Chapter 3

The mean value theorem and curve sketching

3.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST

In sketching the graph of a function it is very useful to know where it rises and where it falls. The graph shown in Figure1 rises from A to B, falls from B to C, and rises again from C to D.

Figure 1

21 xx )()( 21 xfxf

(1) DEFINITION A function f is called increasing on an interval I if

whenever in I .

A function that is increasing or decreasing on I is called monotonic on I.

TEST FOR MONOTONIC FUNCTIONS Suppose f is continuous on [a, b] and differentiable on (a,b).(a) If f'(x) > 0 for all x in (a,b), then f is increasing on [a, b].(b) If f'(x) < 0 for all x in (a, b), then f is decreasing on [a, b].

PROOF (a) Let x1 and x2 be any two numbers in [a, b] with x1<x2 . Then f is continuous on [x1, x2] and differentiable on (x1, x2), so by the Mean Value Theorem there is a number c between x1 and x2 such that

(2)

Now f '(c) > 0 by assumption and x2 -x1> 0 because x1<x2 . Thus the right side of Equation 2 is positive, and so

This shows that f is increasing on [a,b].

It is similar to prove (b).

))(()()( 1212 xxcfxfxf

).()(or 21 xfxf 0)()( 12 xfxf

Example 1 Find where the function f (x) = 3x4 - 4x3 - 12x2 +5 is increasing and where it is decreasing.

(c) If f ' does not change sign at c (that is f ' is positive on both sides of c or negative on both sides), then f has no local extremum at c.

(b) If f ' changes from negative to positive at c then f has a local minimum at c.

(a) If f ' changes from positive to negative at c, then f has a local maximum at c.

THE FIRST DERIVATIVE TEST Suppose that c is a critical number of a continuous function f.

Example 3 Find the local and absolute extreme values of the function f(x)= x3(x -2)2, -1 3. Sketch its graph.

f(6/5)=1.20592 is a local maximum; f(2)=0 is a local minimum; absolute maximum value is f(3)=27; absolute minimum value is f(-1)=-9. The sketched in Figure 6.

x

3.4 CONCAVITY AND POINTS OF INFLECTION

Figure 1

Figure 2

(a) Concave upward (b) Concave downward

(1) DEFINITION If the graph of f lies above all of its tangents on an interval I, then it is called concave upward on I. If the graph of f lies below all of its tangents on f , it is called concave downward on I.

Figure 3 shows the graph of a function that is concave upward (abbreviated CU) on the intervals (b, c), (d, e), and (e,p) and concave downward (CD) on the intervals (a, b), (c,d), and (p,q)

The equation of this tangent is

we must show that

whenever (See Figure 4.)

THE TEST FOR CONCAVITY Suppose f is twice differentiable on an interval I.(a) If f "(x) > 0 for all x in I, then the graph of f is concave upward on I.(b) If f"(x) < 0 for all x in I, then the graph of f is concave downward on I.

))(()( axafafy

))(()()( axafafxf )( axIx

Figure 4

Proof of (a) First let us take the case where x > a. Applying the Mean Value Theorem to f on the interval [a,.x], we get a number c, with a < c < x, such that(2) f (x) - f(a) == f’ (c) (x - a)

Since f " > 0 on f we know from the Test for Monotonic Functions that f ' is

increasing on I. Thus, since a < c, we have f'(a)<f’(c)

and so, multiplying this inequality by the positive number x -a ,we get (3) Now we add f(a) to both sides of this equality:

f(a) +. f'(a) (x - a) < f(a) + f'(c) (x - a)But from Equation 2 we have f(x) = f(a) + f'(c) (x -a). So this inequality becomes (4) f(x)>f(a) +f '(a)(x-a)

which is what we wanted to prove. .For the case where x < a,we have f'(c) < f'(a), but multiplication by the negative number x - a reverses the inequality, so we get (3) and (4) as before.

))(())(( axcfaxaf

(2) DEFINITION A point P on a curve is called a point of inflection if the curve changes from concave upward to concave downward or from concave downward to concave upward at P.

Example 1 Determine where the curve y=x3 -3x+1 is concave upward and where it is concave downward. Find the inflection points and sketch the curve.

