Calculus with Precalculus

Vladimir V. Kisil

Chapter 0
General Information

This interactive manual is designed for students attending this course in Eastern Mediterranean University. The manual is being regularly updated in order to reflect material presented during lectures.

The manual is available at the moment in HTML with frames (for easier navigation), HTML without frames and PDF formats. Each from these formats has its own advantages. Please select one better suit your needs.

There is on-line information on the following courses:

• Calculus I.

• Calculus II.

• Geometry.

There are other on-line calculus manuals. See for example

E-calculus of D.P. Story.

0.1  Warnings and Disclaimers

• These Web pages are designed in order to help students as a source of additional information. They are NOT an obligatory part of the course.

• The main material introduced during lectures and is contained in Textbook. This interactive manual is NOT a substitution for any part of those primary sources of information.

• It is NOT required to be familiar with these pages in order to pass the examination.

• The entire contents of these pages is continuously improved and updated. Even for material of lectures took place weeks or months ago changes are made.

0.2  Recommended Exercises

Section 1.1. Problems 17, 23, 25

Section 1.2. Problems 17, 27

Section 1.3. Problems 25, 31, 39, 63

Section 1.4. Problems 11, 19, 23, 33

Section 1.5. Problems 15, 49, 55, 65

Section 2.1. Problems 1, 19

Section 2.2. Problems 21, 27

Section 2.3. Problems 21, 53, 55

Section 2.4. Problems 13, 17, 19, 41, 45

Section 2.5. Problems 17, 19, 47, 55

Section 2.6. Problems 11, 17, 27, 33

Section 2.7. Problem 7

Section 2.8. Problems 1, 5

Section 3.1. Problems 7, 25

Section 3.2. Problems 11, 15, 41

Section 3.3. Problems 5, 17, 29

Section 3.4. Problems 7, 13, 23

Section 3.5. Problems 13, 17, 31, 41

Section 3.6. Problems 5, 7, 29

Section 4.1. Problems 9, 17

Section 4.2. Problems 17, 21, 23, 31, 35, 39

Section 4.3. Problems 9, 27, 35

Section 4.4. Problems 9, 27, 33

Section 4.5. Problems 13, 21, 25

Section 4.6. Problems 19, 25, 35, 47, 59

Section 5.1. Problems 15, 23

Section 5.2. Problems 11, 13, 29

Section 5.3. Problems 7, 15, 23, 27, 29

Section 5.4. Problems 3, 14

Section 5.5. Problems 7, 12, 29, 33

Section 6.1. Problems 25, 33

Section 6.2. Problems 37, 41

Section 6.3. Problems 33, 37

Section 6.4. Problems 11, 23, 31, 37

Section 6.5. Problems 35, 43, 45

Section 6.7. Problems 5, 21, 43, 53, 61

Section 6.8. Problems 32, 34

Section 6.9. Problems 9, 17, 63, 65

Section 7.1. Problems 9, 17, 29, 31

Section 7.2. Problems 3, 5, 13, 21

Section 7.3. Problems 5, 9, 11, 17

Section 7.4. Problems 7, 21, 25

Section 7.5. Problems 5, 9, 15

Section 7.7. Problems 13, 15, 53, 63

Section 8.1. Problems 23, 31

Section 8.2. Problems 13, 27, 47

Section 8.3. Problems 5, 11, 15, 25, 27, 39, 43

Section 8.4. Problems 7, 9, 15, 17, 25, 31

Section 8.5. Problems 11, 15, 21, 29

Course Outline

Chapter 1
Limits

1.1  Introduction to Limits

The following is a quote from a letter of Donald E. Knuth to ``Notices of AMS''.

The most important of these changes would be to introduce the O notation and related ideas at an early stage. This notation, first used by Bachmann in 1894 and later popularized by Landau, has the great virtue that it makes calculations simpler, so it simplifies many parts of the subject, yet it is highly intuitive and easily learned. The key idea is to be able to deal with quantities that are only partly specified, and to use them in the midst of formulas.

I would begin my ideal calculus course by introducing a simpler `` A notation,'' which means ``absolutely at most.'' For example, A(2) stands for a quantity whose absolute value is less than or equal to 2. This notation has a natural connection with decimal numbers: Saying that π is approximately 3.14 is equivalent to saying that π = 3.14+A(.005). Students will easily discover how to calculate with A:

10A(2) = A(100) ;   3.14+A(.005)     1+A(0.01)   = 3.14+A(.005)+A(0.0314)+A(.00005) = 3.14+A(0.3645)=3.14+A(.04) .

I would of course explain that the equality sign is not symmetric with respect to such notations; we have 3=A(5) and 4=A(5) but not 3=4, nor can we say that A(5)=4. We can, however, say that A(0)=0. As de Bruijn points out in [1, § 1.2], mathematicians customarily use the = sign as they use the word ``is'' in English: Aristotle is a man, but a man isn't necessarily Aristotle.

The A notation applies to variable quantities as well as to constant ones. For example,

sinx = A(1) ; x = A(x) ; A(x) = xA(1) ; A(x)+A(y) = A(x+y) if x ≥ 0 and y ≥ 0 ;   1+A(t)   2 = 1+3A(t) if t=A(1) .

Once students have caught on to the idea of A notation, they are ready for O notation, which is even less specific. In its simplest form, O(x) stands for something that is CA(x) for some constant C, but we don't say what C is. We also define side conditions on the variables that appear in the formulas. For example, if n is a positive integer we can say that any quadratic polynomial in n is O(n²). If n is sufficiently large, we can deduce that

  n+O(√n )     lnn+γ+O(1/n)   = nlnn+γn+O(1)+O(√nlnn)+O(√n )+O(1/√n ) = nlnn+γn+O(√nlnn) .

I'm sure it would be a pleasure for both students and teacher if calculus were taught in this way. The extra time needed to introduce O notation is amply repaid by the simplifications that occur later. In fact, there probably will be time to introduce the `` o notation,'' which is equivalent to the taking of limits, and to give the general definition of a not-necessarily-strong derivative:

f(x+ε)=f(x)+f′(x)ε+o(ε) .

