The Derivative in Graphing and Applications

By: Tair Akhmejanov

Section 5.1:Analysis of Functions

• Increasing• Decreasing• Concavity• Inflection Points

To describe functions we need a few definitions A function is increasing on an interval if f(x1)<f(x2) when x1<x2

A function is decreasing on an interval if f(x1)>f(x2) when x1>x2

A function is constant on an interval if f(x1)=f(x2) for all x1 and x2

Thus, a function is increasing on an interval where it has a positive slope, decreasing on an interval where it has a negative slopes, and constant on an interval where its graph has zero slope.

Let f be a function that is continuous on a closed interval [a,b] and differentiable on the open interval (a,b).

If for every value of x in (a,b), then f is increasing on [a,b] If for every value of x in (a,b), then f is decreasing on [a,b] If for every value of x in (a,b), then f is constant on [a,b]

( )f x ( )f x ( )f x

f If f is differentiable on an open interval I, then f is concave up on I if is increasing on I, and f is concave down on I if is decreasing on I.

Let f be differentiable twice on an open interval I. If on I, then f is concave up on I. If on I, then f is concave down on I.

If f is continuous on an open interval containing a value xo, and if f changes the direction of its concavity at the point (xo,f(xo)), then we say that f has a n inflection point at xo, and we call the point (xo,f(xo)) on the graph of f an inflection point of f.

f

( )f x ( )f x

Concavity

3 26 9 4y x x x State when is increasing, decreasing, concave up, concave down, and all of the inflection points.

23 12 9y x x

3

1

x

x

6 12y x

( 1)x

13

2x

( 3 9)( 1)y x x ( 3 9)x

( 3 9)( 1)x x

+ - -

- - +

- + -

f(x) is decreasing on (- ,-3] U [-1, )f(x) is increasing on [-3,-1]f(x) is concave down on (-2, )f(x) is concave up on ( , -2)f(x) has an inflection point at (-2,-2)

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State when is increasing, decreasing, concave up, concave down, and all of the inflection points.

3( )

8

xf x

x

f(x) is increasing on (- ,-8)U(-8, )f(x) is never decreasingf(x) is concave up on (- , -8)f(x) is concave down on (-8, )f(x) has no inflection points

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Section 5.2

• Relative Extrema• First Derivative Test• Second Derivative Test

Suppose that Suppose that ff is a function defined on an open interval containing is a function defined on an open interval containing the number xthe number xoo. If f has a relative extremum at x=x. If f has a relative extremum at x=xoo then either then either or or ff is not differentiable at x is not differentiable at xoo..

First Derivative TestFirst Derivative Test

If on an open interval extending left from x and If on an open interval extending left from x and on an open interval extending right from xon an open interval extending right from xoo, then , then ff has a relative has a relative maximum at xmaximum at xoo..

If on an open interval extending left from x and If on an open interval extending left from x and on an open interval extending right from xon an open interval extending right from xoo, then , then ff has a relative has a relative minimum at xminimum at xoo..

If has the same sign on an open interval extending left from xIf has the same sign on an open interval extending left from xo o

and on an open interval extending right from xand on an open interval extending right from xoo, then , then ff does not have a relative extremum at xdoes not have a relative extremum at xoo..

( )f x

( )f x

( )f x

( )f x

( )f x

( )f x

Second Derivative TestIf and > 0, then f has a relative minimum at xo.If and < 0, then f has a relative maximum at xo.If and = 0, then the test is inconclusive; that is, f may have a relative maximum, a relative minimum, or neither at xo.

