applications of differentiation curve sketching. why do we need this? the analysis of graphs...

Applications of DifferentiationCurve Sketching

Why do we need this?The analysis of graphs involves looking at

“interesting” points and intervals and at horizontal and vertical asymptotes. We use calculus techniques to help us find all of the important aspects of the graph of a function so that we don’t have to plot a large number of points.

Curve Sketching

Graphing Skills AKA skills you will have when this chapter has ended

Domain and Range Symmetry x and y intercepts Asymptotes Extrema Inflection Points

Curve Sketching

4

Extrema on an IntervalExtrema of a Function ● Relative Extrema and Critical Numbers ● Finding Extrema on a Closed Interval

Curve Sketching

Global Extrema Example

Curve Sketching

Lets look at the following function:

How many horizontal tangent lines does this curve have?

What is the slope of a horizontal tangent?

Curve Sketching

Finding Horizontal Tangents

Differentiate:

Set the derivative equal to zero (slope =0):

Solve for x:

This is the x coordinate where you have a horizontal tangent on your graph.

3f x x x

23 1f x x

23 1 0f x x

2 13 1 0

3x x

Curve Sketching


Differentiate:


Solve for x:


4 33 4f x x x

3 212 12f x x x

3 212 12 0f x x x

3 2 212 12 0 12 1 0 0,1x x x x x

Curve Sketching


Differentiate:


Solve for x:


2

3

9 3xf x

x

3 2 2 2

6 4

18 9 3 3 9 81x x x x xf x

x x

2

4

9 810

xf x

x

22

4

9 810 9 81 0 3

xx x

x

Curve Sketching

Vertical Tangents Keep in mind where the derivative is

defined:

When x = 0 we have slope that is undefined

This is where we have a vertical tangent line.

This counts as a critical number as well and should be considered as such.

2

4

9 81xf x

x

Curve Sketching

Definition of ExtremaLet f be defined on an interval I containing c.

1. f(c) is the maximum of f on I if f(c) > f(x)

for all x on I.2. f(c) is the minimum of f on I if f(c) < f(x)

for all x on I.The maximum and minimum of a function on an

interval are the extreme values, or extrema, of the function on the interval.

The maximum and minimum of a function on an interval are also called the absolute maximum and absolute minimum on the interval, respectfully.

Curve Sketching

Definition of a Critical Number

Critical numbers are numbers you check to locate any extrema.Let f be defined at c.

If f ‘(c) = 0 or if f is undefined at c, then c is a critical number

Curve Sketching

Finding Extrema on a Closed Interval Step 1: find the critical numbers of f in

(a,b) Step 2: Evaluate f at each critical

number in (a,b) Step 3: Evaluate f at each endpoint of

[a,b] Step4: The least of the values is the

minimum, the greatest is the maximum.

Curve Sketching

Theorem: Relative Extrema Occur only at Critical Numbers

If f has a relative minimum or relative maximum at

x = c, then c is a critical number of f.

These are also known as locations of horizontal tangents

Curve Sketching

Critical NumbersLet f be defined at c. if f’(c)=0

or if f is not differentiable at c, then c is a critical number of f.

Relative extrema (max or min) occur only at a critical number.

Curve Sketching

Relative Extrema

Curve Sketching

Additional Extrema on a Closed Interval-examples:

Curve Sketching

Locate the Absolute Extrema

undefined. is or where 0any Find

5] [3, interval on the 2

Given

cfcft

txf

undefined. is or where 0 wherelocated are numbers critical

2

22

ff

ttf

.any for 0 not does xf

.2at undefined is xf

extrema.an oflocation thebecannot it so 5] [3, interval in thenot is 2

.3

55 and 33 interval. theof endpoints Check the ff

5] [3, interval on the minimum a is 5/3) (5, and maximum a is 3) (3,

19

Finding Extrema on a Closed Interval

Left Endpoint

Critical Number

Critical Number

Right Endpoint

f(-1)=7 f(0)=0 f(1)=-1Minimum

f(2)=16Maximum

Find the extrema:

Differentiate the function:

Find all values where the derivative is zero or UND:

2]. 1,[ interval on the 43 34 xxxf

23 1212 xxxf

01212 23 xxxf

0112 xx

1,0 when 0 xxfGiven endpoints, consider them too!!!

