CHAPTER 4: Applications of Derivatives · Applications of Derivatives 4.1: Extrema of a Function...

CHAPTER 4:

Applications of Derivatives

4.1: Extrema of a Function and the “EVT” 4.2: Rolle’s Theorem and the “MVT” for Derivatives 4.3: The First Derivative Test (“1st DT”) 4.4: The Second Derivative and the Second Derivative

Test (“2nd DT”) 4.5: Graphing a Function 4.6: Optimization 4.7: Applications to Rectilinear Motion and

Economics 4.8: Newton’s Method

• We use critical numbers (“CN”s) to find local extrema of a function. • We sometimes use the “EVT Method” to find absolute extrema. • We use the “1st DT” and “2nd DT” to classify points at CNs as local maximum

points, local minimum points, or (using the 1st DT) neither. • A function’s first derivative may tell us where the function is increasing or

decreasing. • A function’s second derivative may tell us where the function’s graph is concave

up or concave down. • Other applications include Newton’s Method, which numerically approximates a zero of a function.

(Section 4.1: Extrema of a Function and the “EVT”) 4.1.1

SECTION 4.1: EXTREMA OF A FUNCTION AND THE “EVT”

LEARNING OBJECTIVES

• Identify absolute and local extrema (maxima and minima) for a function’s graph. • Know the Extreme Value Theorem (“EVT”) and when it applies. • Use the “EVT Method” to find absolute extrema of a continuous function on a

closed, bounded interval. • Know what critical numbers (“CNs”) are, and find them for various functions. PART A: TERMINOLOGY

The extrema of a function f are the maxima and minima of f , if any. Extrema is the plural of extremum. There are different types of maxima (plural of maximum) and minima (plural of minimum).

Example 1 (Defining and Classifying Extrema of Functions)

Here is the graph of y = f x( ) for a function f : y

Point P 5,10( ) is an absolute maximum point (“A.Max.Pt.”) on the graph, because it is the highest point on the entire graph. It is the “King of the World” … apologies to Jim Cameron. More precisely…

• 10, or y = 10 , is the highest function (y) value on Dom f( ) .

• 10 is the absolute maximum value (“A.Max. Value”) of f . • 5, or x = 5 , is the absolute maximum (“A.Max.”) of f . • Sources vary on the terminology. Some would say 5 is an absolute maximum point on the real number line.


Point P is also a local maximum point (“L.Max.Pt.”), because it is the highest point locally on its “hill.” It is the “King of Its Hill.”

• There exists a neighborhood of x = 5 , say 4.9, 5.1( ) , on which 10 is the highest function (y) value.

• 5 is a local maximum (“L.Max.”) of f .

• 10 is a local maximum value (“L.Max. Value”) of f .

Point Q 8, 3( ) is not an absolute minimum point (“A.Min.Pt.”), because it is not the lowest point on the entire graph. Point S, for instance, is lower.

Point Q is a local minimum point (“L.Min.Pt.”), because it is the lowest point locally in its “valley.”

• There exists a neighborhood of x = 8 , say 7.9, 8.1( ) , on which 3 is the lowest function (y) value.

• 8 is a local minimum (“L.Min.”) of f .

• 3 is a local minimum value (“L.Min. Value”) of f .

Point R 12, 7( ) is not an A.Max.Pt., because it is not the highest point on the entire graph. Point P, for instance, is higher.

Point R is a L.Max.Pt. on the graph, because it is the highest point locally on its “hill.” Point S 14,1( ) is not a L.Max.Pt., because the points “immediately” to its left are higher.

• There does not exist a neighborhood of x = 14 on which 1 is the highest function (y) value.

Point S is a L.Min.Pt., because the points “immediately” to its left and to its right are not lower.

• There exists a neighborhood of x = 14 , say 13.9, 14.1( ) , on which 1 is the lowest function (y) value.