The function f is continuous at x if

f(x+ε)=f(x)+o(1) ;

and so on. But I would not mind leaving a full exploration of such things to a more advanced course, when it will easily be picked up by anyone who has learned the basics with O alone. Indeed, I have not needed to use ``o'' in 2200 pages of The Art of Computer Programming, although many techniques of advanced calculus are applied throughout those books to a great variety of problems.

Students will be motivated to use O notation for two important reasons. First, it significantly simplifies calculations because it allows us to be sloppy-but in a satisfactorily controlled way. Second, it appears in the power series calculations of symbolic algebra systems like Maple and Mathematica, which today's students will surely be using.

[1]: N. G. de Bruijn, Asymptotic Methods in Analysis (Amsterdam: North-Holland, 1958).

[2]: R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics (Reading, Mass.: Addison-Wesley, 1989).

1.2  Definition of Limit

Definition 1 Let a function f be defined on an open interval containing a, except possible at a itself, and let L be a real number. The statement

lim x → a  f(x) = L

(1.1)

(L is the limit of function f at a) means that for every ε > 0, there is a δ > 0 such that if 0 < | x−a | < δ, then | f(x)−L | < ε.

We introduce a special kind of limits:

Definition 2 We say that variable y is o(z) ( o notation) in the neighborhood of a point a if for every ε > 0 there exists δ > 0 such that if 0 < | x−a | < δ, then | y | < ε| z |.

Then we could define limit of a function as follows

Definition 3 A function f(x) has a limit L at point a if for x ≠ 0

f(a+x)=L+o(1).

Theorem 4 If

lim x → a  f(x)=L

and L > 0, then there is an open interval (a−δ,a+δ) containing a such that f(x) > 0 for every x in (a−δ,a+δ), except possibly x=a.

Exercise 5 Verify limits using Definition

lim x → 3  (5x+3) = 18; lim x → 2  (x2+1) = 5.

1.3  Techniques for Finding Limits

Theorem 1 The following basic limits are:

lim x → a  c = c; (1.2) lim x → a  x = a. . (1.3)

Theorem 2 If both limits

lim x → a  f(x) and lim x → a  g(x)

exist, then

lim x → a  [ f(x)±g(x)] = lim x → a  f(x) ± lim x → a  g(x); (1.4) lim x → a  [ f(x)·g(x)] = lim x → a  f(x) · lim x → a  g(x). (1.5) lim x → a     f(x) g(x)   = lim x → a  f(x) lim x → a  g(x) , (1.6)

provided

lim x → a  g(x) ≠ 0.

Corollary 3

1. From formulas (2.2) and (2.5) follows
   

   lim x → a  [ cf(x)]=c lim x → a  f(x).

2. If a, m, b are real numbers then
   

   lim x → a  (mx+b)=ma+b.

3. If n is a positive integer then
   

   lim x → a  xn = an. lim x → a  [f(x)]n =   lim x → a  f(x)   n   ,

   provided there exists the limit
   

   lim x → a  f(x).

4. If f(x) is a polynomial function and a is a real number, then
   

   lim x → a  f(x) = f(a).

5. If q(x) is a rational function and a is in the domain of q, then
   

   lim x → a  q(x) = q(a).

Exercise 4 Find limits

lim x → 2  √3; lim x → −3  x; lim x → 4  (3x+1); lim x → −2  (2x−1)15; lim x → 1/2   2x2+5x−3 6x2−7x+2 ; lim x → −2   x2+2x−3 x2+5x+6 ; lim x → π  5  √      x−π x+π   ; lim h → 0     1 h       1 √ 1+h −1   

Theorem 5 If a > 0 and n is a positive integer, or if a ≤ 0 and n is an odd positive integer, then

lim x → a  n √   x   = n √   a   .

Theorem 6 [Sandwich Theorem] Suppose f(x) ≤ h(x) ≤ g(x) for every x in an open interval containing a, except possibly at a. If

lim x → a  f(x) = L= lim x → a  g(x),

then

lim x → a  h(x) = L.

Exercise 7 Find limits

lim x → 0  x2 sin  1 x2 ;        lim x → π/2  (x−  π 2 ) cosx.

Remark 8 All such type of result could be modified for one-sided limits.

1.4  Limits Involving Infinity

Definition 1 Let a function f be defined on an infinite interval (c,∞) (respectively (−∞, c)) for a real number c, and let L be a real number. The statement

lim x → ∞  f(x) = L        ( lim x → −∞  f(x) = L)

means that for every ε > 0 there is a number M such that if x > M (x < M), then | f(x)−L | < ε.

Theorem 2 If k is a positive rational number and c, then

lim x → ∞   c xk =0 and lim x → −∞   c xk =0

Definition 3 Let a function f be defined on an open interval containing a, except possibly at a itself. The statement

lim x → a  f(x) = ∞

means that for every M > 0, there is a δ > 0 such that if 0 < | x−a | < δ, then f(x) > M.

Exercise 4 Find limits

lim x → −1   2 x2 x2−x−1 lim x → 9/2   3x2 (2x−9)2 (1.7) lim x → ∞   −x3+2x 2x2−3 lim x → −∞   2x2−x+3 x3+1 (1.8)

1.5  Continuous Functions

Definition 1 A function f is continuous at a point c if

lim x → c  f(x) = f(c).

If function is not continuous at c then it is discontinuous at c, or that f has discontinuity at c. We give names to the following types of discontinuities:

1. Removable discontinuity:
   

   lim x → c  f(x) ≠ f(c).

2. Jump discontinuity:
   

   lim x → +c  f(x) ≠ lim x → −c  f(x).

3. Infinite discontinuity:
   

   lim x →±c  f(x) = ±∞.

Exercise 2 Classify discontinuities of

f(x) =      −x2 if x < 1 2 if x=1 (x−2)−1 if x > 1.

Theorem 3

1. A polynomial function f is continuous at every real point c.

2. A rational function q=f/g is continuous at every number except the numbers c such that g(c)=0.

PROOF. The proof follows directly from Corollary 2.3.3. ^[¯]

Exercise 4 Find all points at which f is discontinuous

f(x)=  x−1 x2+x−2 .