Curve Sketching

Homework for 3.1 Page 169-171#1-46 #61-64 are physics problems #65-70 are good thinking problems

Curve Sketching

First Derivative Test (Ch.3 S.3) Finding the critical numbers:

C is called a critical number for f if f’(c)=0 (horizontal tangent line) or f’(c) does not exist

Curve Sketching

Increasing or Decreasing functions A function f is called increasing on a

given interval (open or closed) if for x1 any and x2 in the interval, such that x2> x1, f(x2)>f(x1)

A function f is called decreasing on a given interval (open or closed) if for x1 any and x2 in the interval, such that x2> x1, f(x2)<f(x1)

Curve Sketching

Maximums, Minimums, Increasing, Decreasing Let the graph represent some function f.

1

2

3

1

1

2

2

2

3

a b d e fc

inimumxfef

xfdf

inimumxfcf

caxfbf

mfd, interval on the

maximumec, interval on the

mdb, interval on the

maximum, interval on the

decreasing is 0 .3

maxor min a has 0 .2

increasing is 0 .1

fxf

fxf

fxf

Curve Sketching

Copyright © Houghton Mifflin Company. All rights

reserved.

3-24

Definitions of Increasing and Decreasing Functions and Figure 3.15


reserved.

3-25

Theorem 3.6 The First Derivative Test

Finding Intervals Where the Function is Increasing/Decreasing

A. 4 3( ) 2 1f x x x B. 2

8 16( )

xf x

x

C. 2 3( ) 2 3f x x x

Curve Sketching

Increasing and Decreasing Example

1. In each case, sketch a graph of a continuous function with the given properties. -- + -- A. ( 1) 0f and (3) 0f ( )f x | | -1 3

+ -- -- B. (1) 0g and (4)g is undefined ( )g x | | 1 4

+ -- + -- C. ( 2) 0h and (2) 0h ( )h x | | | (0)h is undefined -2 0 2

Curve Sketching

LOCAL (RELATIVE) EXTREME VALUES

If x = c is not at an endpoint, then f (c) is a local (relative) maximum value for f(x) if f(c) > or = f(x) for all x-values in a small open interval around x = c.

If x = c is not at an endpoint, then f (c) is a local (relative) minimum value for f(x) if f(c) < or = f(x) for all x-values in a small open interval around x = c.

Curve Sketching

Finding local extrema

A. 4 3( ) 2 1f x x x B. 2

8 16( )

xf x

x

C. 2 3( ) 2 3f x x x

Curve Sketching

Local Extrema Example See text for additional examples

Curve Sketching

Homework for 3.3

pg. 186 #9-38, 43-50, 65-84 (good for AP)

Curve Sketching

3.4 Concavity and the Second Derivative Test

Second Derivative Test


reserved.

3-35

Definition of Concavity and Figure 3.24

The graph of f is called

concave up

on a given interval if for any two

points on the interval, the graph

of f lies below the chord that

connects these points.

Concavity

Concavity

The graph of f is called

concave down

on a given interval if for any two

points on the interval, the graph

of f lies above the chord that

connects these points.


reserved.

3-38

Theorem 3.7 Test for Concavity


reserved.

3-39

Definition of Point of Inflection and Figure 3.28


reserved.

3-40

Theorem 3.8 Points of Inflection


reserved.

3-41

Theorem 3.9 Second Derivative Test and Figure 3.31

Second Derivative Test

Finding the points of inflection:

C is called a point of interest for f if

f’’(c)=0 or f’(c) does not exist.

NOTE: The fact that f’’(x) = 0 alone does not

mean that the graph of f has an

inflection point at x.

We must find the points where the concavity

changes.

Points of Inflection

A point of inflection is a point

where the concavity of the graph

changes from concave up to

down or vice-versa.

Concavity and the Second Derivative Test

A. 4 3( ) 2 1f x x x B. 2

8 16( )

xf x

x

C. 2 3( ) 2 3f x x x

Curve Sketching

Homework for 3.4 P. 195 #1-52, 61-68 (good for AP prep)

Curve Sketching

Limits at Infinity (3.5)

Curve Sketching

What happens at x = 1?

What happens near x = 1?

As x approaches 1, g increases without bound, or g approaches infinity.

As x increases without bound, g approaches 0.

As x approaches infinity g approaches 0.