0( )f x

0( )f x

0( )f x

0( ) 0f x

0( ) 0f x

0( ) 0f x

Find the relative extremum of Find the relative extremum of

Find the relative extremum ofFind the relative extremum of

42( ) 2

4

xf x x

3

2

( ) 4

0 ( 4)

0

2

2

f x x x

x x

x

x

x

2

2

2

2

( ) 3 4

(0) 3(0) 4 4

(2) 3(2) 4 8

( 2) 3( 2) 4 8

f x x

f

f

f

By second derivative test f(x) has a relative maximum at (0,0), and relative minimums at (2,-4) and (-2,-4)

3( ) 9 6f x x x By second derivative test f(x) has a relative minimum at and a relative maximum at

( 3, 6 3 6) ( 3,6 3 6)

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Section 5.4

• Rectilinear motion

Rectilinear MotionRectilinear Motion

( ) ( )ds

v t s tdt

( ) ( )ds

v t s tdt

If s(t) is the position function of a particle moving on a coordinate line, then the instantaneous velocity of the particle at time t is defined by:

Instantaneous speed at time t:

If s(t) is the position function of a particle moving on a coordinate line, then the instantaneous acceleration of the particle at time t is defined by:

2

2

( ) ( )

( ) ( )

dva t v t

dt

d sa t s t

dt

*Note*: A particle in rectilinear motion is speeding up when its velocity and acceleration have the same sign and slowing down when they have opposite signs.

Reminders about particlesReminders about particles When the velocity is negative, the particle is moving to the left.When the velocity is negative, the particle is moving to the left.

When the velocity is positive, the particle is moving to the right.When the velocity is positive, the particle is moving to the right.

When the velocity and acceleration of the particle have the same When the velocity and acceleration of the particle have the same signs, the particle’s speed is increasing.signs, the particle’s speed is increasing.

When the velocity and acceleration of the particle have opposite When the velocity and acceleration of the particle have opposite signs, the particle’s speed is decreasing (or slowing down).signs, the particle’s speed is decreasing (or slowing down).

When the velocity is zero and the acceleration is not zero, the When the velocity is zero and the acceleration is not zero, the particle is momentarily stopped and changing direction.particle is momentarily stopped and changing direction.

2( ) 6 12 12v t t t

• If the position function of a particle is , t>0

find when the particle is changing directions.

• If the position function of a particle is , t>0 when is the particle speeding up?

3 2( ) 2 6 12 18x t t t t

24 144 288

12

Velocity is never 0 so particle never changes direction

2( ) 8x t t t

(4, )

Try ItTry It

Section 5.5

• Absolute Maxima and Minima

► A function A function ff is said to have an absolute maximum on an interval is said to have an absolute maximum on an interval I I at xat xoo if f(x if f(xoo) is the largest value of ) is the largest value of ff on on II; ; ff(x(xoo) ) ff(x) for all x in the (x) for all x in the domain of domain of ff that are in that are in II..A function A function ff is said to have an absolute minimum on an interval I is said to have an absolute minimum on an interval I at xat xoo if f(x if f(xoo) is the smallest value of ) is the smallest value of ff on on II; ; ff(x(xoo) ) ff(x) for all x in the (x) for all x in the domain of domain of ff that are in that are in II..

Extreme-Value TheoremExtreme-Value Theorem► If a function If a function ff is continuous on a finite closed interval [a,b], then is continuous on a finite closed interval [a,b], then ff

has both an absolute maximum and an absolute minimum on has both an absolute maximum and an absolute minimum on [a,b].[a,b].

► If If ff has an absolute extremum on an open interval (a,b), then it has an absolute extremum on an open interval (a,b), then it must occur at a critical number of must occur at a critical number of ff..

► *Note*:*Note*: Calculate all critical values of Calculate all critical values of f f and and calculate values at calculate values at endpointsendpoints.. Largest value is the absolute maximum and smallest value is the Largest value is the absolute maximum and smallest value is the

absolute minimum.absolute minimum.

► Suppose Suppose ff is continuous and has exactly one relative extremum on is continuous and has exactly one relative extremum on an interval an interval II at x at xo.o.

If If ff has a relative minimum at x has a relative minimum at xoo, then f(x, then f(xoo) is the absolute ) is the absolute minimum of minimum of ff on on I.I.If If ff has a relative maximum at x has a relative maximum at xoo, then f(x, then f(xoo) is the absolute ) is the absolute maximum of maximum of ff on on II..