• WARNING 1: Ties are OK. The fact that there are other x-values in 13.9, 14.1( ) that share that same function value (1) does not take away the status of x = 14 as a L.Min.


Point T is a L.Min.Pt. Point T is a L.Max.Pt., because the points “immediately” to its left and to its right are not higher.

To recap:

y

Point A.Max.Pt.? L.Max.Pt.? A.Min.Pt.? L.Min.Pt.? P YES YES No No Q No No No YES R No YES No No S No No No YES T No YES No YES

§

Absolute extrema are also called global extrema. Local extrema are also called relative extrema. Some students may prefer the word-pairings in these columns:

Absolute (A.) Global

Relative Local (L.)

We try to identify:

Maxima Minima Absolute A.Max. A.Min.

Local L.Max. L.Min.


Example 2 (Classifying Extrema of Functions)

Here is the graph of y = f x( ) for a function f :

y

Maxima Minima Absolute There are 2. None.

Local There are 3. There are 2. §

Example 3 (“Graphical Endpoints” Cannot be L.Max./Min. Points)

Here is the graph of y = f x( ) for a function f :

y

Point P 6, 2( ) is an A.Min.Pt.

However, P is not a L.Min.Pt., because it has no neighboring points to the right.

• There does not exist a neighborhood of x = 6 throughout which f is defined.

More generally, “graphical endpoints” such as Point P cannot be L.Max.Pts. or L.Min.Pts., though sources may differ on this. §

A note on continuity. We will require continuity at local extrema. Other sources do not, but some require that f at least be defined on a neighborhood of a local extremum; these sources would say that x = 0 is a local maximum for the function from Section 2.1, Example 10.

(Section 4.1: Extrema of a Function and the “EVT”) 4.1.5 PART B: EXTREME VALUE THEOREM (“EVT”)

Extreme Value Theorem (“EVT”)

(Assume a < b .)

If a function f is continuous on a closed, bounded interval a, b⎡⎣ ⎤⎦ ,

then f has an absolute maximum and an absolute minimum on a, b⎡⎣ ⎤⎦ . • Think of a, b⎡⎣ ⎤⎦ as the restricted domain of f . We ignore the function outside of this interval. • We specify a closed, bounded interval, because we are not guaranteed absolute extrema on , or −∞,∞( ) , which is both an open and closed (though unbounded) interval.

Example 4 (When EVT Applies)

Let f x( ) = 1

x with restricted domain 2, 5⎡⎣ ⎤⎦ . Here is the graph of y = f x( ) :

y

• The EVT applies here to the closed, bounded interval 2, 5⎡⎣ ⎤⎦ , because f is

continuous on 2, 5⎡⎣ ⎤⎦ . • Therefore, f has an absolute maximum and an absolute minimum on

2, 5⎡⎣ ⎤⎦ . • Here, the absolute extrema of f are at the interval endpoints. The absolute maximum is at x = 2 , and the absolute minimum is at x = 5 . §


Example 5 (When EVT Does Not Apply)

Let g x( ) = 1

x with restricted domain 2, 5( ) . Here is the graph of y = g x( ):

y

• The EVT does not apply here to the interval 2, 5( ) , because it is open.

• g may or may not have absolute extrema on 2, 5( ) .

•• WARNING 2: Sufficient (but not necessary) conditions. The EVT provides sufficient conditions for the existence of absolute extrema; they are not necessary conditions. Even if conditions do not hold true, conclusions might hold true, anyway.

• Here, g has no absolute extrema on 2, 5( ) . Why is there no absolute maximum? Imagine a competition between Tweedledee and Tweedledum, who are trying to pick the highest function (y) value on the x-interval 2, 5( ) . They can go back and forth endlessly. They can alternate between y = 0.49 ,

y = 0.499 , y = 0.4999 , y = 0.49999 , etc. Neither can pick y = 0.5, because

x = 2 is not in the open interval 2, 5( ) . Similarly, there is no absolute minimum. §


Example 6 (When EVT Does Not Apply)

Consider h x( ) = 1

x on the interval −1,1⎡⎣ ⎤⎦ . Here is the graph of y = h x( ) :

• The EVT does not apply here to the interval −1,1⎡⎣ ⎤⎦ , because h is not

continuous on −1,1⎡⎣ ⎤⎦ .