2

1

1g x

x

asymptotes

2

1

1g x

x

x x-1 1/(x-1)^2

0.9 -0.1 100.00

0.91 -0.09 123.46

0.92 -0.08 156.25

0.93 -0.07 204.08

0.94 -0.06 277.78

0.95 -0.05 400.00

0.96 -0.04 625.00

0.97 -0.03 1,111.11

0.98 -0.02 2,500.00

0.99 -0.01 10,000.00

1 0 Undefined

1.01 0.01 10,000.00

1.02 0.02 2,500.00

1.03 0.03 1,111.11

1.04 0.04 625.00

1.05 0.05 400.00

1.06 0.06 277.78

1.07 0.07 204.08

1.08 0.08 156.25

1.09 0.09 123.46

1.10 0.1 100.00

vertical asymptotes

2

1

1g x

x

x x-1 1/(x-1)^2

1 0 Undefined

2 1 1

5 4 0.25

10 9 0.01234567901234570

50 49 0.00041649312786339

100 99 0.00010203040506071

500 499 0.00000401604812832

1,000 999 0.00000100200300401

10,000 9999 0.00000001000200030

100,000 99999 0.00000000010000200

1,000,000 999999 0.00000000000100000

10,000,000 9999999 0.00000000000001000

100,000,000 99999999 0.00000000000000010

horizontal asymptotes

AKA Limits at infinity We know that

And we know that

So we can expand on this…


10lim

x x

2

10lim

x x

Theorem: If r is a positive rational number and c is

any real number, then

Furthermore, if is defined when x<0, then


0lim rx

c

x

0lim rx

c

x

rx

Theorem: The line y=L is a horizontal

asymptote of the graph of f if

or

Limits at Infinity

limx

f x L

limx

f x L

Examples: Finding a limit at infinity

Limits at Infinity

2

2lim 5x x

2

2lim 5 limx x x

5 0 5

Finding a limit at infinity

Direct substitution yields indeterminate form! Now we need to do algebra to evaluate the limit:

Divide each term by x:

Simplify:

Evaluate(keep in mind each 1/x0)

Limits at Infinity

2 1lim

1x

x

x

2 1

lim1x

x

x xx

x x

12

lim1

1x

x

x

2 0 2lim 2

1 0 1x

A comparison (AKA the shortcut!)

Direct substitution yields infinity/infinity! Do algebra:

Notice a pattern?????

Limits at Infinity

2

2 5lim

3 1x

x

x

2

2

2 5lim

3 1x

x

x

3

2

2 5lim

3 1x

x

x

2 2

2

2 2

2 5

lim3 1x

xx xxx x

2

2

2 5

lim1

3x

x x

x

00

3

2

2 2

2

2 2

2 5

lim3 1x

x

x xx

x x

3

3 3

2

3 3

2 5

lim3 1x

x

x xx

x x

2

2

52

lim1

3x

x

x

2

3

2!!

0undefined

3

3

52

lim3 1x

x

x x

A comparison (AKA the shortcut!)

Direct substitution yields infinity/infinity! Do algebra:

Bottom Heavy Balanced Top Heavy

Limits at Infinity

2

2 5lim

3 1x

x

x

2

2

2 5lim

3 1x

x

x

3

2

2 5lim

3 1x

x

x

02

3 !!undefined

Limits with 2 horizontal asymptotes

Note: for

Note: for

Homework Page 205 #2-38E

Curve Sketching

Curve Sketching 3.6 The following slides combine all of our

new graphing and analysis skills with our precalculus skills

Example:

Curve Sketching

3 2( ) 6 9 1f x x x x

Asymptotes:

Polynomial functions do not have asymptotes.

a) Vertical: No vertical asymptotes because f(x) is continuous for all x.

b) Horizontal: No horizontal asymptotes because f(x) is unbounded as x goes to positive or negative infinity .

3 2( ) 6 9 1f x x x x

Curve Sketching

Intercepts:

a) y-intercepts: f(0)=1 y-intercept: (0,1)

b) x-intercepts: difficult to find – use TI-89

Curve Sketching

3 2( ) 6 9 1f x x x x

Critical numbers:

a) Take the first derivative: b) Set it equal to zero:

c) Solve for x:

x=1, x=3

Curve Sketching

3 2( ) 6 9 1f x x x x

Critical Points:

a)Critical numbersx=1 and x=3

b) Find corresponding values of y:

c) Critical points: (1,5) and (3,1)

Curve Sketching

3 2( ) 6 9 1f x x x x

Increasing/Decreasing:

a) Take the first derivative: b) Set it equal to zero:

c) Solve for x:

x=1, x=3

Curve Sketching

3 2( ) 6 9 1f x x x x

d) Where is f(x)undefined? Nowheree) Sign analysis: Plot the numbers found above on a number line.

