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Extreme Values

Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

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Page 1: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Extreme Values

Page 2: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Notation Set Notation Picture

(a,b)[a,b][a,b)(a,b](a,∞)[a, ∞)

a < x < ba < x < ba < x < ba < x < bx > ax > a

Intervals

Inequalities 1. If a < b then a + c < b + c2. If a < b and c < d then a + c < b + d3. If a < b and c > 0 then ac < bc4. If a < b and c < 0 then ac > bc5. If 0 < a < b then >

a1

b1

Page 3: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Solve x2 + 2x > 0

Factor: x ( x + 2) > 0 which has the solutions x = 0 and x = -2.

These numbers divide the number line into three intervals:(-∞, -2) (-2, 0) and (0,∞)

On each of the intervals, we can determine the signs of the factors, and hence deduce the sign of the product.

-2 0+ +-

We can read from the number line that x(x+2) is positive when x < -2 or when x >0

Page 4: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

A

B

C

D

Increasing/Decreasing Functions

decreasing decreasing

increasing

Can you suggest a test for determining whether the function is increasing/decreasing?

1. If f’(x) > 0 for all x in an interval I, then f is increasing on I.2. If f’(x) < 0 for all x in an interval I, then f is decreasing on I.

Page 5: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

The Blue textbook gives the following example at the start of chapter 4:

The mileage of a certain car can be approximated by:

3 20.00015 0.032 1.8 1.7m v v v v

At what speed should you drive the car to obtain the best gas mileage?

Of course, this problem isn’t entirely realistic, since it is unlikely that you would have an equation like this for your car.

Page 6: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

3 20.00015 0.032 1.8 1.7m v v v v

We could solve the problem graphically:

Page 7: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

3 20.00015 0.032 1.8 1.7m v v v v

We could solve the problem graphically:

On the TI-83, we use Math, 7: fMaxHow it works: fMax(function, variable, lowerbound, upperbound).Enter: fMax(Y1,X,0,50)

Page 8: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

3 20.00015 0.032 1.8 1.7m v v v v

We could solve the problem graphically:

The car will get approximately 32 miles per gallon when driven at 38.6 miles per hour.

Page 9: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

3 20.00015 0.032 1.8 1.7m v v v v

Notice that at the top of the curve, the horizontal tangent has a slope of zero.

Traditionally, this fact has been used both as an aid to graphing by hand and as a method to find maximum (and minimum) values of functions.

Page 10: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Maximum and Minimum Functions

(a, f(a))

(b, f(b))

(c, f(c))

(d, f(d))

(e, f(e))

absolute maximum

local maximum

local minimum

Fermat’s TheoremIf f has a max or min at p, then either f’(p) = 0 or f’(p) does not exist.A critical number of a function is a number c in the domain of f such that eitherf’(c) = 0 or f’(c) does not exist.

Page 11: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Local maximum

Local minimum

Notice that local extremes in the interior of the function

occur where is zero or is undefined.f f

Absolute maximum

(also local maximum)

Page 12: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Absolute extreme values are either maximum or minimum points on a curve.

Even though the graphing calculator and the computer have eliminated the need to routinely use calculus to graph by hand and to find maximum and minimum values of functions, we still study the methods to increase our understanding of functions and the mathematics involved.

They are sometimes called global extremes.

They are also sometimes called absolute extrema.(Extrema is the plural of the Latin extremum.)

Page 13: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Local Extreme Values:

If a function f has a local maximum value or a local minimum value at an interior point c of its domain,

and if exists at c, then

0f c

f

Page 14: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Critical Point:

A point in the domain of a function f at which

or does not exist is a critical point of f .0f

f

Note:Maximum and minimum points in the interior of a function always occur at critical points, but critical points are not always maximum or minimum values.

Page 15: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Find the absolute(global) and local(relative) max/min values of the function

f(x) = x3 + 6x2 + 9x + 2, -3.5 < x < 1

Solution:f’(x) = 3x2 + 12x + 9 = 3(x+1)(x+3)

Since derivative exists for all x, f’(x) = 0 at x = -1 and at x = -3These are the critical values.

For max/min:f(-3.5) = 1.125f(-3) = 2f(-1) = -2f(1) = 18

The endpoints of the interval must always be considered for max/min

The absolute max is f(1) = 18 and the absolute min is f(-1) = -2A local max occurs at f(-3) = 2 and a local min occurs at f(-3.5) = 1.125

f(-3) = 2

f(-1) = -2

Page 16: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Finding Maximums and Minimums Analytically:

1 Find the derivative of the function, and determine where the derivative is zero or undefined. These are the critical points.

2 Find the value of the function at each critical point.

3 Find values or slopes for points between the critical points to determine if the critical points are maximums or minimums.

4 For closed intervals, check the end points as well.

Page 17: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

Critical points are not always extremes!

3y x

0f (not an extreme)

Page 18: Extreme Values. NotationSet NotationPicture (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) a < x < b x > a Intervals Inequalities 1.If a < b then a + c < b + c

1/3y x

is undefined.f

(not an extreme)