Extreme Values

Intervals

Notation (a,b) [a,b] [a,b) (a,b] (a,∞) [a, ∞) Set Notation a < x < b a < x < b a < x < b a < x < b x > a x > a Picture

Inequalities

1. If a < b then a + c < b + c 2. If a < b and c < d then a + c < b + d 3. If a < b and c > 0 then ac < bc 4. If a < b and c < 0 then ac > bc 5. If 0 < a < b then >

a

1

b

Solve x 2 + 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

Increasing/Decreasing Functions

A decreasing C B increasing decreasing D Can you suggest a test for determining whether the function is increasing/decreasing?

1.

2.

If f’(x) > 0 for all x in an interval

I

, then f is increasing on

I.

If f’(x) < 0 for all x in an interval

I

, then f is decreasing on

I

.

The Blue textbook gives the following example at the start of chapter 4: The mileage of a certain car can be approximated by: 

0.00015

v

3 

0.032

v

2 

1.8

v



1.7

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.



We could solve the problem graphically: 

0.00015

v

3 

0.032

v

2 

1.8

v



1.7



We could solve the problem graphically: 

0.00015

v

3 

0.032

v

2 

1.8

v



1.7

On the TI-83, we use Math, 7: fMax How it works: fMax(function, variable, lowerbound, upperbound).

Enter: fMax(Y 1 ,X,0,50) 

We could solve the problem graphically: 

0.00015

v

3 

0.032

v

2 

1.8

v



1.7

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





0.00015

v

3 

0.032

v

2 

1.8

v



1.7

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.



Maximum and Minimum Functions

(a, f(a)) local maximum (c, f(c)) (e, f(e)) absolute maximum (d, f(d)) (b, f(b)) local minimum Fermat’s Theorem If 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 f’(c) = 0 or f’(c) does not exist.

c in the domain of f such that either

Absolute maximum (also local maximum) Local maximum Local minimum Notice that local extremes in the interior of the function 

f

 is undefined.



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.

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

They are sometimes called global extremes.

They are also sometimes called absolute extrema.

(

Extrema

is the plural of the Latin

extremum

.) 

Local Extreme Values:

If a function

f

has a local maximum value or a local minimum value at an interior point

f



c

, then

c

of its domain,

f

  0 

Critical Point:

A point in the domain of a function

f f

 at which

critical point

of

f

.

f

 

0

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.



Find the absolute(global) and local(relative) max/min values of the function f(x) = x 3 + 6x 2 + 9x + 2, -3.5 < x < 1 Solution: f’(x) = 3x 2 + 12x + 9 = 3(x+1)(x+3) Since derivative exists for all x, f’(x) = 0 at x = -1 and at x = -3 These are the critical values.

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

f(-3) = 2 f(-1) = -2 f(1) = 18 The endpoints of the interval must always for max/min be considered f(-1) = -2 The absolute max is f(1) = 18 and the absolute min is f(-1) = -2 A local max occurs at f(-3) = 2 and a local min occurs at f(-3.5) = 1.125

1

Finding Maximums and Minimums Analytically:

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

For closed intervals, check the end points as well.



y

Critical points are not always extremes!



x

3 2 1 -2 -1 -1 0

f

1  

0

2 (not an extreme) -2 

y



x

1/ 3 2 1 -2 -1 0 -1 -2

f

1 2 

is undefined.

(not an extreme) p