
 2. Find all critical numbers, then determine the test intervals
 3. Determine the sign of  at an arbitrary number in each test intervals
 4. Apply the first derivative test


Exercises: Find all relative extrema of the functions below

1)

Solution:  
Step 1: ,
Step 2: Critical number is , so the test intervals are and
Step 3: Choose for the interval , and for the interval .
Then we have: and
Step 4: By the first derivative test, is negative to the left of and positive to the right of , then is a relative minimum
Therefore, Relative minimum:
(Note: in this case is a parabola so our answer makes sense)

2)

Solution:  
Step 1: ,
Step 2: Critical number is 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 first derivative test, is positive to the left of and negative to the right of , then is a relative maximum,
and is negative to the left of and positive to the right of , then is a relative minimum.
Therefore, Relative minimum: and Relative maximum:

Absolute Extrema

 Let  be defined on an interval  containing .
 1.  is an absolute minimum of  on  when  for every  in 
 2.  is an absolute maximum of  on  when  for every  in 

Extreme Value Theorem

 If  is continuous on a closed interval , then  has both a minimum value and a maximum value on .

Guidelines for Finding Extrema on a Closed Interval

 To find the extrema of a continuous function  on a closed interval , use the following steps.
 1. Find all critical numbers of 
 2. Evaluate  at each of its critical number
 3. Evaluate  at each end point  and 
 4. The least of these values is the absolute minimum, and the greatest is the maximum.

Exercises: Find all absolute extrema of the function below

1) on

Solution:  
Step 1: , So, critical number is
Step 2:
Step 3: and
Step 4: Absolute Maximum is at
and absolute minimum is at

Return to Topics Page

This page were made by Tri Phan