View source for Math 22 Extrema and First Derivative Test

==Relative Extrema==
  Let <math>f</math> be a function defined at <math>c</math>.
  1. <math>f(c)</math> is a relative maximum of <math>f</math> when there exists an interval <math>(a,b)</math> containing <math>c</math> such that <math>f(x)\le f(c)</math> for all <math>x</math> in <math>(a,b)</math>.
  2. <math>f(c)</math> is a relative minimum of <math>f</math> when there exists an interval <math>(a,b)</math> containing <math>c</math> such that <math>f(x)\ge f(c)</math> for all <math>x</math> in <math>(a,b)</math>.

If <math>f</math> has a relative minimum or relative maximum at <math>x=c</math>, then <math>c</math> is a critical number of <math>f</math>. That is, either <math>f'(c)=0</math> or <math>f'(c)</math> is undefined.

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

[[File:Relative extrema.png]]

==The First-Derivative Test==

  Let <math>f</math> be continuous on the interval <math>(a,b)</math> in which <math>c</math> is the only critical number, then
  
  On the interval <math>(a,b)</math>, if <math>f'(x)</math> is negative to the left of <math>x=c</math> and positive to the right of <math>x=c</math>, then <math>f(c)</math> is a relative minimum.
  
  On the interval <math>(a,b)</math>, if <math>f'(x)</math> is positive to the left of <math>x=c</math> and negative to the right of <math>x=c</math>, then <math>f(c)</math> is a relative maximum.

==Guidelines for Finding Relative Extrema==
  1. Find the derivative of <math>f</math>
  2. Find all critical numbers, then determine the test intervals
  3. Determine the sign of <math>f'(x)</math> at an arbitrary number in each test intervals
  4. Apply the first derivative test


'''Exercises:''' Find all relative extrema of the functions below

'''1)''' <math>f(x)=x^2+2x</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: &nbsp;
|-
|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 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>
|-
|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>(1,3)</math>
|-
|(Note: <math>f(x)</math> in this case is a parabola so our answer makes sense)
|}

'''2)''' <math>f(x)=6x^{\frac{2}{3}}+4x^2+3</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!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 2: Critical number is <math>x=0</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>.
|-
|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
|}

==Absolute Extrema==
The page is under Construction

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'''This page were made by [[Contributors|Tri Phan]]'''