View source for Math 22 Concavity and the Second-Derivative Test

==Formal Definition of Concavity==
  Let <math>f</math> be differentiable on an open interval <math>I</math>. The graph of <math>f</math> is
  1. Concave upward on <math>I</math> when <math>f'(x)</math> is increasing on the interval.
  2. Concave downward on <math>I</math> when <math>f'(x)</math> is decreasing on the interval.

==Test for Concavity==
  Let <math>f</math> be a function whose second derivative exists on an open interval <math>I</math>
  1. If <math>f''(x)>0</math> for all <math>x</math> in <math>I</math>, then the graph of <math>f</math> is concave upward on <math>I</math>.
  2. If <math>f''(x)<0</math> for all <math>x</math> in <math>I</math>, then the graph of <math>f</math> is concave downward on <math>I</math>.

==Guidelines for Applying the Concavity Test==
  1. Locate the <math>x</math>-values at which <math>f''(x)=0</math> or <math>f''(x)</math> is undefined.
  2. Use these <math>x</math>-values to determine the test intervals.
  3. Determine the sign of <math>f'(x)</math> at an arbitrary number in each test intervals
  4. Apply the concavity test


'''Exercises:''' Find the second derivative of <math>f</math> and discuss the concavity of its graph.

'''1)''' <math>f(x)=x^3+2x^2</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: &nbsp;
|-
|'''Step 1''': <math>f'(x)=3x^2+4x</math>, so <math>f''(x)=6x=0</math>
|-
|'''Step 2''': So <math>x=0</math>, so the test intervals are <math>(-\infty,0)</math> and <math>(0,\infty)</math>
|-
|'''Step 3''': Choose <math>x=-1</math> for the interval <math>(-\infty,0)</math>, and <math>x=1</math> for the interval <math>(0,\infty)</math>.
|-
|Then we have: <math>f''(-1)=-6<0</math> and <math>f''(1)=6>0</math>
|-
|'''Step 4''': By the concavity test, <math>f(x)</math> is concave up in <math>(0,\infty)</math> and <math>f(x)</math> is concave down in <math>(-\infty,0)</math>
|}

'''2)''' <math>f(x)=x^4-2x^3+10</math>
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution: &nbsp;
|-
|'''Step 1''': <math>f'(x)=4x^3-6x^2</math>, so <math>f''(x)=12x^2-12x=12x(x-1)=0</math>
|-
|'''Step 2''': So, <math>x=0</math> and <math>x=1</math>, so the test intervals are <math>(-\infty,0),(0,1)</math> and <math>(1,\infty)</math>
|-
|'''Step 3''': Choose <math>x=-1</math> for the interval <math>(-\infty,0)</math>, <math>x=\frac{1}{2}</math> for the interval <math>(0,1)</math> and <math>x=2</math> for the interval <math>(1,\infty)</math>.
|-
|Then we have: <math>f''(-1)=24>0</math>, <math>f''(\frac{1}{2})=-3<0</math> and <math>f''(2)=24>0</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>
|}
==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.
[[Math_22| '''Return to Topics Page''']]

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