Choose test values for each interval created and evaluate the first derivative:

Increasing/Decreasing:

2'(0) 3(0) 12(0) 9 9f Positive ( )f x is increasing on ( ,1) .

2'(2) 3(2) 12(2) 9 3f Negative ( )f x is decreasing on (1,3) .

2'(4) 3(4) 12(4) 9 9f Positive ( )f x is increasing on (3, ) .

f '(x)1 3

+ _ +

3 2( ) 6 9 1f x x x x

Extrema

f’(x) changes from positive to negative at x=1 and from negative to positive at x=3 so and are local extrema of f(x).

Note: Values of corresponding to local extrema of must: Be critical values of the first derivative – values at which

equals zero or is undefined, Lie in the domain of the function, and Be values at which the sign of the first derivative

changes.

Curve Sketching

3 2( ) 6 9 1f x x x x

f(x) is increasing before x=1 and decreasing after x=1.

So (1,5) is a maximum.

f(x) is decreasing before x=3 and increasing after x=3.

So (3,1) is a minimum.

First Derivative Test

Curve Sketching

3 2( ) 6 9 1f x x x x

SECOND DERIVATIVE TEST

Alternate method to find relative max/min:a) Take the second derivative:

b) Substitute x-coordinates of extrema: (negative local max)(positive local min)

c) Label your point(s): local max: (1,5) local min: (3,1)

Curve Sketching

3 2( ) 6 9 1f x x x x

Concave Up/Down

6. Concave up/Concave down: a) Set ( )f x equal to zero: 6 12 0x b) Solve for x : 2x c) Where is ( )f x undefined? Nowhere d) Sign analysis: Plot the numbers found above on a number line. Choose test values for each interval created and evaluate the second derivative.

''(1) 6(1) 12 6 0f ( )f x is concave down on ( ,2) . ''(3) 6(3) 12 6 0f ( )f x is concave up on (2, ) .

f ''(x)

+

2

_

Curve Sketching

3 2( ) 6 9 1f x x x x

Inflection Points

7. Find any inflection points: a) For which values of x (found in 6) is ( )f x defined? 2x b) Find corresponding value of y : 3 2(2) (2) 6(2) 9(2) 1 3f c) ( )f x changes from concave up to concave down at 2x , so (2,3) is an inflection point. Inflection point: (2,3)

Note: Values of x corresponding to inflection points of f must: • be critical values of the second derivative – values at which ( )f x equals zero or is undefined, • lie in the domain of f , and

• be values at which the sign of the second derivative changes

Curve Sketching

3 2( ) 6 9 1f x x x x

Create a table of the values obtained:

Curve Sketching

x 0 1 2 3

f(x)

f’(x)

f’’(x)

9. Plot all points on the coordinate plane and sketch in the rest of the graph. Be sure to include all maximum points, minimum points, and inflection points:

_ _ _ _ _ _

5 4 2 1 -1

| | | | | | | | | -3 -2 -1 1 2 3 4 5 6

f(x) = x - 6x + 9x + 13 2

Curve Sketching

2 2 4( )

2

x xf x

x

Graphing Functions

Domain and Range Domain: evaluate where the function is

defined and undefined, values of discontinuity (under radicals, holes, etc.), nondifferentiability (sharp turns, vertical tangent lines and points of discontinuity).

Range: evaluate the values of y after graphing. Look for minimum and maximum values and discontinuities vertically.

Domain and Range

Finding the Range of a FunctionWe will look at this AFTER we graph the

equation using calculus, you can graph it on the calculator now if you like.

Curve Sketching

SymmetryAbout the y-axis:

Replace x with –x and you get the same equation after you work the algebra

About the x-axis:Replace y with –y and you get the same equation after you work the algebra

About the origin:Replace x with –x and y with –y and get the same equation after you work the algebra

Curve Sketching

Graphing Functions

2

2

2

2 2

2 4( )

2

2 4( )

2

2 4

2

2 4 2 4? ?

2 2

x xf x

x

x xf x

x

x x

x

x x x x

x x

Do we have symmetry about the y-axis?

replace x with –x and we get:

Are these equivalent equations?

NO! So we do not have symmetry about the y-axis.

Curve Sketching

Graphing Functions2

2

2

2 2

2 4

2

2 4

2

2 4

2

2 4 2 4? ?