2( ) 8 176 1800f x x x • At what value of x does f(x) have an absolute minimum given that

• At what point does f(x) have an absolute minimum and at what point does f(x) have an absolute maximum given that

on the interval 4≤x≤9

( ) 16 176 0f x x ( ) 16f x

11x

2 4( )

3

xf x

x

There is an absolute minimum at (5.24,10.47)

Try It

By the second derivative test f(x) has an absolute minimum at

Section 5.6

• Applied Maximum and Minimum Problems

Suggestions for Applied Maximum and Suggestions for Applied Maximum and Minimum ProblemsMinimum Problems

Draw a figure and label the quantities relevant to the problem.Draw a figure and label the quantities relevant to the problem.

Find a formula for the quantity to be maximized or minimizedFind a formula for the quantity to be maximized or minimized

Using the conditions stated in the problem to eliminate variables, express Using the conditions stated in the problem to eliminate variables, express the quantity to be maximized or minimized as a function of one variable.the quantity to be maximized or minimized as a function of one variable.

Find the interval of possible values for this variable from the physical Find the interval of possible values for this variable from the physical restrictions in the problem.restrictions in the problem.

If applicable, use the techniques of the preceding section to obtain the If applicable, use the techniques of the preceding section to obtain the maximum or minimum.maximum or minimum.

Note: Profit = Revenue - CostNote: Profit = Revenue - Cost

P(x) = R(x) - C(x)P(x) = R(x) - C(x)

Find the point on closest to point (4,0)Find the point on closest to point (4,0)

2 2 28 16 7 16D x x x x x 2( ) 7 16L x x x

( ) 2 7L x x 7 / 2

7 / 2

x

y

y x2 2 2 2 2( 4) ( 0) 8 16D x y x x y Use the distance formula.

Plug in for y.Let L(x)=D^2 because the minimum value of D^2 will occur at the same value of x as the minimum value of D.

x

(7 / 2, 7 / 2)

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Given a rectangular sheet of cardboard 18 inches by 24 inches a Given a rectangular sheet of cardboard 18 inches by 24 inches a box is made by cutting a square with sides x inches from each box is made by cutting a square with sides x inches from each corner and bending the resulting sides up. Find the value of x that corner and bending the resulting sides up. Find the value of x that maximizes the volume of the box.maximizes the volume of the box.

Width: 18-2xWidth: 18-2x

Length: 24-2xLength: 24-2x

Depth: xDepth: x

Volume=x(18-2x)(24-2x)Volume=x(18-2x)(24-2x)

Volume of the box maximized at x=3.4

Try It

3.4,10.6x

Eliminate x=10.6 because it does not work in the context of the problem.

3 24 84 432V x x x

Section 5.7

• Newton’s Method

1

( )

( )n

n nn

f xx x

f x

, n=1,2,3…

Newton’s Method

1

( ), 1, 2,3...

( )n

n nn

f xx x n

f x

Newton’s Method is used to find the point at which the graph of a function intersects the x-axis.Usually you are given a starting value, allowed to use a calculator to pick a starting value, or required to make an educated guess about the root to pick a good starting value.

5 3( ) 5f x x x

Given a function , x>0 find its root using Newton’s Method. Start at x1=2.1

Given a function , x>0 find its root using Newton’s Method (you may use a calculator to get your starting value for x)

3 2

2 2

2.1 (2.1) 6(2.1)2.1

3(2.1) 2(2.1) 6x

3 2( ) 6f x x x x

2 2.006299x

The more times this method is repeated by plugging in the new value for x, the more accurate the approximation of the root will be.In this case the root is x=2.

X=1.5395501 A question will usually tell you to run the test a certain amount of times.

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Section 5.8

• Mean-Value Theorem• Rolle’s Theorem

The Mean Value Theorem (for derivatives)

• Let f be differentiable on (a,b) and continuous on [a,b]. Then there is at least one number c in (a,b) such that:

Another way of putting it:• Given a function f

there is a point c in an interval (a,b) where the slope of the tangent line equals the slope of the secant line of the endpoints of the interval.