THE SECOND DERIVATIVE TEST Suppose f " is continuous on an open interval that contains c. (a) If f '(c)=0 and f "(c) > 0, then f has a local minimum at c.(b) If f '(c)=0 and f "(c) < 0, then f has a local maximum at c.

Since f '(3) = 0 and f "(3) >0,f (3) = -27 is a local minimum. The point (0,0) is an inflection point since the curve changes from concave upward to concave downward there. Also (2, -16) since the curve changes from concave downward to concave upward there.

Example 2 Discuss the curve y = x4 - 4x3 with respect to concavity, points of inflection, and local extrema. Use this information to sketch the curve.

3.5 Limits at infinity, horizontal asymptotes

Figure 1

(2) DEFINITION Let f be a function defined on some interval ( , a), Then

means that the values of f(x) can be made arbitrarily close to L by taking x sufficiently large negative.

(1) DEFINITION Let f be a function defined on some interval , Then means that the values of f (x) can be made arbitrarily close to L by taking x

sufficiently large. Lxfx

)(lim

Lxf

x

)(lim

(3) DEFINITION The line y = L is called a horizontal asymptote of the curve y = f(x) if either lim f (x) = L or lim f (x) = L

(4) THEOREM If r >0 is a rational number, then

If r > 0 is a rational number such that is defined for all x, then

x x

01

lim rx x

01

lim rx x

)(lim xfx

)(lim xfx

)(lim xfx

)(lim xfx

Infinite limits at infinity

The notation

is used to indicate that the values of f(x) become large as x

becomes large.Similar meanings are attached to the following

symbols:

Example 1 Example 2

(5) DEFINITION Let f be a function defined on some

interval . Then

means that for every there is a corresponding number

N such that whenever x > N.Lxf

x

)(lim

Lxf )(0

),( a

(6) DEFINITION Let f be a function defined on some

interval . Then

means that for every there is a corresponding number N

such that whenever x < N.

),( aLxf

x

)(lim

0

Lxf )(

(7) DEFINITION Let f be a function defined on some interval . Then

means that for every positive M there is a corresponding number N such that whenever x > N.

),( a

)(lim xf

x

M x f) (

3.6 Curve sketching

(1) This section is to discuss how to sketch the graph of a curve. In high school we learnt to plot points and join the points in order.

Example y=x2

(2) In this section we will draw the graph by following

steps:A. Domain The first step is to determine the domain

of the function f.

B. Intercept Find the x- and y-intersepts .

C. Symmetry (i) even

(ii) odd

(iii) periodic

D. Asymptotes: (i) Horizontal asymptotes

(ii)Vertical asymptotes

H. Sketch the curve Using the information in A---G , draw the graph.

(iii)slant asymptotes

E. Intervals of increase or decrease Use the test for

monotonic functions.

F. Local maximum and minimum valuves Find the

critical numbers of f.

G. Concavity and points of inflection Compute f"(x)

and use the test for concavity.

Example Discuss the curve

1

22

2

x

xy

H. Sketch the curve

SolutionA.Domain B. Intercept C. Symmetry

D. Asymptotes E. Intervals of increase or decrease

F. Local maximum and minimum values

G. Concavity and points of inflection

Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical.

If , then the line y=mx+b is called a slant asymptotes, because the vertical distance between the curve y=f(x) and the line y=mx+b approaches 0.

0)]()([lim

bmxxfx

Slant asymptotes

Example Sketch the graph of

1)(

2

3

x

xxf

3.9 Applications to Economics

Cost function )(xC Average function

If the average cost is a minimum, then marginal cost = average cost

(a) Find the cost, average cost, and marginal cost of producing 1000 items, 2000 items, and 3000 items.

Example 1 A company estimates that the cost (in dollars) of producing x items is

(b) At what production level will the average cost be lowest, and what is this minimum average cost?

