Lial/Hungerford/Holcomb/Mullins: Mathematics with Applications 11e Finite Mathematics with Applications 11e Copyright ©2015 Pearson Education, Inc. All

Lial/Hungerford/Holcomb/Mullins:Mathematics with Applications 11e

Finite Mathematics with Applications 11e

Copyright ©2015 Pearson Education, Inc. All right reserved.


Chapter 12

Applications of the Derivative


Section 12.1

Derivatives and Graphs

Slide 1 - 4Copyright ©2015 Pearson Education, Inc. All right reserved.








Determine from the graph whether f has a local minimum on (a, b)



Find the critical numbers of the given function.

Solution:

Example:3 2( ) 2 3 72 15f x x x x

We have so exists for every x. Setting shows that

2( ) 6 6 72,f x x x ( )f x( ) 0f x

2 6 72 06 xx 26( 12) 0x x

2 12 0x x ( 3)( 4) 0x x 3 0 4 0 or

or 3 4.

x x

x x

Therefore −3 and 4 are the critical numbers of f; these are the only places where local extrema could occur.

The graph at the right shows that there is a local maximum at and a local minimum at

3x4.x


f is continuous on [a, b].


Find the local extrema of function:


Find the local extrema of function:


Review: Find the local extrema of function


Review: given the graph of function f(x) = -x2 + 5 and two points A(-1, 4) and B (2, 1) on the graph.Find equation of the tangent line to the graph that is parallel to AB.


Review: given the graph of function f(x) = 5 – 4/x (x>0) and two points A(1, 1) and B (4, 4) on the graph.Find equation of the tangent line to the graph that is parallel to AB.


Section 12.2

The Second Derivative


The second derivative of function f is the derivative of the first derivative of f, denoted f’’(x).


Find the second derivative of the given functions.

Solution:

Example:

Here, The second derivative is the derivative of

2( ) 24 18 6.x x xf ( ),orf x

( ) 48 18.f x x

Solution:

xy xe

1 .x x x xdyx e e xe e

dx

2

2) ( 2)( 2 .x x x x x xd y

xe e xe ex

e x ed

Using the product rule gives

Differentiate this result to get 2

2.

d y

dx

(a)

(b)

f(x) = 8x3 – 9x2 + 6x + 1



Example:








f’f’’

f’f’’

f’f’’

f’f’’



Find a, b, c, d so that the graph of function f(x) = ax3+bx2+cx+d satifies the conditions below:• Local maximum: (3, 3)• Local minimum: (5, 1)• Inflection point: (4, 2)


Section 12.3

Optimization Applications


Absolute Extrema

Let f be a function defined in the closed interval [a, b]. Let c be a number in the interval.

f has absolute maximum on the interval [a, b] at c if f(x) f(c) for all x [a, b]

f has absolute minimum on the interval [a, b] at c if f(x) f(c) for all x [a, b]



Find absolute extrema of the function on indicated interval.



Consider the function Without graphing, show that f has an absolute minimum on the interval (0, 2).

Solution:

Example: 3( ) 3 1.f x x x

The derivative is which is defined everywhere, so the critical numbers are the solutions of

2( ) 3 3,xf x ( ) 0 :f x

2( ) 3 3f x x

23 3 0x 23 3x 2 1x

1 or 1.x x

The only critical number in the interval (0,2) is Use the second-derivative test to determine whether there is a local extremum at

1.x

1:x

( ) 6f x x

(1) 6(1) 6 0.f

Hence, f has a local minimum at Therefore, by the critical-point theorem, the absolute minimum of f on the interval (0, 2) occurs at

1.x

1.x



Example: A land has the shape of a right triangle with dimension 8m by 10m. We want to build a rectangle house with a corner on the hypotenuse and the opposite corner is a the right angle. What is the dimension of the house that has maximum area?

8m

10m


Example:An open box is to be made by cutting a square from each corner of a 12 12 cm piece of metal and then folding up the sides. The finished box must be at least 1.5cm deep, but not deeper than 3 cm. What size square should be cut from each corner in order to produce a box of maximum volume?

x=2, V(2) = 128 cm3


x = 700m, y = 350m




9 x 6 in.



x = 0


Section 12.4

Implicit Differentiation


Implicit form


Find given that

Solution:

Example: dy

dx1 5 7 8 .x ye e x y

Step 1 Take the derivative of both sides of the equation with respect to x:

Simplify.