Definition 5 If a function f is continuous at every number in an open interval (a,b) we say that f is continuous on the interval (a,b). We say also that f is continuous on the interval [a,b] if it is continuous on (a,b) and

lim x →±+a  f(x) = a        lim x →±−b  f(x)=b.

Theorem 6 If two functions f and g are continuous at a real point c, the following functions are also continuous at c:

1. the sum f+g.

2. the difference f−g.

3. the product fg.

4. the quotient f/g, provided g(c) ≠ 0.

PROOF. Proof follows directly from the Theorem 2.3.2. ^[¯]

Exercise 7 Find all points at which f is continuous

f(x) = √ 9−x √ x−6        f(x)=  x−1 √ x2−1 .

Theorem 8

1. If
   

   lim x → c  g(x) = b

   and f is continuous at b, then
   

   lim x → c  f(g(x)) = f(b) = f( lim x → c  g(x)).

2. If g is continuous at c and if f is continuous at g(c), then the composite function f °g is continuous at c.

Exercise 9 Suppose that

f(x) =    c2x, if x < 1 3cx−2 if x ≥ 1.

Determine all c such that f is continuous on R.

Theorem 10 [Intermediate Value Theorem] If f is continuous on a closed interval [a,b] and if w is any number between f(a) and f(b), then there is at least one point c ∈ [a,b] such that f(c)=w.

Corollary 11 If f(a) and f(b) have opposite signs, then there is a number c between a and b such that f(x)=0.

Exercise 12 Let f(x)=x⁷+3x+2 and g(x)=−10x⁶+3x²−1. Show that there is a solution of the equation f(x)=g(x) on the interval (−1,0).

Chapter 2
Derivative

2.1  Tangent Lines and Rates of Changes

If a body pass a distance d within time t then average velocity is defined as

vav =  d t .

If the time interval t → 0 then we obtain instantaneous velocity

va = lim t → 0   s(a+t) − s(a) t . (2.4)

These and many other examples lead to the notion of derivative.

2.2  Definition of Derivative

I would define the derivative by first defining what might be called a ``strong derivative'': The function f has a strong derivative f′(x) at point x if

f(x+ε)=f(x)+f′(x)ε+O(ε2)

whenever ε is sufficiently small. The vast majority of all functions that arise in practical work have strong derivatives, so I believe this definition best captures the intuition I want students to have about derivatives.

Definition 1 The derivative of a function f is the function f′ whose value at x is given by

f′(x) = lim x → 0   f(x+h)−f(x) h , (2.5)

provided the limit exists.

2.3  Techniques of Differentiation

It is also follows that

Theorem 1 If a function f is differentiable at a, then f is continuous at a.

Exercise 2 Give an example of a function f which is continuous at point x=0 but is not differentiable there.

It could be similarly proven the following

Theorem 3 Let f and g be differentiable functions at point c then the following functions are differentiable at c also and derivative could be calculated as follows:

1. sum and difference (f±g)′(x) = f′(x) ±g′(x).

2. product (fg)′(x)=f′(x)g(x)+f(x)g′(x).

3. fraction
   

      f g   ′(x) =  f′(x)g(x)−f(x)g′(x) g2(x) .

2.4  Derivatives of the Trigonometric Functions

Theorem 1 The following limits are:

lim φ→ 0  sinφ = 0; (2.8) lim φ→ 0  cosφ = 1; (2.9) lim φ→ 0   sin φ φ = 1. (2.10)

PROOF. Only the last limit is non-trivial. It follows from obvious inequalities

sinφ < φ < tanφ

and the Sandwich Theorem 2.3.6. ^[¯]

Corollary 2 The following limit is

lim x → 0   1−cosφ φ =0.

Exercise 3 Find following limits if they exist:

lim t → 0   4t2+3t sint t2 ; lim t → 0   cost 1−sint ; lim x → 0  x cotx; lim x → 0   sin3x sin5x .

We are able now to calculate derivatives of trigonometric functions

Theorem 4

sin′x = cosx; cos′x = sinx; (2.11) tan′x = sec2 x; cot′x = csc2 x; (2.12) sec′x = secx tanx; csc′x = −cscx cotx. (2.13)

PROOF. The proof easily follows from the Theorem 3.4.1 and trigonometric identities on page pageref. ^[¯]

Exercise 5 Find derivatives of functions

y=  sinx x ; y=  1−cosz 1+cosz ; y=  tanx 1+x2 ; y=csct sint.

2.5  The Chain Rule

Exercise 1 Find derivatives

y=   z2−  1 z2   3   ; y=  x4−3x2+1 (2x+3)4 ; y= 3 √   8r3+27   ; y=(7x+ √   x2+3   )6.

2.6  Implicit Differentiation

Exercise 1 Find the slope of the tangent lines at given points

1. x²y+siny=2π at P(1,2π).

2. 2x³−x²y+y³−1=0 at P(2,−3).

2.7  Related Rates

Exercise 1

1. If S=z³ and dz/dt=−2 when z=3, find dS/dt.

2. If x²+3y²+2y=10 and dx/dt=2 when x=3 and y=−1, find dy/dt.

Exercise 2 Suppose a spherical snowball is melting and the radius is decreasing at a constant rate, changing from 12 in. to 8 in. in 45 min. How fast was the volume changing when the radius was 10 in.?

2.8  Linear Approximations and Differentials

An application of this formulas connected with estimation of errors of measurements:

	Exact value	Approximate value
Absolute error	∆y = y−y0	dy=f′(x0) ∆x
Relative error	[ ∆y/(y0)]	[ dy/(y0)]
Percentage error	[ ∆y/(y0)] ×100%	[ d y/(y0)] ×100%

Exercise 1 Use linear approximation to estimate f(b):

1. f(x)=−3x²+8x−7; a=4, b=3.96.

2. f(φ)=cscφ+cotφ, a=45^°, b=46^°.

Exercise 2 Find ∆y, dy, dy−∆y for y=3x²+5x−2.

Chapter 3
Appliactions of Derivative

3.1  Extrema of Functions

Definition 1 Let f is defined on S ⊂ R and c ∈ S.

