Applications of the Derivative

4• Applications of the First Derivative

• Applications of the Second Derivative

• Curve Sketching

• Optimization I

• Optimization II

Increasing/Decreasing

A function f is increasing on (a, b) if f (x1) < f (x2) whenever x1 < x2.

A function f is decreasing on (a, b) if f (x1) > f (x2) whenever x1 < x2.

Increasing IncreasingDecreasing

Increasing/Decreasing/Constant

.,on increasing is then

,, intervalan in of each valuefor 0 If

baf

baxxf

.,on decreasing is then

,, intervalan in of each valuefor 0 If

baf

baxxf

.,on constant is then

,, intervalan in of each valuefor 0 If

baf

baxxf

Sign Diagram to Determine where f (x) is Inc./Dec.

).(cf

)(or 0)( xfxf

,0)( cf

Steps:

1. Find all values of x for which is discontinuous and identify open intervals with these points. These points are called critical points.

2. Test a point c in each interval to check the sign of

a. If

b. If ,0)( cf

f is increasing on that interval.

f is decreasing on that interval.

Example16)( 23 xxxf

xxxf 123)( 2 0123 2 xx

Determine the intervals where

0)4(3 xx04or 03 xx

4,0x

0 4

+ - +

f is increasing

on

is increasing and where it is decreasing.

,0 4,

f is decreasing

on 0,4

x

Relative Extrema

)()( cfxf

A function f has a relative maximum at x = c if there exists an open interval (a, b) containing c such that for all x in (a, b).

A function f has a relative minimum at x = c if there exists an open interval (a, b) containing c such that

)()( cfxf

for all x in (a, b).

Relative Maxima

Relative Minima

x

y

Critical Numbers of f

A critical number of a function f is a number in the domain of f where

exist.not does )(or 0)( xfxf

(horizontal tangent lines, vertical tangent lines and sharp corners)

x

y

The First Derivative Test

1. Determine the critical numbers of f.

2. Determine the sign of the derivative of f to the left and right of the critical number.

left right

f(c) is a relative maximum

f(c) is a relative minimum

No change No relative extremum

Example

.16)( 23 xxxf

xxxf 123)( 2

0123 2 xx

Find all the relative extrema of

0)4(3 xx04or 03 xx

4,0x

0 4

+ - +

Relative max.

f (0) = 1 Relative min. f (4) = -31

f x

Example3 3( ) 3f x x x

2

233

1( )

3

xf x

x x

2 1 0x

Find all the relative extrema of

3 3 0x x

0, 1, 3x

-1 0 1

Relative max. Relative min.

or

33

+ + - - + +

3( 1) 2f 3(1) 2f

( ) undefinedf x( ) 0f x

x

Concavity

( ) 0f x

Let f be a differentiable function on (a, b).

1. f is concave upward on (a, b) if is increasing on (a, b). That is, for each value of x in (a, b).

2. f is concave downward on (a, b) if is decreasing on (a, b). That is, for each value of x in (a, b).

f ( ) 0f x

f

concave upward concave downward

Determining the Intervals of Concavity1. Determine the values for which the second

derivative of f is zero or undefined. Identify the open intervals with these points.

2. Determine the sign of in each interval fromf step 1 by testing it at a point, c, on the interval.

( ) 0,f c f is concave up on that interval.

f is concave down on that interval.( ) 0,f c

Example

( ) 6 12 6( 2)f x x x

Determine where the functionis concave upward and concave downward.

3 2( ) 6 1f x x x

2( ) 3 12f x x x

2

– +f

f concave down

on

f concave up on

, 2 2,

x

Inflection PointA point on the graph of a continuous function f where the tangent line exists and where the concavity changes is called an inflection point.

To find inflection points, find any point, c, in the domain where ( ) 0 or ( )f x f x

changes sign from the left to the right of c,

is undefined.

f IfThen (c, f (c)) is an inflection point of f.

The Second Derivative Test

( ) 0f c

1. Compute ( ) and ( ).f x f x

2. Find all critical numbers, c, at which ( ) 0.f c

f has a relative maximum at c.

f has a relative minimum at c.

The test is inconclusive.

( ) 0f c If Then

( ) 0f c

Example 1

3 2( ) 4 12 8f x x x x

2( ) 12 24 8f x x x

Classify the relative extrema of using the second derivative test.

4 3 2( ) 4 4 5f x x x x

Critical numbers: x = 0, 1, 2

4 2 1x x x

(0) 8 0f (1) 4 0f (2) 8 0f

Relative max.

Relative minima

(1) 4f (0) (2) 5f f

Example 2

3 2( ) 6 15 0 4N t t t t t

An efficiency study conducted for Elektra Electronics showed that the number of Space Commander walkie-talkies assembled by the average worker t hr after starting work at 8 A.M. is given by

At what time during the morning shift is the average worker performing at peak efficiency?

2

Step 1. Find the first and the second derivatives

of ( ).

( ) 3 12 15 and ( ) 6 12

. . . . . .

N t

N t t t N t t

Example 2 (cont.)

Step 2.

Peak efficiency means that the rate of growth is maximal, that occurs at the point of inflection.

( ) 0 2.N t t

At 10:00 A.M. during the morning shift, the

average worker is performing at peak efficiency.

Vertical Asymptote

Horizontal Asymptote

The line x = a is a vertical asymptote of the graph of a function f if either

The line y = b is a horizontal asymptote of the graph of a function f if

lim ( ) or lim ( )x a x a

f x f x

is infinite.

lim ( ) or lim ( )x x

f x b f x b

Finding Vertical Asymptotes of Rational Functions

If

is a rational function, then x = a is a vertical asymptote if Q(a) = 0 but P(a) ≠ 0.

