Extrema and the Extreme Value Theorem

AP Calculus Mrs. Jo Brooks

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Extrema and the Extreme Value Theorem Local and Absolute Extrema. Extrema are the points where we will find a maximum or minimum on the curve.

If they are local or relative extrema, then they will be the highest or lowest points in that vicinity. If they are absolute extrema, they will be the highest or lowest points anywhere on the curve or within a closed interval.

When you do the problems, be sure to be aware of the difference between the two types of extrema! An important Theorem is the Extreme Value Theorem. It states that if f is a continuous function on a closed interval [a, b], then the function f has both a minimum and a maximum on the interval. That makes sense. If I put my pencil on the coordinate plane and trace out a curve within an interval, somewhere there will be a highest point and somewhere there will be a lowest point. If you have two or more points that have the same highest/lowest y-value then you have more than one absolute max or min. A horizontal line, for instance, has an infinite number of both max and min points. Every point on the line is one!! Note that there are two critical conditions of this theorem: the function must be continuous and the interval must be closed. If the interval is not closed or you have a "hole" on the curve at a maximum or minimum, then we will not be able to find a max or min.


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Look at these figures: 1 - Here we have a min at the bottom of the parabola but no max because the curve goes up infinitely at the ends.

2 - Here the domain of the parabola is restricted to only use x-values of 0 to 2, so this time we still have the min at the vertex and we also have a max at the top right endpoint.


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3 - Here, though, the bottom endpoint has become open, so we no longer have a minimum. We can never reach the bottom point of (0, 0), we just keep getting closer and closer to the y-coordinate of 0. You give me one "min" point and I can give you one that is lower. The absolute minimum does not exist.

4 - In a similar manner, in this figure both endpoints are open circles, so we will have neither a max nor a min.


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Critical Values and Relative Extrema Critical numbers (or critical values, CVs). These are the points where the derivative of the function is either equal to zero or undefined. To understand why these points are important, let's think about what it means to have an undefined or equal-to-zero first derivative. The first derivative gives us the slope of the tangent line to a curve. If the first derivative equals zero, then the slope of the tangent line equals zero and therefore must be horizontal. If you think about a curve and its tangent lines, you will usually only have a horizontal tangent at a "hump" or a "dip" in the curve (it sometimes occurs where the concavity changes - something we will explore later). When you are sketching a curve, these hills and valleys are important areas.

If the derivative is undefined, then you are either at a point with a vertical tangent, a break in the curve, an endpoint, or a cusp - all important areas as well.

To find critical numbers, set the first derivative equal to zero and solve for x. Also look to see if there are places where the derivative (and therefore the slope of the tangent line) is undefined. Because these points will give us all cases of horizontal or vertical tangents, we know that if there is a relative minimum or maximum on function f, then it must occur at a critical number of f.


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Finding Extrema on a Closed Interval To find extrema on a closed interval, you will have to check more than the critical values of the derivative. Because we have endpoints at the end of the closed interval, those points might end up being a max or min within the interval, even though they are not relative extrema of the curve and therefore will not satisfy the condition that f '(c) = 0 or f '(c) is undefined. The steps to finding the extremes on a closed interval are: 1. Find the critical numbers of the function within the open interval. (where the derivative is either equal to zero or undefined) 2. Evaluate the function at each of those critical numbers. (find the y-values at each of the x-values you found in step #1) 3. Evaluate the function at each of the endpoints of the closed interval. (find the y-values at the x-values that are the endpoints of the interval) 4. The least of the y-values found in steps (2) and (3) is the absolute minimum, the greatest is the absolute maximum. Remember that you will have more than one answer for the absolute max and min if two points have the same y-values. Be sure to remember this. Tons of mistakes are made by students who forget the difference between relative extrema (you only have to do step 1) and absolute extrema on a closed interval (all 4 steps required). This is a prime distinguisher between the 3’s and the 4’s or 5’s on the AP Exam!


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Here are a few examples: A: Let f (x) = -1/x, with a domain of 2 1x− ≤ ≤ − . Find the extrema of f on the closed interval. Since this is a closed interval and f (x) is continuous on this interval (the only place where the function is undefined is at x = 0, but that is not within our interval), we know that the Extreme Value Theorem guarantees that there will be an absolute minimum and an absolute maximum within the interval.

f (x) = -1/x = -x-1 f '(x) = x-2

We take the derivative of f (x). First I rewrote it with negative exponent so that I would not have to use the quotient rule.

f '(x) = 0 = x-2 = 1/x2

There are no values of x for which the function equals 0.

When x = 0 the function is undefined, but this value is outside of the given domain and therefore cannot be used.

There are no CVs. We will only be able to find extrema at the endpoints.

We set the derivative equal to 0, because when the slope of the tangent line is zero, we will have a horizontal tangent and therefore a "hump"/max or a "valley"/min.