1. f(c) is the maximum value of f on S if f(x) ≤ f(c) for every x ∈ S.

2. f(c) is the minimum value of f on S if f(x) ≥ f(c) for every x ∈ S.

Maximum and minimum are called extreme values, or extrema of f. If S is the domain of f then maximum and minimum are called global or absolute .

Exercise 2 Give an examples of functions which do not have minimum or maximum values.

The important property of continuous functions is given by the following

Theorem 3 If f is continuous on [a,b], then f takes on a maximum and minimum values at least once in [a,b].

Sometimes the following notions are of great importance

Definition 4 Let c be a number in domain of f.

1. f(c) is the local maximum f if there is an open interval (a,b) such that c ∈ (a,b) and f(x) ≤ f(c) for every x ∈ (a,b) in the domain of f.

2. f(c) is the local minimum f if there is an open interval (a,b) such that c ∈ (a,b) and f(x) ≥ f(c) for every x ∈ (a,b) in the domain of f.

The local extrema could be determined from values of derivative:

Theorem 5 If f has a local extremum at a number c in an open interval, then either f′(c)=0 or f′(c) do not exist.

PROOF. The proof follows from the linear approximation of ∆f by f′(c)∆x: if f′(c) exists and f′(c) ≠ 0 then in an open interval around c there is values of f whose greater and less than f(c). ^[¯]

The direct consequence is:

Corollary 6 if f′(c) exists and f′(c) ≠ 0 then f(c) is not a local extremum of c.

Critical numbers of f are whose points c in the domain of f where either f′(c)=0 or f′(c) does not exist.

Theorem 7 If a function f is continuous on a [a,b] and has its maximum or minimum values at a number c ∈ (a,b), then either f′(c)=0 or f′(c) does not exist.

So to determine maximum and minimum values of f one should accomplish the following steps

1. Find all critical points of f on (a,b).

2. Calculate values of f in all critical points from step 1.

3. Calculate the endpoint values f(a) and f(b).

4. The maximal and minimal values of f on [a,b] are the largest and smallest values calculated in 2 and 3.

Exercise 8 Find extrema of f on the interval

1. y=x⁴−5x+4; [0,2].

2. y=(x−1)^2/3−4; [0,9].

Exercise 9 Find the critical numbers of f

y = 4x3+5x2−42x+7; y = 3 √   x2−x−2   ; y = (4z+1) √   z2−16   ; y = 8cos3t−3sin2x−6x.

3.2  The Mean Value Theorem

Theorem 1 [Rolle's Theorem] If f is continuous on a closed interval [a,b] and differentiable on the open interval (a,b) and if f(a)=f(b), then f′(c)=0 for at least one number c in (a,b).

PROOF. There is two possibilities

1. f is constant on [a,b] then f′(x)=0 everywhere.

2. f(x) is not constant then it has at least one extremum point c (Theorem 4.1.3) which is not the end point of [a,b], then f′(c)=0 (Theorem 4.1.5).

Exercise 2 Shows that f satisfy to the above theorem and find c:

1. f(x)=3x²−12x+11; [0,4].

2. f(x)=x³−x; [−1,1].

Rolle's Therem is the principal step to the next

Theorem 3 [Mean Value Theorem or Lagrange's Theorem] If f is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c ∈ (a,b) such that

f′(c)=  f(b)−f(a) b−a

or, equivalently,

f(b)−f(a)=f′(c)(b−a).

PROOF. The proof follows from applivcation of the Rolle's Theorem 4.2.1 to the function

g(x)=f(x) −  f(b)−f(a) b−a (x−a).

^[¯]

We start applications of Mean Value Theorem by two corrolaries:

Corollary 4 If f′(x)=0 for all x in some interval I, then there is a constant C such that f(x)=C for all x in I.

Corollary 5 If f′(x)=g′(x) foar all x ∈ I, then there is a constant C such that f(x)=g(x)+C.

Exercise 6 Shows that f satisfy to the above theorem and find c:

1. f(x)=5x²−3x+1; [1,3].

2. f(x)=x^2/3; [−8,8].

3. f(x)=x³+4x; [−3,6].

Exercise 7 Prove: if f continuous on [a,b] and if f′(x)=c there, then f(x)=cx+d for a d ∈ R.

Exercise 8 Prove:

| sinu− sinv | ≤ | u−v |.

3.3  The First Derivative Test

Theorem 1 Let f be continuous on [a,b] and differentiable on (a,b).

1. If f′(x) > 0 for every x in (a,b), then f is increasing on [a,b].

2. If f′(x) < 0 for every x in (a,b), then f is decreasing on [a,b].

PROOF. For any numbers x₁ and x₂ in (a,b) we could write using the Mean Value Theorem :

f(x1)−f(x2) = f′(c) (x1−x2).

Then for x₁ > x₂ we will have f(x₁) > f(x₂) if f′(c) is positive and f(x₁) < f(x₂) if f′(c) is negative. ^[¯]

To check the sign of continuous derivative in an interval [a,b] which does not contain critical points it is enough to verify it for a single point k ∈ (a,b) (See Intermediate Value Theorem ). We shall call f′(k) a test value.

Test 2 [First Derivative Test] Let c is a critical number for f, f is continuous in an open interval I containing c and differentiable in I, except possibly at c itself. Then from the above theorem it follows that

1. If f′ changes from positive to negative at c, then f(c) is a local maximum of f.

2. If f′ changes from negative to positive at c, then f(c) is a local minimum of f.

3. If f′(x) > 0 or f′(x) < 0 for all x ∈ I, x ≠ c, then f(c) is not a local extremum of of c.

Exercise 3 Find the local extrema of f and intervals of monotonicity, sketch the graph

1. y=2x³+x²−20x+1.

2. y=10x³(x−1)².

3. y=x(x²−9)^1/2.

4. y=x/2−sinx.

5. y=2cosx+cos2x.

Exercise 4 Find local extrema of f on the given interval

1. y=cot² x+2cotx [π/6,5π/6].

2. y=tanx − 2secx [−π/4, π/4].

