dp s tudies y2 chapter 21 application of differential calculus

DP STUDIES Y2Chapter 21 Application of Differential Calculus

CONTENTS

A. Increasing and decreasing functionsB. Stationary pointsC. Rates of changeD. Optimization

A: INCREASING AND DECREASING FUNCTIONS

Quick reminderof notations forinequalities an intervals.


We can determine intervals where a curve is increasing or decreasing by considering f'(x) on the interval in question. For most functions that we deal with in this course:


Many functions are either increasing or decreasing for all x є R . We say these functions are monotone increasing or monotone decreasing.


for a strictly increasing function, an increase in x produces an increase in y

“Monotone increasing is an uphill battle all the way.”


for a strictly decreasing function, an increase in x produces a decrease in y.

“It’s all downhill with monotone decreasing.”

EXAMPLE 1

Find the intervals where f(x) is:a. Increasingb. decreasing


a. The increasing part is as x increase, y also increase.

b. The decreasing part is as x increase, y decrease.

xandx 21

21 x

SIGN DIAGRAMS

Sign diagrams for the derivative are extremely useful for determining intervals where a function is increasing or decreasing.

derivative = slope

If the derivative is negative, the function is decreasing.If the derivative is positive, the function is increasing.

The critical values for f’(x) are the values of x for which f’(x) = 0 or f’(x) is undefined.

When f’(x) = 0, the critical values are shown on a number line using tick marks.When f’(x) is undefined, the critical values are shown with a vertical dotted line.

SIGN DIAGRAMS

turning point (derivative =0)

SIGN DIAGRAMS

SIGN DIAGRAMSExample 2:Find the intervals where f(x) = 2x3 + 3x2 – 12x – 5 is increasing or decreasing.

SIGN DIAGRAMSSolution to example 2:f(x) = 2x3 + 3x2 – 12x – 5f(x) = 6x2 – 6x – 12

Set f(x) = 0 and solve by factoring or quadratic formula

6(x2 – x – 2) = 06(x – 2)(x + 1) = 0 => x = 2, x = -1

C: STATIONARY POINTS

A stationary point of a function is a point where f’ (x) = 0.

It could be a local maximum, local minimum, or stationary inflection.

EXAMPLE 3

Find and classify all stationary points of f(x) = x3 – 3x2 – 9x + 5

1,3

)1)(3(30

)32(30

963)('

593)(

2

2

23

xx

xx

xx

xxxf

xxxxf

-1 3

+ + –

Stationary points at x = 3 and x = -1Determine the interval of increasing and decreasing bychoosing x-values less than -1, between -1 and 3, andgreater 3.

x < -1: (-2 – 3)(-2 + 1) = (–)(–)=+-1 < x < 3: (0 – 3)(0 + 1) = (–)(+)=–x > 3: (4 – 3)(4 + 1) = (+)(+) = +

increasing decreasing increasing

local maximum local minimum

-1 3

+ + –



Example 4:Find the greatest and least value of y = x3 –

6x2 + 5 on the interval -2 < x < 5.

1. Take the derivative of the function.2. Find the stationary points.3. Draw the sign diagram.4. Determine the local max. and local min. within

the given interval.

xxy

xxy

123'

56

derivative the take:1 Step

2

23

4,0

)4(30

1230

123'

for x solve and 0 set :2 Step

2

2

xandx

xx

xx

xxy

0 4

positive

xchoose

negative

xchoose

positive

xchoose

xx

)45)(5(3

4

)41)(1(3

40

)41)(1(3

0

)4(30

++–



3: Draw the sign diagram

CHECK VALUESCheck y-values at the stationary points and the endpoints.

Stationary point, x = 0y = 03 – 6(0)2 + 5 = 5Stationary point, x = 4y = 43 – 6(4)2 + 5 = -27

Left endpoint, x = -2y = 23 – 6(-2)2 + 5 = -27Right endpoint, x = 5y = 53 – 6(5)2 + 5 = -20

The greatest value, 5, occurs at x = 0.The least value, -27, occurs at x = 4 and at x = -2

C. RATE OF CHANGE

Example 5:

C. RATE OF CHANGE

Solution to example 5:

C. RATE OF CHANGE

C. RATE OF CHANGE

Example 6

C. RATE OF CHANGE

D. OPTIMIZATION

There are many problems for which we need to find the maximum or minimum value of a function.The solution is often referred to as the optimum solution and the process is called optimization.

D. OPTIMIZATIONWe can find optimum solutions in several ways:using technology to graph the function and search for the maximum or minimum value

using analytical methods such as the formula x =-b/2a for the vertex of a parabola

using differential calculus to locate the turning points of a function.

These last two methods are useful especially when exact solutions are required.

WARNING!!

The maximum or minimum value does not always occur when the first derivative is zero. It is essential to also examine the values of the function at the endpoint(s) of the interval under consideration for global maxima and minima.

D. OPTIMIZATION

D. OPTIMIZATION

Example 7:A 4 liter container must have a square base, verticalsides, and an open top. Find the most economicalshape which minimizes the surface area of materialneeded.

D. OPTIMIZATION

Example 8:A square sheet of metal 12 cm x 12 cm has smallersquares cut from its corners as shown.What sized square should be cut out so that when the sheet is bent into an open box it will hold the maximum amount of liquid?

V’ = 12x2 – 96x + 144 V’ = 12(x2 – 8x + 12)V’ = 12(x – 6)(x – 2)Set the derivative to 0 and solve

x – 6 = 0 x = 6x – 2 = 0 x = 2

If we put the x-values back into the original equation we get

V = x(12 – 2x)2 at x = 6, V = 6(12 – 2(6)2) = 0at x = 2, V = 2(12 – 2(2)2) = 8

Therefore, the only value for x is 2. The square we are cutting out is a 2cm x 2cm.

Example 9:

Solution to example 9:

dp s tudies y2 chapter 21 application of differential calculus

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