(a) The average cost function is

The marginal cost function is

C'(x) = 2 + 0.002x

xxx

xCxc 001.02

2600)()(

Solution

(b) To minimize the average cost we must have marginal cost = average cost C'(x) = c(x)

2 + 0.002x = 2600/x+ 2 + 0.001x This equation simplifies to

0.001x = 2600/x so x2 =2600/0.001 = 2,600,000 and To see that this production level actually gives a minimum we note thatc"(x) =5200/x3> 0, so c is concave upward on its entire domain. The minimum

average cost is c(1612) = 2600/1612+2+0.001(1612) = $5.22/item

1612000,600,2 x

Let p(x) be the price per unit that the company can charge if it sells x units. Then p is called the demand function (or price function) and we would expect it to be a decreasing function of x. If x units are sold and the price per unit is p(x), then the total revenue is

R(x) = x p(x) and R is called the revenue function (or sales function). The

derivative R' of the revenue function is called the marginal revenue function and is the rate of change of revenue with respect to the number of units sold.

If x units are sold, then the total profit is P(x) = R(x) - C(x)

and P is called the profit function. The marginal profit function is P', the derivative of the profit function. In order to maximize profit we look for the critical numbers of P, that is, the numbers where the marginal profit is 0. But if

P'(x) = R'(x) - C'(x) = 0 then

R'(x) = C'(x) Therefore

If the profit is a maximum, then marginal revenue = marginal

cost

To ensure that this condition gives a maximum we could use the Second Derivative Test. Note that P"(x) = R"(x) - C"(x) < 0 when R"(x) < C"(x) and this condition says that the rate of increase of marginal revenue is less than the rate of increase of marginal cost. Thus the profit will be a maximum when R'(x) = C'(x) and R"(x) < C"(x)

Example 2 Determine the production level that will maximize the profit for a company with cost and demand functions C(x) = 3800 + 5x - x2/1000 p(x) = 50 - x/100

SOLUTION The revenue function is R(x) = x p(x) = 50x -x2/100 so the marginal revenue function is R'(x) = 50 -x/50 and the marginal cost function is C'(x)=5-x/500Thus marginal revenue is equal to marginal cost when

50-x/50=5-x/500Solving, we get x = 2500To check that this gives a maximum we compute the second derivatives:

R"(x) = -1/50 C"(x) = -1/500Thus R"(x) < C"(x) for all values of x. Therefore a production level of 2500 units will

maximize profits.

3.10 Antiderivatives

We know how to solve the derivative problem: given a function, find its derivative. But many problems in mathematics and its applications require us to solve the inverse of the derivative problem: given a function, find a function F whose derivative is f. If such a function F exists, it is called an antiderivative of f.

(1) DEFINITION A function F is called an antiderivative of f on an interval I if F '(x) = f (x) for all x in I.

(2) THEOREM If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is F(x) + Cwhere C is an arbitrary constant.

(3) Table of antidifferntiation formulas

Some of the most important applications of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing something. In many cases these problems can be reduced to finding the maximum or minimum values of a function. Let us first explain exactly what we mean by maximum and minimum values.

3.1 MAXIMUM AND MINIMUM VALUES

Figure 1

Minimum value f(a) Maximum value f(d)

In Figure 1, if we consider only values of x near b [for instance, if we restrict our attention to the interval (a,c)], then f(b) is the largest of those values of f(x) and is called a local maximum value of f. Likewise, f(c) is called a local minimum value of f because f(c) f(x) for x near c [in the interval (b, d), for instance]. The function f also has a local minimum at e. In general we have the following definition.

Example 1 f(x) = x2

Figure 2 Minimum value, no maximum

Example 2 Please find the absolute maximum, the absolute minimum, local maximum and local minimum of the function

41 18163)( 234 xxxxxf

We have seen that some functions have extreme values, while others do not. The following theorem gives conditions under which a function is guaranteed to possess extreme values.

(3) THE EXTREME VALUE THEOREM If f is continuous on a closed interval [a,b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers c and d in [a, b].

The extreme value theorem is illustrated in the following.

Figure 5

21if 0

10if )(

2

x

xxxf

Example 3 Find the extreme values of

2)( xxf

, 2 0 x, 2 0 x

Example 4 Find the extreme values of

on the different intervals

and

2 0 x . 2 0 x

(4) FERMAT' S THEOREM If f has a local extremum (that is, maximum or minimum) at c, and if f '(c) exists, then f '(c) =0.