3.4  Concavity and the Second Derivative Test

Definition 1 Let f be differentiable on an open interval I. The graph of f is

1. concave upward on I if f′ is increasing on I;

2. concave downward on I if f′ is decreasing on I.

If a graph is concave upward then it lies above any tangent line and for downward concavity it lies below every tangent line.

Test 2 [Test for Concavity] If the second derivative f′′ of f exists on an open interval I, then the graph of f is

1. concave upward on I if f′′(x) > 0 on I;

2. concave downward on I if f′′(x) < 0 on I.

Definition 3 A point (c,f(c)) on the graph of f is a point of inflection if the following conditions are satisfied:

1. f is continuous at c.

2. There is an open interval (a,b) containing c such that the graph has different types of concavity on (a,c) and (c,b).

Test 4 [Second Derivative Test] Suppose that f is differentiable on an open interval containing c and f′(c)=0.

1. If f′′(c) < 0, then f has a local maximum at c.

2. If f′′(c) > 0, then f has a local minimum at c.

Exercise 5 Find the local extrema of f, intervals of concavity and points of inflections.

1. y=2x⁶−6x⁴.

2. y=x^1/5−1.

3. y=6x^1/2−x^3/2.

4. y=cosx −sinx.

5. y=x+2cosx.

3.5  Summary of Graphical Methods

1. Find domain of f.

2. Estimate range of f and determine region where f is negative and positive.

3. Find region of continuity and classify discontinuity (if any).

4. Find all x- and y-intercepts.

5. Find symmetries of f.

6. Find critical numbers and local extrema (using the First of Second Derivative Test), region of monotonicity of f.

7. Determine concavity and points of inflections.

8. Find asymptotes.

Exercise 1 Sketch the graphs:

f(x) =  2x2−x−3 x−2 ; f(x) =  8−x3 2x2 ; f(x) =  −3x √ x2+4 ; f(x) = x3+  3 x ; f(x) =  −4 x2+1 ; f(x) = | x3−x |; f(x) = | cosx | + 2.

3.6  Optimization Problems. Review

Exercise 1

1. Find the minimum value of A if A=4y+x², where (x²+1)y=324.

2. Find the minimum value of C if C=(x²+y²)^1/2, where xy=9.

Exercise 2 A metal cylindrical container with an open top is to hold 1 m³. If there is no waste in construction, find the dimension that require the least amount of material.

Exercise 3 Find the points of the graph of y=x³ that is closest to the point (4,0).

Chapter 4
Integrals

4.1  Antiderivatives and Indefinite Integrals

Definition 1 A function F is an antiderivative of the function f on an interval I if F′(x)=f(x) for every x ∈ I.

It is obvious that if F is an antiderivative of f then F(x)+C is also antiderivative of f for any real constant C. C is called an arbitrary constant. It follows from Mean Value Theorem that every antiderivative is of this form.

Theorem 2 Let F be an antiderivative of f on an interval I. If G is any antiderivative of f on I, then

G(x)=F(x)+C

for some constant C and every x ∈ I.

Exercise 3 The above Theorem may be false if the domain of f is different from an interval I. Give an example.

Definition 4 The notation

⌠ ⌡ f(x) dx = F(x) + C,

where F′(x) = f(x) and C is an arbitrary constant, denotes the family of all antiderivatives of f(x) on an interval I and is called indefinite integral .

Theorem 5

⌠ ⌡  d dx (f(x)) dx = f(x)+C; (4.1)  d dx   ⌠ ⌡ f(x) dx   = f(x). (4.2)

The above Theorem allows us to construct a primitive table of antiderivatives from the tables of derivatives:

Derivative Indefinite Integral  d dx (x) = 1 ⌠ ⌡ 1 dx = ⌠ ⌡  dx = x+C  d dx (  xr+1 r+1 ) = xr    (r ≠ −1) ⌠ ⌡ xr dx =  xr+1 r+1 +C    (r ≠ −1)  d dx (sinx ) = cosx ⌠ ⌡ cosx dx = sinx +C  d dx (−cosx ) = sinx ⌠ ⌡ sinx dx = −cosx +C  d dx (tanx) = sec2 x ⌠ ⌡ sec2 x dx = tanx +C  d dx (−cotx) = csc2 x ⌠ ⌡ csc2 x  dx = −cotx +C  d dx (secx ) = secx tanx ⌠ ⌡ secx tanx  dx = secx +C  d dx (−cscx ) = cscx cotx ⌠ ⌡ cscx cotx  dx = −cscx +C

(4.3)

Theorem 6

⌠ ⌡ c f(x) dx = c ⌠ ⌡ f(x)  dx (4.4) ⌠ ⌡ [f(x) ±g(x)] dx = ⌠ ⌡ f(x) dx± ⌠ ⌡ g(x) dx (4.5)

Exercise 7 Find antiderivatives of the following functions

9t2−4t+3; 4x2−8x+1;  1 z3 −  3 z2 ; √   u3   −  1 2 u−2 +5; (3x−1)2;  (t2+3)2 t6 ;  7 cscx ; −  1 5 sinx;  1 sin2 t .

A Differential equation is an equation that involves derivatives or differentials of an unknown function. Additional values of f or its derivatives are called initial conditions.

Exercise 8 Solve the differential equations subject to the given conditions

f′(x) = 12x2−6x+1,        f(1)=5; (4.6) f′′(x) = 4x−1,        f′(2)=−2,     f(1)=3; (4.7)  d2y dx2 = 3sinx−4cosx,        y=7, y′=2 if x=0. (4.8)

4.2  Change of Variables in Indefinite Integrals

Theorem 1 If F is an antiderivative of f, then

⌠ ⌡ f(g(x))g′(x) dx = F(g(x))+C.

If u=g(x) and du=g′(x) dx, then

⌠ ⌡ f(u) du = F(u) +C.

Exercise 2 Find the integrals

⌠ ⌡ x (2x2+3)10 dx; ⌠ ⌡ x2 3 √   3x3+7    dx; (4.9) ⌠ ⌡   1+  1 x   −3      1 x2   dx; ⌠ ⌡  t2+t (4−3t2−2t3)4 dt; (4.10) ⌠ ⌡  sin2x √ 1−cos2x dx; ⌠ ⌡ sin3 x cosx dx. (4.11)

4.3  Summation Notation and Area

Definition 1 We use the following summation notation:

n ∑ k=1  ak = a1+a2+…+an.

