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- s tutoring book

7477=167

- tax + b

2ax + b +

2ax + b

- ifb2-4ac>0

- 4ac < 0

- 4ac = 0

- 2ax+bC ZalnIax'+bx+c,
- -2aJax2+bx+c

2ax + b (r

- 1) (4ac

- - b2) (ax2 + bx + c)'-

- 3)a + (r

- 1) (4ac

- - b2) J (ax 2 + bx + cy-i 1
- (axe + bx + c)'

(2c + bx)

- 1) (4ac

- - b2) (ax2 + bx + c)'-'
- - b2) I(ax2 + bx + c)'

Ja+bu 6

- = b2[J(a + bu)2
- - 2a(a + bu)1 + a2 Inca + bud] + C a

7' J (a

- r Ju(a+bu)a

In au + a2

_ 1 1 du 1 u(a + bu)2 a(a + bu)

- - 2a lnja + bul] + C
- (au+dbur = bs I a + bu
- a + bu In I

Forms Involving 12

JuVa + bu du

- + bu)" + C

=2(bu-2a)

- 2(Sbu5- 2a) (a

-+IraI+C

- ifa>0 ifa 0 for all x

(b) f(x) > x for all x (for example,

f(5) is a number that must be larger than 5)

(a) Sketch the power functions x-',

- x'12 on the same set of axes

(b) Sketch the power functions x,

- x' on the same set of axes

- for example,
- f (7) = 3
- and f(-7) = 3,
- f(-4) =
- -2 and f(4) =
- and so on

- -f(x) for all x
- for example,
- -12 and f(-3) = 12,f(-6) _
- -2 and f(6) = 2,
- and so on

- sin x and x' are odd,
- 2x + 3 and x2 + x are neither

(a) Figure 17 shows the graph of a function f(x) for x plete the graph for x 0

- (b) Complete the graph in Fig
- 17 if f is odd

Letf(t) be the position of a particle on a number line at time I

(a) f is a constant function (c) f is a decreasing function (d) f(t) > 0 for all t (b) f(i) = t

The Trigonometric Functions

- 1 by considering the six trigonometric functions

two of which (vibrations and electron flow) are described later in the section

We assume that you have studied trigonometry before starting calculus and therefore this section contains only a summary of the main results

A list of trigonometric identities and formulas is included at the end of the section for reference

- cosine and tangent Using Fig
- we define (1)
- tang=r x

Figure 1 shows a positive 0 corresponding to a counterclockwise rotation away from the positive x-axis

A negative 0 corresponds to a clockwise rotation

The distance r is always positive,

but the signs of x and y depend on the quadrant

so that 0 is a second quadrant angle,

x is negative and y is positive

- thus sin 0 is positive,

while cos 9 and tan 0 are negative

- 2 indicates the sign of sin 0,

cos 8 and tan 0 for 0 in the various quadrants

- 516N OFcOS 9 41( N OF tan 6

z Degrees versus radians An angle of 180° is called it radians

More generally,

- to convert back and forth use

number of radians _ IT number of degrees 180

- x number of radians
- number of degrees =

number of radians = 180 x number of degrees

- in both radians and degrees,

and the corresponding functional values

Table 2

Radians

Radians

- cos IV-2

1' N/2-

- 270° 360°

3ar/2 2w

- tan 1/Vs 1

In most situations not involving calculus,

it makes no difference whether we use radians or degrees,

- but it turns out (Section 3
- 3) that for the calculus of the trigonometric functions,

it will be better to use radian measure

- on a circle

- the indicated

arc length s'is just a fraction of the entire circumference,

- namely,

the fraction 9/360 if 0 is measured in degrees,

and 8/21r if 8 is measured in radians

- with 0 in radian measure,

2ar = r8

- 1/Functions
- 21rr = 180r8,

which is not as If degrees are used,

the formula is s'= 360 attractive as (5)

Reference angles Trig tables list sin 0,

cos 0 and tan 0 for 0 < 0 < 90°

we use knowledge of the appropriate signs given in Fig

- 2 plus reference angles,

as illustrated in the following examples

If 0 is a second quadrant angle,

- its reference angle is 180°

so 150° has reference angle 30° (Fig

- cos 150° =
- -cos 30° =
- sin 150° = sin 30° = 1,
- tan 150° _

- its reference angle is 0

- 180°,

so 210° has reference angle 30° (Fig

- sin 210° =
- -sin 30° =
- cos 210° =
- -cos 30° =

-,11\/-3,

- tan 210° = tan 30° = 1/V3_

- its reference angle is 360°

so 330° has reference angle 30° (Fig

- sin 330° =
- -sin 30° =
- cos 330° = cos 30° _ Imo,
- tan 330° =
- -tan 30° =

- sin 0 =
- opposite leg hypotenuse '
- tan 0 =

cos 0 = opposite leg adjacent leg

- adjacent leg hypotenuse '

The Trigonometric Functions

cos x and tan x Figures 8-10 give the graphs of the functions,

- with x measured in radians

The graphs show that sin x and cos x

- have period 21r (that is,
- they repeat every 2ir units),
- while tan x has period it

Furthermore,

- -1 s'sin x s'1 and
- -1 s'cos x < 1,

so that each function has amplitude 1

On the other hand,

the tangent function assumes all values,

- that is,

has range Note that sin x and cos x are defined

- for all x,

but tan x is not defined at x = tv/2,

- a sin bx,
- for positive a and b,

has amplitude a and period 21r/b

- 5 sin Ix has period 4ar and amplitude 5
- 1/Functions

and amplitude as a sin bx but is also shifted

As an example,

- consider sin(2x

first plot a few points to get your bearings

the most convenient values of x are those which

- make the angle 2x
- - 3r a multiple of r/2
- the table in Fig
- 12 chooses angles 0 and r/4 to produce points (0,
- 0) on the graph