(5) DEFINITION A critical number of a function f is a number c in the domain of f such that either f '(c) = 0 or f '(c) does not exist.

(7) To find the absolute maximum and minimum values of a continuous function f on a closed interval [a, b]:

1. Find the values of f at the critical numbers of f in (a, b).

2. Find the values of f(a) and f(b).

3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.

(6) If f has a local extremum at c, then c is a critical number

of f .

Example 5 Find the absolute maximum and minimum values of the function

42

113)( 23 xxxxf

3.2 THE MEAN VALUE THEOREM

We will see that many of the results of this chapter depend on one central fact, which is called the Mean Value Theorem. But to arrive at the Mean Value Theorem we first need the following result.

ROLLE'S THEOREM Let f be a function that satisfies the following three hypotheses:

1. f is continuous on the closed interval [a, b].

2. f is differentiable on the open interval (a,b).

3. f(a)=f(b)

Then there is a number c in (a,b) such that f '(c) = 0.

Figure 1

PROOF There are three cases:

CASE lll f(x) < f(a) for some x in (a,b) [as in Figure l (c) or (d)]

By the Extreme Value Theorem, f has a minimum value in [a, b] and, since f(a) = f(b), it attains this minimum value at a number c in (a,b). Again f '(c) = 0 by Fermat's Theorem.

CASE I f(x) = k, a constant. Then f '(x) = 0, so the number c can be

taken any number in (a,b).

CASE II f(x)> f(a) for some x in (a, b) [as in Figure l(b) or (c)]

By the Extreme Value Theorem (which we can apply by hypothesis 1) f has a maximum value somewhere in [a, b]. Since f (a) = f (b), it must attain this maximum value at a number c in the open interval (a, b). Then f has a local maximum at c and, by hypothesis 2, f is differentiable at c. Therefore f '(c) = 0 by Fermat's Theorem.

SOLUTION First we use the Intermediate Value Theorem (1,5.9) to show that a root exists.

To show that the equation has no other real root we use Rolle's Theorem and argue by contradiction.

Example 1 Prove that the equation x3 + x -1 = 0 has exactly one real root.

THE MEAN VALUE THEOREM

Let f be a function that satisfies the following hypotheses:1. f is continuous on the closed interval [a, b].

2. f is differentiable on the open interval (a, b).

Then there is a number c in (a,b) such that

(1)

or, equivalently,

(2)

ab

afbfcf

)()()(

))(()()( abcfafbf

we can see that it is reasonable by interpreting it geometrically. The slope of the secant line AB is

ab

afbfmab

)()(

PROOF The equation of the line AB can be written as

)()()(

)(asor)()()(

)( axab

afbfafyax

ab

afbfafy

Figure 5

)()()(

)()()()3( axab

afbfafxfxh

ab

afbfxfxh

)()()()(

ab

afbfcfch

)()(

)()(0

ab

afbfcf

)()()(

First we must verify that h satisfies the three hypotheses of Rolle's Theorem

1. The function h is continuous on [a, b] because it is the sum of f and a first-degree polynomial, both are continuous.

3. h(a) =h(b).

Since h satisfies the hypotheses of Rolle's Theorem, that theoremsays there is a number c in (a, b) such that h’ (c) =0. Therefore

and so

2. The function h is differentiable on (a, b) because both f and

the first-degree polynomial are differentiable. In fact, we can

compute h' directly from Equation 4:

Figure 6 illustrates this calculation: the tangent line at this value of c is parallel to the secant line OB.

Figure 6

Example 2 To illustrate the Mean Value Theorem with a specific function, let us consider f (x) = x3 - x , a = 0, b = 2.

The Mean Value Theorem can be used to establish some of the basic facts of differential calculus. One of these basic facts is the following

theorem.

(3) THEOREM If f ' (x) = 0 for all x in an interval (a, b), then f is constant on (a, b)

(4) COROLLARY If f '(x) = g'(x) for all x in an interval (a, b),

then f - g is constant on (a,b); that is , f (x) = g(x) + c where c is a constant.