• Here, h has no absolute extrema on −1,1⎡⎣ ⎤⎦ . §

Example 7 (When EVT Does Not Apply … but the Conclusions are True)

Let f x( ) = sin x with restricted domain 0, 2π( ) . Here is the graph of

y = f x( ) :

y

• The EVT does not apply to the interval 0, 2π( ) , because it is open.

• In spite of this, the conclusions happen to hold true, anyway. f has an absolute maximum and an absolute minimum on 0, 2π( ) . §


PART C: CRITICAL NUMBERS (“CNs”) The Derivative at a Local Extremum

If a function f has a L.Max. or L.Min. at x = c , then ′f c( ) = 0 or

′f c( ) does not exist (“DNE”).

y y

• WARNING 3: The converse is not true. If ′f c( ) = 0 or “DNE,” then c is not necessarily a local extremum.

y y

′f c( ) = 0 ′f c( ) DNE c is not a local extremum. c is not a local extremum.

Critical Numbers (“CNs”)

The critical numbers (“CNs”) of a function f are the numbers in Dom f( ) where ′f is 0 or “DNE.”

Local extrema can only appear at CNs.


PART D: THE “EVT METHOD”

Assume that a function f is continuous on a closed, bounded interval a, b⎡⎣ ⎤⎦ , where a < b .

By the EVT, f has an absolute maximum and an absolute minimum on a, b⎡⎣ ⎤⎦ . To locate these, we will use the “EVT Method,” which hinges on the “EVT Principle.”

“EVT Principle”

Assuming continuity, the absolute extrema of f on a, b⎡⎣ ⎤⎦ must be at CNs or interval endpoints (a or b).

Example 8 (“EVT Principle”)

Consider the graph of y = f x( )

here. On the x-interval 1, 8⎡⎣ ⎤⎦ , f has an absolute maximum at x = 5 (a CN) and an absolute minimum at x = 1, an interval endpoint.

§

“EVT Method”

Assuming continuity, this method is used to locate the absolute extrema of f on a, b⎡⎣ ⎤⎦ .

Step 1. Find all CNs of f in a, b( ) .

• Why not a, b⎡⎣ ⎤⎦ ? We will examine the endpoints, anyway, and two-sided derivatives are automatically “DNE” there.

Step 2. Evaluate f at the CNs from Step 1. That is, for each CN c from Step 1, evaluate f c( ) .

Step 3. Evaluate f a( ) and f b( ) , the function values at the endpoints.

Step 4. The candidates for absolute extrema are the CNs (if any) from Step 1 and the endpoints a and b. A candidate with the highest function value is an A.Max. of f on a, b⎡⎣ ⎤⎦ . A candidate with the lowest function

value is an A.Min. of f on a, b⎡⎣ ⎤⎦ .


Example 9 (“EVT Method”)

Let f x( ) = x1/3 − x + 4 on the restricted domain −1, 8⎡⎣ ⎤⎦ .

§ Solution

Step 1. Find all CNs of f in −1, 8( ) .

• ′f x( ) = 1

3x−2/3 −1 = 1

3x2/3 −1 = 1− 3x2/3

3x2/3

• ′f does not exist (DNE) at x = 0 . 0 is a CN in −1, 8( ) .

•• WARNING 4: Interval check. Make sure to check that the CNs are in the intended interval, here −1, 8( ) .

•• WARNING 5: Hiding restrictions. When simplifying

′f x( ) , beware of hiding domain restrictions. For example, if

1

x−1 is rewritten as x, the restriction x ≠ 0( ) is hidden.

• Solve ′f x( ) = 0 on −1, 8( ) .