2 2

x xy

x

x xy

x

x xy

x

x x x x

x x

Do we have symmetry about the x-axis?

replace y with –y and we get:


NO! So we do not have symmetry about the x-axis.

Curve Sketching

Graphing Functions

2

2

2

2

2 2

2 4

2

2 4

2

2 4

2

2 4

2

2 4 2 4? ?

2 2

x xy

x

x xy

x

x xy

x

x xy

x

x x x x

x x

Do we have symmetry about the origin?

replace x with –x and y with –y and we get:


NO! So we do not have symmetry about the origin.

Curve Sketching

x and y-intercepts x-intercepts: Set top of the fraction

equal to 0 in the original equation and solve for x values.

y-intercepts: Substitute 0 in for x and solve for y values.

x-intercepts

2

2

2

2

2

2 4( )

2

2 40

2

0 2 4

4

2

2 2 4 1 4

2 1

2 4 16

2 1

???

x xf x

x

x x

x

x x

b b acx

a

x

x

Replace y with 0 and we get ???

so do we cross the x-axis?

y-intercepts

2

2

2 4( )

2

0 2 0 4(0)

0 2

42

2

x xf x

x

f

Replace x with 0 and we get y = -2

so we cross the y-axis at (0,-2)

Curve Sketching

Horizontal AsymptotesThe line y=L is called a horizontal asymptote for f if the limit as x approaches infinity from both the left and/or right is equal to L

In other words, find the original equation as x approaches positive or negative infinity.

Limits at Infinity

Curve Sketching

Horizontal Asymptotes2

2 2

2

2

2

2 4( )

2

( ) ?

12 4

12

2 41

1 12

1

0

x xf x

xx

f x

x x xx

x

x x

x x

As x goes to infinity what happens to our equation?

It goes to infinity as well. So we have no horizontal asymptotes

Curve Sketching

Vertical Asymptotes The line x=c is called a vertical asymptote

for f at x=c if the limit goes to positive or negative infinity as x approaches c from the positive and/or negative side.

This is the complicated way of saying set the denominator (if there is one) to zero and find the x values.

Infinite Limits

Curve Sketching

Vertical Asymptotes

2 2 4( )

22 0

2

x xf x

xx

x

We set the denominator to

equal zero

and find that we have an asymptote

at x = 2

Curve Sketching

Slant AsymptotesGiven a function that consists of a

composition of functions:

If there are no common factors in the numerator and denominator, perform long division. The NONfractional part gives you the equation of the slant asymptotes.

Curve Sketching

Graphing Functions2

2

2 4( )

24

22 2 4

x xf x

x

xx

x x x

Performing long division:

we get a non-fractional part of

the answer

of y = x for our slanted asymptote

Curve Sketching

First Derivative Test

212

1 12 2

1 22

2

2

2 2

2 2

22

2

2 4( ) 2 4 2

2

( ) 2 4 2 2 4 2

2 2 2 2 4 1 2

2 2 2 4

2 2

2 2 2 2 4 4

2 2

40 4 0

2

( 4)

x xf x x x x

xd d

f x x x x x x xdx dx

x x x x x

x x x

x x

x x x x x x

x x

x xx x

x

x x

0

0, 4x x

Take the derivative of the function,

set it equal to 0 and find the corresponding values of x.

These are our CRITICAL POINTS

(Potential extrema)

Graphing Functions

2

2

2

2 2

4

2

4

2 2

4

3

2 4( )

2

4( )

2

2 4 2 4 2 2( )

2

2 4 2 4 2

2

2 6 8 2 8

2

2 80 2 8 0 4

2

x xf x

x

x xf x

x

x x x x xf x

x

x x x x

x

x x x x

x

xx x

x

Take the second derivative

Set it equal to 0

Solve for x

This is our potential point of inflection

Graphing Functions2 2 4

( )2

x xf x

x

Here we compile all of the information to enable us to give a pretty accurate graph WITHOUT technology

Organize the Informationx f(x) f’(x) f’’(x) Conclusion

-∞<x<0 + - Increasing, concave down

X=0 -2 0 - Relative Max

0<x<2 - - Decreasing, concave down

X=2 Undef Undef Undef Vertical asymptote

2<x<4 - + Decreasing, concave up

X=4 6 0 + Relative min

4<x<∞ + + Increasing, concave up

Homework: Page 215 #1-45, (67-71 are good thinking problems)

Curve Sketching

applications of differentiation curve sketching. why do we need this? the analysis of graphs...

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