We also check for values for which the function is undefined.

f (-2) = 1/2 absolute minimum f (-1) = 1 absolute maximum

We evaluate the endpoints of the interval to find the values of y for each point. The highest is our absolute maximum and the lowest is the absolute minimum.


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B: Let ( ) 2/33h θ θ= with a domain of 27 8θ− ≤ ≤ . Find the absolute extrema.

( ) 1/32h θ θ −′ = Find the derivative of h(θ).

( )3

1/3 22 0h θ θθ

−′ = = =

There are no values of θ for which the function equals 0.

When θ = 0 the function is undefined. CV = 0

Now we will set it equal to zero to find critical values and check for values which make the function undefined.

Our critical value is θ = 0. We will check that point as well as the endpoints to find our absolute extrema.

h(0) = 0 absolute minimum h(-27) = 27 absolute maximum h(8) = 12

The greatest y-value is at θ = -27 so 27 is our absolute maxima. The least y-value is at θ = 0, so 0 is our absolute minimum. Remember that the max or min value of a function is the y-coordinate of our point.

C: Let ( ) 23 12 5f x x x= − + on the interval [1, 4]. Find the absolute extrema.

( ) 6 12f x x′ = − Find the derivative of f (x).

( ) 6 12 0f x x′ = − = When x = 2

The function is never undefined. CV = 2

Now we will set it equal to zero to find critical values and check for values which make the function undefined.

Our critical value is x = 2. We will check that point as well as the endpoints to find our absolute extrema.

f (2) = -7 absolute minimum f (1) = -4 f (4) = 5 absolute maximum

The least y-value is at x = 2 so -7 is our absolute minimum. The greatest y-value is at x = 4, so 5 is our absolute maxima. Remember that the max or min value of a function is the y-coordinate of our point.

Finding Absolute Extremaon a closed interval

1. Take the derivative of the function

6 12y x

23 12 5 on the interval [1, 4]y x x


2. Set the derivative equal to zero and solve for x.

6 12 0y x 2x

There are no values for which y ‘ is undefined.

: 2CV x

3. Also look for values of x where the derivative is undefined.

4. Name your CVs.


5. Find the function values for all of the CVs.

: 2, 7CV

6. Find the function values for the endpoints on the interval.


Endpoints: 1, 4 , 4,5


7. Give the answers for the absolute max and min.This means they want the points, so if they ask for the absoluteextrema, give the answer like this:

absolute min : 2, 7


absolute max : 4,5

8. Caution: if they ask for “the extreme values of the function”,then give the answer like this because it means they want the y-values:

the absolute min of the function is -7 at x = 2the absolute max of the function is 5 at x = 4

Finding Absolute Extrema on a closed interval

1. Take the derivative of the function

6 12y x′ = −

23 12 5 on the interval [1, 4]y x x= − +


2. Set the derivative equal to zero and solve for x.

6 12 0y x′ = − =2x =

There are no values for which y ‘ is undefined.

: 2CV x =

3. Also look for values of x where the derivative is undefined.

4. Name your CVs.


5. Find the function values for all of the CVs.

( ): 2, 7CV −

6. Find the function values for the endpoints on the interval.


( ) ( )Endpoints: 1, 4 , 4,5−


7. Give the answers for the absolute max and min. If they ask for the absolute extrema, give the answer like this:

( )absolute min : 2, 7−


( )absolute max : 4,5

8. Caution: if they ask for “the extreme values of the function”, then give the answer like this:

the absolute min of the function is -7 at x = 2the absolute max of the function is 5 at x = 4


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Homework Examples #1, 3, 5 #1: Look at the graph and identify each x-value at which any absolute extreme value occurs. Explain why the answer is consistent with the Extreme Value Theorem. The curve is in the book. I will not reproduce the curve here, since it is in the book. Absolute max at x = b Absolute min at x = c2

Absolute extrema will be the highest and lowest points anywhere on the curve. The points at x = a and x = c1 are relative extrema. They are the highest and lowest points in the vicinity, but not on the entire curve.

It fits the EVT because we have a continuous curve on a closed interval, the Extreme Value Theorem guarantees that we will have extremes.

Explain why it fits the EVT.

#3: Look at the graph and identify each x-value at which any absolute extreme value occurs. Explain why the answer is consistent with the Extreme Value Theorem. The curve is in the book. I will not reproduce the curve here, since it is in the book. Absolute max at x = c Absolute min - none

We do not have an absolute min because the bottom “points” at x = a and x = b are holes. There is no way to reach the y-values to get our min value.

It fits the EVT because our continuous curve is on an open interval, not on a closed interval, the Extreme Value Theorem does not apply.



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#5: Look at the graph and identify each x-value at which any absolute extreme value occurs. Explain why the answer is consistent with the Extreme Value Theorem. The curve is in the book. I will not reproduce the curve here, since it is in

the book. Absolute max at x = c Absolute min at x = a

The highest and lowest points will be our absolute extremes.