Theorem 2

n ∑ k=1  c = cn; (4.12) n ∑ k=1  (ak ±bk) = n ∑ k=1  ak ± n ∑ k=1  bk; (4.13) n ∑ k=1  cak = c   n ∑ k=1  ak   . (4.14)

Theorem 3

n ∑ k=1  k =  n(n+1) 2 ; (4.15) n ∑ k=1  k2 =  n(n+1)(2n+1) 6 ; (4.16) n ∑ k=1  k3 =    n(n+1) 2   2   . (4.17)

This sum will help us to find inscribed rectangular polygon and circumscribed rectangular polygon.

Exercise 4 Find the area under the graph of the following functions:

1. y=2x+3, from 2 to 4.

2. y=x²+1, from 0 to 3.

4.4  The Definite Integral

4.5  Properties of the Definite Integral

Theorem 1 If c is a real number, then

⌠ ⌡ b a  c dx = c(b−a).

Theorem 2 If f is integrable on [a,b] and c is any real number, then cf is integrable on [a,b] and

⌠ ⌡ b a  cf(x) dx = c ⌠ ⌡ b a  f(x) dx.

Theorem 3 If f and g are integrable on [a,b], then f±g is also integrable on [a,b] (a > b) and

⌠ ⌡ b a  [f(x)±g(x)] dx = ⌠ ⌡ b a  [f(x)±g(x)] dx.

Theorem 4 If f is integrable on [a,b] and f(x) ≥ 0 for all x ∈ [a,b] then

⌠ ⌡ b a  f(x) dx ≥ 0.

Corollary 5 If f and g are integrable on [a,b] and f(x) ≥ g (x) for all x ∈ [a,b] then

⌠ ⌡ b a  f(x) dx ≥ ⌠ ⌡ b a  g(x) dx.

Theorem 6 [Mean Value Theorem for Definite Integrals] If f is continuous on a closed interval [a,b], then there is a number z in the open interval (a,b) such that

⌠ ⌡ b a  f(x) dx = f(z)(b−a).

Definition 7 Let f be continuous on [a,b]. The average value f_av of f on [a,b] is

fav =  1 b−a ⌠ ⌡ b a  f(x) dx.

Once it is known that integration is the inverse of differentiation and related to the area under a curve, we can observe, for example, that if f and f′ both have strong derivatives at x, then

f(x+ε)−f(x) = ⌠ ⌡ ε 0  f′(x+t) dt = ⌠ ⌡ ε 0    f′(x)+f′′(x) t+O(t2)    dt = f′(x)ε+f′′(x)ε2/2+O(ε3) . (4.18)

4.6  The Fundamental Theorem of Calculus

Theorem 1 [Fundamental Theorem of Calculus] Suppose f is continuous on a closed interval [a,b].

1. If the function G is defined by
   

   G(x)= ⌠ ⌡ x a  f(t) dt

   for every x in [a,b], then G is an antiderivative of f on [a,b].

2. If F is any antiderivative of f on [a,b], then
   

   ⌠ ⌡ b a  f(x) dx=F(b)−F(a).

PROOF. The proof of the first statement follows from the Mean Value Theorem for Definite Integral . End the second part follows from the first and initial condition

⌠ ⌡ a a  f(x) dx=0.

^[¯]

Corollary 2

1. If f is continuous on [a,b] and F is any antiderivative of f, then
   

   ⌠ ⌡ b a  f(x) dx= F(x) ]ab=F(b)−F(a).

2. 
   

   ⌠ ⌡ b a  f(x) dx =   ⌠ ⌡ f(x) dx   b a  .

3. Let f be continuous on [a,b]. If a ≤ c ≤ b, then for every x in [a,b]
   

   d dx ⌠ ⌡ x c  f(t) dt = f(x).

Theorem 3 If u=g(x), then

⌠ ⌡ b a  f(g(x)) g′(x) dx= ⌠ ⌡ g(b) g(a)  f(u) du.

Theorem 4 Let f be continuous on [−a,a].

1. If f is an even function,
   

   ⌠ ⌡ a −a  f(x) dx = ⌠ ⌡ a −a  f(x) dx.

2. If f is an odd function,
   

   ⌠ ⌡ a −a  f(x) dx = 0.

Exercise 5 Calculate integrals

⌠ ⌡ −1 −2    x−  1 x   2    dx; ⌠ ⌡ 4 1  √   5−x    dx; ⌠ ⌡ 1 −1  (v2−1)3v dv; ⌠ ⌡ π/2 0  3sin(  1 2 x) dx; ⌠ ⌡ −π/6π/6(x+sin 5x) dx; ⌠ ⌡ π/3 0   sinx cos2 x dx; ⌠ ⌡ 5 −1  | 2x−3 | dx;

Chapter 5
Applications of the Definite Integral

5.1  Area

Exercise 1 Find areas bounded by the graphs:

1. x=y², x−y=2.

2. y=x³, y=x².

3. y=x^2/3, x=y².

4. y=x³−x, y=0.

5. x=y³+2y²−3y, x=0.

6. y=4+cos2x, y=3 sin[ 1/2]x.

Exercise 2 Express via sums of integrals areas:

1. y = √x , y=−x, x=1, x=4.

5.2  Solids of Revolution

Theorem 1 Let f be continuous on [a,b], and let R be the region bounded by the graph of f, the x-axis, and the vertical lines x=a and x=b. The volume V of the solid of revolution generated by revolving R about the x-axis is

V = ⌠ ⌡ b a  π[f(x)]2 dx.

Theorem 2 [Volume of a Washer]

V = ⌠ ⌡ b a  π( [f(x)]2 −[g(x)]2 )  dx.

Exercise 3

1. y=1/x, x=1, x=3, y=0; x-axis;

2. y=x³, x=−2, y=0, x-axis;

3. y=x²−4x, y=0; x-axis;

4. y=(x−1)²+1, y=−(x−1)²+3; x-axis;

Exercise 4 Find volume of revolution for y=x³, y=4x rotated around x=4.