( )( )

( )

P xf x

Q x

Ex.3 1

( )5

xf x

x

f has a vertical asymptote at x = 5.

Find the vertical asymptote for the function2

2

4 5( ) .

25

x xf x

x

f has a vertical asymptote at x = 5.

Example

2

2

4 5 ( 1)( 5)( )

25 ( 5)( 5)

1( ) when 5

5

x x x xf x

x x x

xf x x

x

Factoring

Finding Horizontal Asymptotes of Rational Functions

Ex.2

2

3 2 1( )

5

x xf x

x x

f has a horizontal asymptote at

2

2

3 2 1lim

5x

x x

x x

2

2 13

lim1

5x

x x

x

3.

5y

0

00

2

Divide

by

x

Curve Sketching Guide1. Determine the domain of f.2. Find the intercepts of f if possible.

4. Find all horizontal and vertical asymptotes.3. Look at end behavior of f.

5. Determine intervals where f is inc./dec.6. Find the relative extrema of f. 7. Determine the concavity of f. 8. Find the inflection points of f. 9. Sketch f, use additional points as needed.

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

1. Domain: (−∞, ∞).2. Intercept: (0, 1)

3. lim ( ) and lim ( )x x

f x f x

4. No Asymptotes

5. 2( ) 3 12 9;f x x x f inc. on (−∞, 1) U (3, ∞); dec. on (1, 3).

6. Relative max.: (1, 5); relative min.: (3, 1)

7. ( ) 6 12;f x x f concave down (−∞, 2); up on (2, ∞).

8. Inflection point: (2, 3)

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

0,1

x

y

ExampleSketch:

2 3( )

3

xf x

x

1. Domain: x ≠ −3

2. Intercepts: (0, −1) and (3/2, 0)

3. 2 3 2 3

lim 2 and lim 23 3x x

x x

x x

4. Horizontal: y = 2; Vertical: x = −3

5. 2

6( ) ;

( 3)f x

x

f is increasing on (−∞,−3) U (−3, ∞).

6. No relative extrema.

7. 3

18( ) ;

( 3)f x

x

f is concave down on (−3, ∞) and concave up on (−∞, −3).

8. No inflection points

Sketch:2 3

( )3

xf x

x

y = 2

x = −3

0, 1

3,0

2

x

y

Absolute Extrema

)()( cfxf

A function f has an absolute maximum at x = c if for all x in the domain of f.

A function f has a absolute minimum at x = c if

)()( cfxf

for all x in the domain of f.

Absolute Maximum

Absolute Minimum

x

y

Absolute ExtremaIf a function f is continuous on a closed interval [a, b], then f attains an absolute maximum and minimum on [a, b].

a b a ba b

Attains max. and min.

Attains min. but not max.

No min. and no max.

Interval open Not continuous

x x x

y

y

y

Finding Absolute Extremaon a Closed Interval

1. Find the critical numbers of f that lie in (a, b).

2. Compute f at each critical number as well as each endpoint.

Largest value = Absolute Maximum

Smallest value = Absolute Minimum

ExampleFind the absolute extrema of 3 2 1

( ) 3 on , 4 .2

f x x x

2( ) 3 6 3 ( 2)f x x x x x

Critical values at x = 0, 2

(0) 0

(2) 4

1 7

2 8

4 16

f

f

f

f

Absolute Min.

Absolute Max.

Evaluate

ExampleFind the absolute extrema of

1( ) on 3, .

2f x

x

Notice that the interval is not closed. Look graphically:

Absolute Max.

(3, 1)x

y

Optimization Problems1. Assign a letter to each variable mentioned in the

problem. Draw and label figure as needed.

2. Find an expression for the quantity to be optimized.

3. Use conditions to write expression as a function in one variable (note any domain restrictions).

4. Optimize the function.

ExampleAn open box is formed by cutting identical squares from each corner of a 4 in. by 4 in. sheet of cardboard. Find the dimensions of the box that will yield the maximum volume.

xx

x

4 – 2x

4 – 2xx

(4 2 )(4 2 ) ; in 0,2 ......V lwh x x x x

2( ) 16 32 12V x x x 4(2 3 )(2 )x x

Critical points: 2

2,3

x

3

(2) 0

(0) 0

24.74 in

3

V

V

V

The dimensions are 8/3 in. by 8/3 in. by 2/3 in. giving a maximum box volume of 4.74 in3.

2 316 16 4V x x x x

ExampleAn metal can with volume 60 in3 is to be constructed in the shape of a right circular cylinder. If the cost of the material for the side is $0.05/in.2 and the cost of the material for the top and bottom is $0.03/in.2 Find the dimensions of the can that will minimize the cost.

2 60V r h 2(0.03)(2) (0.05)2C r rh

top and bottom

sidecost

……

2 60V r h

22

60(0.03)(2) (0.05)2r r

r

2

60h

rSo

2(0.03)(2) (0.05)2C r rh

2 60.06 r

r

2

60.12C r

r

2

60 gives 0.12C r

r

36

2.52 in.0.12

r

which yields 3.02 in.h

Sub. in for h

……

So with a radius ≈ 2.52 in. and height ≈ 3.02 in., the cost is minimized at ≈ $3.58.

Graph of cost function to verify absolute minimum:

2.5r

C