It fits the EVT because our curve is not continuous on the closed interval, the Extreme Value Theorem does not apply, even though we do have absolute extrema in this case.



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Homework Examples #13, 19, 21, 25 #13: Use analytic methods to find the extreme values of the function on the interval and where they occur.

( ) ln( 1)h x x 0 3x

This is our function and interval.

1( ) 01

f xx

at no value of x

f ‘ (x) is undefined when x = -1.

Extreme values on a closed interval will only be found at the critical values or at the endpoints. First find the CVs by finding where f ‘ = 0 or where it is undefined.

f (0) =0 f (-1) is undefined f (3) = ln 4

Check the y-values of the function at the CVs and the endpoints.

Absolute max = ln 4 at x = 3 Absolute min = 0 at x = 0

State your conclusions. Be careful about how you give the answer here!! They asked for the extreme values of the function, so they really want the y-values as the answer.

#19: Find the Critical Values of the function, then sketch your curve and identify the CVs where the function has extreme values.

22 8y x x 9 This is our equation.

y ‘ = 4x - 8 = 0 when x = 2 CV: x = 2

To find the CVs, we set the derivative equal to zero and solve. This tells us where the slope of the tangent line is horizontal (its slope is zero) and where we might have a max or min.

The sketch should be on your paper. Min value is 1 at x = 2.

Now graph the function to tell if it is a max or min or neither. Not every horizontal tangent will be at an extreme value. Find the value of the min by putting x = 2 into the function.


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3 2 8 5y x x x This is our equation.

y ‘ = 3x2 + 2x - 8 = 0 y ‘ = (3x - 4)(x + 2) = 0 when x = -2, 4/3 CV: x = -2, 4/3

To find the CVs, we set the derivative equal to zero and solve. This tells us where the slope of the tangent line is horizontal (its slope is zero) and where we might have a max or min.

The sketch should be on your paper. Local max at (-2, 17) Local min at (4/3, -41/27)

Now graph the function to tell if it is a max or min or neither. Not every horizontal tangent will be at an extreme value. Find the value of the min and max by putting the CVs into the function equation.


1/ 22

2

1 11

yx

x

This is our equation. I rewrote it so that it would be easier to take the derivative.

3/ 221 1 2

2y x x

3/ 223/ 22

1 01

xy x xx

when x = 0 and is undefined when x = 1 (out of the domain) CV: x = 0

To find the CVs, we set the derivative equal to zero and solve. This tells us where the slope of the tangent line is horizontal (its slope is zero) and where we might have a max or min. In this case, the function y is also undefined at x = 1, so the CV is out of the domain.

The sketch should be on your paper. Absolute min at (0, 1)

Now graph the function to tell if it is a max or min or neither. Not every horizontal tangent will be at an extreme value. Find the value of the min and max by putting the CVs into the function.


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Homework Examples #43, 45, 47 #43: Find the Critical Values of the function, then sketch your curve and identify the CVs where the function has extreme values.

2

2

2 4,6 4,

x x xy

x x x

This is our equation. There will be two parts to the derivative because it is defined piece-wise.

2 2,2 6,

11

x xy

x x

y ‘ = 0 when x = -1, 3 y’ is undefined at x = 1 CV: x = -1, 1, 3

To find the CVs, we set the derivative equal to zero and solve. In this case, y ‘ is undefined at x = 1 (the slopes of each piece are different).

The sketch should be on your paper. Local min at (1, 1) Absolute max at (-1, 5) and (0, 5)

Now graph the function to tell if it is a max or min or neither. Not every horizontal tangent will be at an extreme value. Find the value of the min and max by putting the CVs into the function.

#45: Match the table with a graph of f (x). x f ‘ (x) a 0 b 0 c 5

This is our table of values of the derivative at different values of x. Since the derivative gives us the slope of the tangent line, it can tell us what the curve has to look like at those values of x.

Graph (c) matches The first two tell us that at x = a and x = b we have a tangent line with a slope of zero. That means that the tangent line will be horizontal and we will probably have a smooth valley or hump. The last point tells us that at x = c, the tangent line has a slope of m = 5. That means it is positive and rather steep.


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#47: Match the table with a graph of f (x). x f ‘(x) a does not exist b 0 c -2

This is our table of values of the derivative at different values of x. Since the derivative gives us the slope of the tangent line, it can tell us what the curve has to look like at those values of x.

Graph (d) matches The first one tell us that at x = a the derivative does not exist. That means that we will have a cusp or sharp corner. The second tells us that at x = b we have a tangent line with a slope of zero. That means that the tangent line will be horizontal and we will probably have a smooth valley or hump. The last point tells us that at x = c, the tangent line has a slope of m = -2. That means it is negative, downward sloping.

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Extrema and the Extreme Value Theorem