5.3  Volumes by Cylindrical Shells

Exercise 1 Find volumes:

1. y=√x, x=4, y=0, y-axis.

2. y=x², y²=8x, y-axis.

3. y=2x, y=6, x=0, x-axis.

4. y=√{x+4}, y=0, x=0, x-axis.

5.4  Volumes by Cross Section

Theorem 1 [Volumes by Cross Sections] Let S be a solid bounded by planes that are perpendicular to the x-axis at a and b. If, for every x in [a,b], the cross-sectional area of S is given by A(x), there A is continuous on [a,b], then the volume S is

V = ⌠ ⌡ b a  A(x) dx.

Corollary 2 [Cavalieri's theorem] If two solids have equal altitudes and if all cross sections by planes parallel to their bases and at the same distances from their bases have equal areas, then the solids have the same volume.

Exercise 3 Let R be the region bounded by the graphs of x=y² and x=9. Find the volume of the solid that has R as its base if every cross section by a plane perpendicular to the x-axis has the given shape.

1. Rectangle of height 2.

2. A quartercircle.

Exercise 4 Find volume of a pyramid if its altitude is h and the base is a rectangle of dimensions a and 2a.

Exercise 5 A solid has as its base the region in xy-plane bounded by the graph of y²=4x and x=4. Find the volume of the solid if every cross section by a plane perpendicular to the y-axis is semicircle.

5.5  Arc Length and Surfaces of Revolution

Theorem 1 Let f be smooth on [a,b]. The arc length of the graph of f from A(a, f(a)) to B(b,f(b)) is

Lba = ⌠ ⌡ b a  √   1+[f′(x)]2    dx

We could introduce the arc length function s for the graph of f on [a,b] is defined by

s(x)= ⌠ ⌡ x a  √   1+[f′(t)]2   dt.

Theorem 2 Let f be smooth on [a,b], and let s be the arc length function for the graph of y=f(x) on [a,b]. if ∆x is an increment in the variable x, then

ds dx = √   1+[f′(x)]2   ; (5.1) ds = √   1 + [f′(x)]2   ∆x. (5.2)

Exercise 3 Find the arc length:

1. y = 2/3 x^2/3; A(1, 2/3), B(8, 8/3);

2. y=6 ³√{x²}+1; A(−1,7), B(−8,25);

3. 30xy³−y⁸=15; A(8/15,1), B([ 271/240],2);

Exercise 4 Find the length of the graph x^2/3+y^2/3=1.

Theorem 5 If f is smooth and f(x) ≥ > 0 on [a,b], then the area S of the surface generated by revolving the graph of f about the x-axis is

S = ⌠ ⌡ b a  2πf(x) √   1+[f′(x)]2   dx.

Exercise 6 Find the area of the surface generated by revolving of the graph

1. y=x³; A(1,1), B(2,8);

2. 8y=2x⁴+x⁻², A(1,3/8), B(2, 129/32);

3. x=4 √y; A(4,1), B(12,9).

Chapter 6
Trancendential Functions

6.1  The Derivative of the Inverse Function

Definition 1 Let f be a one-to-one function with domain D and range R. A function g with domain R and range D is the inverse function of f, if for all x ∈ D and y ∈ R y=f(x) iff x=g(y).

Theorem 2 Let f be a one-to-one function with domain D and range R. If g is a function with domain R and range D, then g is the inverse function of f iff both the following conditions are true:

1. g(f(x))=x for every x ∈ D.

2. f(g(x))=x for every y ∈ R.

Exercise 3 Find inverse function.

1. f(x)=[(3x+2)/2x−5];

2. f(x)=5x²+2, x ≥ 0;

3. f(x) = √{4−x²}.

Theorem 4 If f is continuous and increasing on [a,b], then f has an inverse function f⁻¹ that is continuous and increasing on [f(a), f(b)].

Theorem 5 If a differentiable function f has an inverse function g=f⁻¹ and if f′(g(c)) ≠ 0, then g is differentiable at c and

g′(c) =  1 f′(g(c)) .

(6.1)

PROOF. The formula follows directly from differentiation by the Chain rule the identity f(g(x))=x (see Theorem 7.1.2). ^[¯]

Exercise 6 Find domain and derivative of the inverse function.

1. f(x) = √{2x+3};

2. f(x) = 4−x², x ≥ 0;

3. f(x) = √{9−x²}, 0 ≤ x ≤ 3.

Exercise 7 Prove that inverse function exists and find slope of tangent line to the inverse function in the given point.

1. f(x) = x⁵+3x³+2x−1, P(5,1);

2. f(x) = 4x⁵−(1/x³), P(3,1);

3. f(x)=x⁵+x, P(2,1).

6.2  The Natural Logarithm Function

Definition 1 The natural logarithm function, denoted by ln, is defined by

lnx = ⌠ ⌡ x 1   1 t dt,

for every x > 0.

From the properties of definite integral follows that

lnx > 0 if x > 1; lnx < 0 if x < 1; lnx = 0 if x=1.

Theorem 2

d d x (lnx ) =  1 x .

Theorem 3 If u=g(x) and g is differentiable, then

d d x (lnu ) =  1 u  d u d x ,     if u > 0;  d d x (ln| u | ) =  1 u  d u d x ,     if u ≠ 0.

Corollary 4 The natural logarithm is an increasing function.

This gives a new way to prove the principal laws of logarithms .

Exercise 5 Prove using laws of logarithms that

lim x → +∞  = +∞    and lim x → 0−  = −∞.

From this Exercise and Corollary 7.2.4 follows

Corollary 6 To every real number x there corresponds exactly one positive real number y such that lny = x.

Exercise 7 Find implicit derivatives:

1. 3y−x²+lnxy=2.

2. y³+x²lny=5x+3.

Another useful application is logarithmic differentiation which is given by the formula:

d d x f(x)=f(x)  d d x ln(f(x)).

(6.2)

It is useful for functions consisting from products and powers of elemntary functions.

