Problem of the Day (Calculator allowed)

Let f be a function that is differentiable on the open interval (1, 10). If f(2) = -5, f(5) = 5, and f(9) = -5, which of the following must be true?

I. f has at least 2 zeroesII. The graph of f has at least one horizontal tangent.III. For some c, 2 < c < 5, f(c) = 3.A) NoneB) I onlyC) I and II onlyD) I and III onlyE) I, II, and III

Problem of the Day (Calculator allowed)

Let f be a function that is differentiable on the open interval (1, 10). If f(2) = -5, f(5) = 5, and f(9) = -5, which of the following must be true?

I. f has at least 2 zeroesII. The graph of f has at least one horizontal tangent.III. For some c, 2 < c < 5, f(c) = 3.A) NoneB) I onlyC) I and II onlyD) I and III onlyE) I, II, and III

I. 2 sign changes implies 2 zeroesII. Rolle's TheoremIII. Intermediate Value Theorem

•Discuss concavity as an indicator of function behavior •Recognize inflection as a change in the rate of change•Use the 2nd Derivative Test

3-4: Concavity & The Second Derivative TestObjectives

:

©2002 Roy L. Gover (www.mrgover.com)

DefinitionConcave up means the graph of f is above the tangent lines.

f

Concave down means the graph of f is below the tangent lines.

f

Definition

Definition

Concave up “holds water”

Definition

Concave down “spills water”

Analysis 31

( )3

f x x x 2'( ) 1f x x

Is there a relationship between the graphs of f(x) & f ’(x)?

Is there a relationship between the concavity of f(x) and f’(x)?

Where does concavity change?Is there a relationship between where concavity changes and f’(x)?

DefinitionGraph of f is concave up on interval I if f’ is increasing on I

Concave Up

f’(x) increasing

DefinitionGraph of f is concave down on interval I if f’ is decreasing on I

Concave Down

f’(x) decreasing

Review•f’ >0: slope of the tangent lines are positive; f is incr.•f’ <0: slope of the tangent lines are negative; f is decr.•f’ =0: slope of the tangent line is zero;f is neither increasing nor decreasing.

Important Ideaf ’’>0:slope

of tangent lines are becoming more positive (less negative) from left to right.

f(x)

f ‘(x)

f”(x)

Important Ideas

f ’’=0:slope of tangent lines are not changing.

f(x)

f ‘(x)

f”(x)

Important Ideas

f ’’<0:slope of tangent lines are becoming more negative (less positive) from left to right.

f(x)

f ‘(x)

f”(x)

Important IdeaLet f be a function such that f” exists on (a,b), then:

•f” (x)>0 for all x in (a,b) f is concave up.•f” (x)<0 for all x in (a,b) f is concave down.

Procedure

1. Locate x values at which f ’’=0 or undefined.2. Use these x values to determine intervals.3. Test the sign of f ’’ in each interval

Determining Concavity:

ExampleDetermine the open intervals on which

is concave up and concave down...

1( )

3

xf x

x

ExampleStep 1:Find the values of x where f ” =0 or undefinedStep 2:Make a table using intervals determined in step 1Step 3:Choose a value in each interval & evaluate the 2nd derivative at the value

Warm-UpDetermine the open intervals on which the graph of

is concave up and concave down.

2

6( )

3f x

x

Solution

IntervalTest Value

-2 0 2

Sign f ”(x)

+ - +

Concl. Up Down

Up

( , 1) ( 1,1) (1, )

Up

Down

Upf(x)2

2 3

36( 1)"( )

( 3)

xf x

x

Definition•Inflection point is the point where concavity changes.•Inflection points occur where f’’(x)=0 or is undefined but f ”(x)=0 or undefined doesn’t guarantee an inflection point.

Important Idea•An inflection point is the point where concavity changes.•An inflection point is where the rate of change changes from increasing to decreasing or vice versa.

4( )f x xTry This

Confirm that has a point of inflection at (0,0).

No inflection point at (0,0)

Definition2nd Derivative Test:Let f be a function such that f’(c)=0 and f” exists:•If f’’(c)>0, then f(c) is a local min•If f’’ (c)<0, then f(c) is a local max•If f” (c)=0, test fails

Procedure

1. Find critical numbers by setting f’(x)=0.

2.Find f’’(c) where c is a critical number.3. f ”(c)>0 local min; f ”(c)<0 local max.

Second Derivative Test:

ExampleFind the relative extrema (max and/or min) of:

using the second derivative test

4 3( ) 4f x x x

Try ThisFind the relative extrema (max and/or min) of:

using the second derivative test

5 3( ) 3 5f x x x

SolutionMax at (1,2)

Min at (-1,-2)

What did you do to determine there was no extrema at (0,0) since f ”(0)=0?

Lesson Close•How do you test for concavity?

•To test for local extrema, do you prefer the 1st derivative or 2nd derivative test? Why?

Assignment

195/1-5 odd, 21-35 odd