( ) ( )( )

f b f af c

b a

Another way of putting it:

The Rolle’s theorem is just an instance of the Mean Value Theorem when the

slope is zero, and therefore the tangent line is a horizontal tangent.

• Let f be differentiable on (a,b) and continuous on [a,b]. If f(a)=f(b)=0, then there is at least one number c in (a,b) such that f’(c)=0.

Rolle’s Theorem

3 2( ) 12 7f x x x x

2(4) ( 4)3 24 ( )

4 ( 4)

f fc c f c

on the interval [-4,4]

Find the values of c that satisfy the Mean Value Theorem

12 8 3

3c

Try It

2284 1003 24 7

8c c

20 3 24 16c c

Try It

• Find values of c that satisfy the Mean Value Theorem for

on the interval [-1,2]

Find the values of c that satisfy Rolle’s Theorem for

on the interval [0,1]

6( ) 3f x

x No Solution

( ) (1 )f x x x C=1/2

3c

1) Find the values of c that satisfy Rolle’s Theorem for on the interval [-1,1]

2) Find the absolute maximum and minimum values of on the interval [-3,3]

3) A rectangle is inscribed in a semicircle, that has a radius of 4, with one side on the semicircle’s diameter. What is the largest area this rectangle can have?

3( )f x x x

A)

QUIZ

B)1

3c C) 3

2c D) 2c

A) 16 B) C) -4 D) 98

3( )f x x x

A) min.=-3 B) min.= C) min=-24 D) min.=-.385max=3 max= max=24 max=.385

1

3

1

3

4) A rectangular field, bounded on one side by a building, is to be fenced in on the other three sides. If 3,000 feet of fence is to be used, find the dimensions of the largest field that can be fenced in.

5) Find the values of c that satisfy the Mean Value Theorem

for on the interval [0,4]

6) Find the point at the relative maximum and the point at the relative minimum of

4

3

A) 750 ft. by 1,000 ft. B)500 ft. by 1,500 ft. C)350 ft. by 2,300 ft. D)750 ft. by 1,500 ft.

A) B) C) D)

3( ) 24 16f x x x 2

3

4

2

2

2

A)max.:(4,0) B)no relative maximum C)max:(3,8) D)max:(5,24) min:(3,-27) min:(3,-27) no relative minimum min:(-1,-14)

4 3( ) 4f x x x

7) A poster is to contain 100 square inches of picture surrounded by a 4-inch margin at the top and bottom and a 2-inch margin on each side. Find the overall dimensions that will minimize the total area of the poster.

8) Given the position function , find when the particle is speeding up.

9) If the position function of a particle is 0<t<4 , find when the particle is changing direction.

4 50A) 5in. by 5in. B) 5in. By 3.5in. C) in. by in. D) in. by in.8 2 50 4 504 50

A) 2<t<7/2 B) 2<t<7/2 and t>5 C)2<t<3 and t>5 D)4>t>7

3 2( ) 2 21 60 3x t t t t

( ) sin( )2

tx t

A)t= ,3 B)t= /2 C)t= , /2 D)4 4 3( ) 4f x x x

10) Given a function state when it is concave up, concave down, and the points of inflection.

A) Cocave up: (-3,0)U(2, ) Concave down: (- ,-3)U(0,2)

Inflection point(s): (-3,189); (0,0); (2, -16)

4 3( ) 4f x x x

D) Concave up: (- ,0)U(2, ) Concave down: (0,2) Inflection point(s): (2,-16)

B) Concave up: (- , )

Concave down: never Inflection point(s): none

C) Concave up: (- ,0)U(2, ) Concave down: (0,2) Inflection point(s): (2,0)

Credits• This Presentation: Tair Akhmejanov

• Fact Checker: Alex Bishop

• Text Book: Anton, Bivens, Davis

• Princeton Review: David S. Kahn

• Barron’s: Shirley O. Hockett, David Bock

• PowerPoint: Microsoft Corp.

Bibliography of Pictures

• www.regonaudio.com/MathPaige.html

• www.karlscalculus.org/calc5_3.html