•• Assume x ≠ 0 , since we already know ′f 0( ) does not exist (DNE). It helped to address the “DNE” case first.

′f x( ) = 0

1− 3x2/3

3x2/3 = 0

1− 3x2/3 = 0 Remember, x ≠ 0.( )1= 3x2/3

x2/3 = 13

x2/3( )3= 1

3⎛⎝⎜

⎞⎠⎟

3

Fortunately, 3 is odd. Raising both sides to an even exponent would have obligated us to check our tentative solutions.


(cont.)

x2 = 127

x = ± 127

Approximate here?( )

x = ± 1

3 3⋅ 3

3

x = ± 39

x ≈ ± 0.192

We will use the approximations x ≈ ± 0.192 , which are both CNs in −1, 8( ) .

• The [approximate] CNs of f in −1, 8( ) are: 0, − 0.192 , and 0.192.

Steps 2-4. Evaluate f at the CNs from Step 1 and the interval endpoints −1 and 8. Compare function values and identify A.Max. and A.Min.

x f x( ) Absolute Extrema

a = −1 f −1( ) = 4

− 0.192 f −0.192( ) ≈ 3.615

0 f 0( ) = 4

0.192 f 0.192( ) ≈ 4.385 A.Max. at about 0.192

b = 8 f 8( ) = −2 A.Min. at 8

The A.Max.Pt. is at about 0.192, 4.385( ) .

The A.Min.Pt. is at 8, −2( ) .

§


PART E: FINDING CRITICAL NUMBERS (CNs)

Example 10 (Finding CNs)

Let f x( ) = 3x +1( ) x2 − 9 . Find the critical numbers (CNs) of f .

§ Solution

Step 1. Find Dom f( ) .

• f x( ) is real ⇔ x2 − 9 ≥ 0 .

• Dom f( ) is the solution set for this inequality.

• See the Precalculus notes, Section 2.7: Nonlinear Inequalities, Example 1.

• Dom f( ) = −∞, −3( ⎤⎦∪ 3,∞⎡⎣ ) = x ∈ x ≥ 3{ } .

If you are unable to find the domain, at least check to see if your proposed CNs are in the domain. That is, for each proposed CN c, check to see that f c( ) is real.

Step 2. Find ′f x( ) .

f x( ) = 3x +1( ) x2 − 9 . Use the Product Rule of Differentiation.

′f x( ) = Dx 3x +1( )⎡⎣ ⎤⎦ ⋅ x2 − 9( ) +

3x +1( ) ⋅ Dx x2 − 9( )1/2⎡⎣

⎤⎦( )

= 3[ ]⋅ x2 − 9( ) +

3x +1( ) ⋅ 12

x2 − 9( )−1/2 2 x( )⎛⎝⎜

⎞⎠⎟

= 3 x2 − 9 +x 3x +1( )x2 − 9

= 3 x2 − 91

⋅ x2 − 9x2 − 9

+x 3x +1( )x2 − 9

=3 x2 − 9( ) + x 3x +1( )

x2 − 9

(Section 4.1: Extrema of a Function and the “EVT”) 4.1.13 (cont.)

= 3x2 − 27 + 3x2 + x

x2 − 9

= 6x2 + x − 27x2 − 9

Step 3. Where in Dom f( ) is ′f “DNE”?

Dom f( ) = −∞, −3( ⎤⎦∪ 3,∞⎡⎣ ) . The only numbers in Dom f( ) that make ′f “DNE” are −3 and 3.

−3 and 3 are CNs of f .

Step 4. Where in Dom f( ) is ′f x( ) = 0?

We will consider the domains of f and ′f later (ignore for now).

′f x( ) = 0

6x2 + x − 27

x2 − 9= 0

6x2 + x − 27 = 0

x = −1± 64912

by the Quadratic Formula( )

x ≈ −2.21 or x ≈ 2.04

However, −2.21 and 2.04 are not in Dom f( ), or −∞, −3( ⎤⎦∪ 3,∞⎡⎣ ) ,

so they cannot be CNs.