Chain rule: ( )u ud due e

du dx


Find given that

Solution:

Example: dy

dx1 5 7 8 .x ye e x y

Step 2 Solve the last equation for :dy

dx

Move terms involving to the left side.

dy

dx

Move terms without to the right side.

dy

dx

Factor out .dy

dx

Divide both sides by 8.ye





Section 12.5

Related Rates


A cancerous tumor in the shape of a sphere is undergoing radiation treatment. The radius of the tumor is decreasing at the rate of 2 millimeters per week. How fast is the volume of the tumor changing when the radius of the tumor is 10 millimeters?

The volume (in cubic millimeters) of a sphere of radius r

(in millimeters) is given by

Solution:

Example:

34.

3V r

The relationship between the volume and radius of the tumor was given:

3.4

3V r

To find differentiate both sides of the given equation with

respect to time (remember to use the chain rule since r is a function of time):

24(3 )

3

dV drr

dt dt

,dV

dt

Chain rule

24dV dr

rdt dt

Simplify.


Solution:

Example:

The volume of the tumor is decreasing at a rate of about 2513 cubic millimeters per week.

24 (10) ( 2) 800 2513.dV

dt Substitute and simplify.

Substitute the given values of (since the radius

is decreasing by 2 mm per week) into the previous equation to find that

and 10 2dr

rdt

A cancerous tumor in the shape of a sphere is undergoing radiation treatment. The radius of the tumor is decreasing at the rate of 2 millimeters per week. How fast is the volume of the tumor changing when the radius of the tumor is 10 millimeters?

The volume (in cubic millimeters) of a sphere of radius r

(in millimeters) is given by

34.

3V r


Section 12.6

Curve Sketching


Continued on next slide


Continued from previous slide


Graph

Solution:

Example: 3 2( ) 2 3 12 1.f x x x x

2

2

6 6 12 0

2 0

( 1)( 2) 0

1 or 2.

x x

x x

x x

x x

Divide both sides by 6.

Factor.

The y-intercept is 3 2(0) 2 0 3 0 12 0 1 1.f Step 1

To find the x-intercepts, we must solve the equation

There is no easy way to do this by hand, so skip this step. Since f(x) is a polynomial function, the graph has no asymptotes, so we can also skip Step 3.

3 22 3 12 1 0.x x x Step 2

The first derivative is and the second derivative is

2( ) 6 6 12,f x x x ( ) 12 6.f x x

Step 4

The first derivative is defined for all x, so the only critical numbers are the solutions of ( ) 0 :f x

Step 5


Graph

Solution:

Example: 3 2( ) 2 3 12 1.f x x x x

Using the second-derivative test on the critical number we have,

1,x

( 1) 12( 1) 6 18 0.f

Hence, there is a local maximum when that is, at the point Similarly,

so there is a local minimum when (at the point ).

( 1, ( 1)) ( 1,8).f 1,x

(2) 12(2) 6 18 0,f

(2, (2)) (2, 19)f 2,x

2 2

( ) 0 and ( ) 0

6 12 0 6 126 6 0.

f f

x

x x

x xx

Next, we determine the intervals on which f is increasing or decreasing by solving the inequalities

Step 5


Graph

Solution:

Example: 3 2( ) 2 3 12 1.f x x x x

The critical numbers divide the x-axis into three regions. Testing a number from each region, as indicated below, we conclude that f is increasing on the intervals and decreasing on , 1) and (2, )( ( 1,2).

The second derivative is defined for all x, so the possible points of inflection are determined by the solutions of

12 6 0

1.

2

x

x

( ) 0 :xf

Step 5

Step 6


Graph

Solution:

Example: 3 2( ) 2 3 12 1.f x x x x

Determine the concavity of the graph by solving

12

( ) 0 and ( ) 0

1 1.

6 0 12 6

2 2

0

f f

x x

x

x x

x

1,

2

Therefore, f is concave upward on the interval and concave

downward on 1, .2

Consequently, the only point of inflection is 1 1 1

, , 5.5 .2 2 2f

Since f is a third-degree polynomial function, we know that when x is very large in absolute value, its graph must resemble the graph of its highest degree term, that is, the graph must rise sharply on the right side and fall sharply on the left.

32 ;x

Step 6

Step 7


Graph

Solution:

Example: 3 2( ) 2 3 12 1.f x x x x

Combining this fact with the information in the preceding steps, we see that the graph of f must have the general shape shown below.

Step 7


Graph

Solution:

Example: 3 2( ) 2 3 12 1.f x x x x

Now we plot the points determined in Steps 1, 5, and 6, together with a few additional points to obtain the graph below.

Step 8




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Lial/Hungerford/Holcomb/Mullins: Mathematics with Applications 11e Finite Mathematics with Applications 11e Copyright ©2015 Pearson Education, Inc. All