Exercise 8 Find derivative of functions using logarithmic differentiation:

f(x) = (5x+2)3 (6x+1)2; (6.3) f(x) =    √ (3x2+2) √ 6x−7   ; (6.4) f(x) =  (x2+3)5 √ x+1 . (6.5)

6.3  The Exponential Function

Definition 1 The natural exponential function, denoted by expx = e^x, is the inverse of the natural logarithm function. The letter e (=2.718281828…) denotes the positive real number such that lne = 1.

By the definition

lnex = x,        x ∈ R (6.6) elnx = x,        x > 0. (6.7)

By the same definition we could derive laws of exponents from the laws of logarithms.

Theorem 2

d d x (ex) = ex (6.8)  d d x (eg(x)) = eg(x)  d g(x) d x (6.9) (6.10)

PROOF. The proof of the first identity follows from differentiation of 7.6 by the chain rule . The second identity follows from the first one and the chain rule . ^[¯]

Exercise 3 Find implicit derivatives

1. xe^y+2x−ln(y+1)=3;

2. e^x cosy = x e^y.

Exercise 4 Find extrema and regions of monotonicity:

1. f(x)=x² e^−2x;

2. f(x) = e^1/x.

6.4  Integration Using Natural Logarithm and Exponential Functions

The following formulas are direct consequences of change of variables in definite integral and definition of logarithmic and exponential functions:

⌠ ⌡  1 g(x) g′(x) dx = ln| g(x) |+C; (6.11) ⌠ ⌡ eg(x) g′(x) dx = eg(x) +C. (6.12)

From here we could easily derive

Theorem 1

⌠ ⌡ tanu  du = − ln| cosu |+C; (6.13) ⌠ ⌡ cotu  du = ln| sinu |+C; (6.14) ⌠ ⌡ secu  du = ln| secu + tanu |+C; (6.15) ⌠ ⌡ cscu  du = ln| cscu − cotu |+C. (6.16)

Exercise 2 Evaluate integrals

⌠ ⌡  x3 x4−5 dx; ⌠ ⌡  (2+lnx)10 x dx; ⌠ ⌡  ex (ex+2)2 dx; ⌠ ⌡ cot 3 √ x 3 √ x2 dx.

6.5  General Exponential and Logarithmic Functions

Definition 1 The exponential function with base a is defined as follows:

f(x)=ax=elnax=exlna.

(6.17)

From this definition the following properties follows

Theorem 2

d d x (ax) = ax lna ;  d d x (au) = au lna  d u d x ; ⌠ ⌡ ax dx =    1 lna   ax + C; ⌠ ⌡ au du =    1 lna   au + C.

Exercise 3 Evaluate integrals

⌠ ⌡ 5−5x dx; ⌠ ⌡  (2x+1)2 2x  dx; ⌠ ⌡ ee dx; ⌠ ⌡ x5 dx; ⌠ ⌡ x√5 dx; ⌠ ⌡ (√5)x dx;

Exercise 4 The region under the graph of y=3^−x from x=1 to x=2 is revolved about the x-axis. Find the volume of the resulting solid.

Having a^x already defined we could give the following

Definition 5 The logarithmic function with base a f(x)=log_a x is defined by the condition y=log_a x iff x=a^y.

The following properties follows directly from the definition

Theorem 6

d d x (loga x) =  d d x    lnx lna   =  1 lna  1 x ;  d d x (loga | u |) =  d d x    ln| u | lna   =  1 lna ·  1 u  d u d dx .

Exercise 7 Find derivatives of functions:

f(x) = log√3cos5x; f(x) = lnlogx.

6.7  Inverse Trigonometric Functions

Exercise 1 Prove that a periodic function could not be a one-to-one function.

A way out could be as follows: we restrict a trigonometric function f to an interval I in such a way that f is one-to-one on I and there is no a bigger interval I′ ⊃ I that f is one-to-one on I′.

Definition 2

1. The arcsine ( inverse sine function) denoted arcsin is defined by the condition y=arcsinx iff x=siny for −1 ≤ x ≤ 1 and −π/2 ≤ y ≤ π/2.

2. The arccosine ( inverse cosine function) denoted arccos is defined by the condition y=arccosx iff x=cosy for −1 ≤ x ≤ 1 and 0 ≤ y ≤ π.

3. The arctangent ( inverse tangent function) denoted arctan is defined by the condition y=arctanx iff x=tany for x ∈ R and −π/2 ≤ y ≤ π/2.

4. The arcsecant ( inverse secant function) denoted arcsec is defined by the condition y=arcsecx iff x=secy for | x | > 1 and y ∈ [0, π/2) or y ∈ [π, 3 π/2).

Exercise* 3 Why there is no a much need to introduce inverse functions for cotangent and cosecant?

Applying Theorem on derivative of an inverse function we could conclude that

Theorem 4

d dx (arcsinu) =  1 √ 1−u2  du dx ;  d dx (arccosu) = −  1 √ 1−u2  du dx ;  d dx (arctanu) =  1 1+u2  du dx ;  d dx (arcsecu) =  1 u √ u2−1  du dx .

As usually we could invert these formulas for taking antiderivatives:

Theorem 5

⌠ ⌡  1 √ a2−u2  du = arcsin  u a +C = −arccos  u a + C; ⌠ ⌡  1 a2+u2  du =  1 a arctan  u a +C; ⌠ ⌡  1 u √ u2−a2  du =  1 a arcsec  u a +C.

And following formulas could be verified by differentiation:

Theorem 6

⌠ ⌡ arcsinu  du = u arcsinu + √   1−u2   +C; ⌠ ⌡ arccosu  du = u arccosu − √   1−u2   +C; ⌠ ⌡ arctanu  du = u arctanu −  1 2 ln(1+u2) +C; ⌠ ⌡ arcsecu  du = u arcsecu ln   u √   u2−1     +C.

6.8  Hyperbolic Functions

The following functions arise in many areas of mathematics and applications.

Definition 1

hyperbolic sine function: sinhx =  ex−e−x 2 ; hyperbolic cosine function: coshx =  ex+e−x 2 ; hyperbolic tangent function: tanhx =  ex−e−x ex+e−x ; hyperbolic cotangent function: cothx =  ex+e−x ex−e−x .