• WARNING 6: Domain check. We must reject any proposed CNs that are not in Dom f( ) .

Step 5. List the CNs of f .

The CNs of f are −3 and 3.


Here is the graph of y = f x( ) :

• WARNING 7: CNs can be at endpoints. Interval endpoints such as 3 and

−3 can be CNs, but graphical endpoints such as 3, 0( ) and −3, 0( ) cannot be L.Max./Min. Pts. (if we require continuity). However, graphical endpoints can be A.Max./Min. Pts. (not here, but see Examples 8 and 9). §


Let f x( ) = sec x . Find the critical numbers (CNs) of f .

§ Solution


• f x( ) = sec x = 1

cos x.

• f x( ) is real ⇔ cos x ≠ 0 .

•• TIP 1: Domain optional? If you are unable to find the domain, this observation turns out to be sufficient for us to proceed.

• Dom f( ) = x ∈ x ≠ π

2+πn n∈( )⎧

⎨⎩

⎫⎬⎭

.



′f x( ) = sec x( ) tan x( ) = 1

cos x⎛⎝⎜

⎞⎠⎟

sin xcos x

⎛⎝⎜

⎞⎠⎟

.

This form will help in Steps 3 and 4!


• ′f x( ) “DNE” ⇔ cos x = 0 ⇔ f x( ) “DNE” (is undefined).

• This means that any real value of x that makes ′f x( ) “DNE” is not

in Dom f( ) ; therefore, it cannot be a CN.

• f has no CNs from this case.


′f x( ) = 0 ⇔ sin x = 0 and cos x ≠ 0( ) ⇔ sin x = 0 See Note below.( ) ⇔ x = πn n∈( )

• Note: If x satisfies sin x = 0 , then it is automatically true that cos x ≠ 0 . sin x and cos x cannot be simultaneously 0; that is, they cannot be 0 for the same real value of x. Here are several reasons:

•• If they could be simultaneously 0, then we would have a counterexample to the Pythagorean Identity sin

2 x + cos2 x = 1.

•• The point 0, 0( ) does not lie on the Unit Circle.

•• The graphs of y = sin x and y = cos x have no x-intercepts (corresponding to real zeros) in common.


Step 5. Write the CNs of f in set-builder form.

The CNs of f are given by:

x ∈ x = πn n∈( ){ } .

Here is the graph of y = sec x :

• The CNs correspond to the indicated L.Min. and L.Max. Pts. • The vertical asymptotes (“VAs”) correspond to the x-values that make f x( ) and ′f x( ) “DNE.”

• The graph of y = sec x has no x-intercepts, indicating that sec x has no real zeros. This fact will prove useful in Example 12. §



Let f x( ) = tan x . Find the critical numbers (CNs) of f .

§ Solution


• f x( ) = tan x = sin x

cos x.

• As in Example 11, f x( ) is real ⇔ cos x ≠ 0 .

• Dom f( ) = x ∈ x ≠ π

2+πn n∈( )⎧

⎨⎩

⎫⎬⎭

.


′f x( ) = sec2 x = 1

cos2 x.

This form will help in Steps 3 and 4!


• ′f x( ) “DNE” ⇔ cos x = 0 ⇔ f x( ) “DNE” (is undefined), as in Example 11. • f has no CNs from this case.


• f has no CNs from this case, either.

• From the end of Example 11, sec x is never 0, so sec2 x is never 0.

• Also, the fraction

1cos2 x

is never 0, because 1 is never 0.

Step 5. State: f has no CNs.


Here is the graph of y = tan x :

• The tangent line at 0, 0( ) is not horizontal, as it was for the graph of

y = x3 . Its slope is 1, because ′f 0( ) = sec2 0( ) = 1 . §

CHAPTER 4: Applications of Derivatives · Applications of Derivatives 4.1: Extrema of a Function...

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