Sin 0 = 0

Application to simple harmonic motion If a cork is pushed down in a bucket of water and then released (or,

- similarly,

a spring is stretched and released),

- it bobs up and down

Experiments show that if a particular cork oscillates between 3 units above and 3 units below the water level with the timing indicated in Fig

its height h at time t is given by h(t) = 3 sin 't

TpMa t = it

i1ME to

- riME t=1T FIG
- flME 1:=2
- tmEt=3'7r
- 3 The Trigonometric Functions

- 14 minutes after time 0
- ) More generally,
- the amplitude,

frequency and shift depend on the cork,

the medium and the size and timing of the initial push down,

- but the oscillation,

called simple harmonic motion,

- always has the
- form a sin(bt + c),
- or equivalently a cos(bt + c)

- ) in a wire

- and if i(t) is the current,
- that is,

the amount of charge per second flowing in a given direction at time t,

then i(t) is of the form a sin(bt + c) or a cos(bt + c)

If i(t) = 10 cos t then at time t = 0,

- 10 units of charge per second flow in the given direction
- at time t = it/2,
- the flow momentarily stops
- at time t = jr,
- 10 units of charge per second flow opposite to the given direction

The graph of f(x) sin x First consider two special cases

although usually we do not actually sketch the lines

which are reflections of one another in the x-axis,

are called the envelope of 2 sin x

- -2 sin x also lies between those lines
- in addition,

the effect of the negative factor

- -2 is to change the signs of y-coordinates,

so the graph is the reflection in the x-axis of the graph of 2 sin x (Fig

I'+ Similarly,

the graph of x' sin x is sandwiched between the curves y = ±x' which we sketch as guides (Fig

called the envelope of x' sin x,

are reflections of one another in the x-axis

- 1/Functions

whenever x' is negative (as it is to the left of the y-axis) we not only change the amplitude but also reflect sine in the x-axis to obtain x' sin x

The result in Fig

- 15 shows unbounded oscillations

to sketch the graph off (x) sin x,

first draw the curve y = f (x) and the curve y = f (x),

- its reflection in the x-axis,
- to serve as the envelope

Then change

the height of the sine curve so that it fits within the envelope,

and in addition reflect the sine curve in the x-axis whenever f(x) is negative

Secant,

cosecant and cotangent By definition,

- sec x =
- csc x =
- cot x =
- cos x sin x

In each case,

the function is defined for all values of x such that the denominator is nonzero

csc x is not defined for x = 0,

- 17 In a right triangle (Fig
- sec 0 = (8)

hypotenuse hypotenuse csc 0 = opposite leg' adjacent leg' adjacent leg cot B = opposite leg

- 3 The Trigonometric functions

- and sin x2 to mean sin(x2)

Standard trigonometric identities Negative angle formulas (9)

- sin(-x) =

-sin x,

- csc(-x) =

-csc x,

- cos(-x) = cos x,
- sec(-x) = sec x,
- tan(-x) _

-tan x,

- cot(-x) _

Addition formulas (10)

sin(x + y) = sin x cos y + cos x sin y sin(x

- - y) = sin x cos y
- - cos x sin y cos(x + y) = cos x cos y
- - sin x sin y cos(x
- - y) = cos x cos y + sin x sin y Double angle formulas
- sin 2x = 2 sin x cos x (11)
- cos 2x = cos2x
- - sin2x = 1
- - 2 sin2x = 2 cos(x

2 tan x

- tan 2x =

- tan2x

- sin2x + cos2x = 1 (12)
- 1 + tan2x = sec2x 1 + Cot2x = csc2x

- sin2yx = cos2gx
- - cos x 2 1
- + cos x 2

- sin xcos y =
- sin(x + y) + sin(x
- 1/Functions
- cos x sin y = cos x cos y =
- sin x sin y =
- sin(x + y)

- sin(x

- cos(x + y) + cos(x
- - cos(x + y) 2

Factoring formulas

- sin x + sin y = 2 cos (15)
- - sin y = 2 cos x 2 y sin cos x + cos y = 2 cos x +YCOSX
- - cos y = 2 sin XY sin

Reduction formulas

- - 0) = sin 0 sin(Tr
- - 0) = cos 0 cos(Tr
- -cos 0 sin(7r
- - 0) = sin 0 Law of Sines (Fig

- 19) 2 = a 2 b 2
- - 2ab cos C c

19) (19)

area of triangle ABC = iab sin C

- (a) it/5 (b) 51r/6 (c)

-7r/3 2

Convert from degrees to radians

- (a) IT (b)
- -90° (c) 100°

Evaluate without using a calculator

(a) sin 210° (b) cos 31r (c) tan 51T/4

- (a) sin
- x (b) tan 4x
- (d) 5 sin(''px + a)
- (e) 2 cos(3x
- (c) 3 cos

Let sin x = a,

cos y = b and evaluate the expression in terms of a and b,

- if possible
- (a) sin(-x) (d)
- -cos y (b) cos(-y) (e) sin2x
- (f) sin x2

use right triangle trigonometry to find an exact answer,

rather than tables or a calculator which will give only approximations

(a) Find cos 0 if 0 is an acute angle and sin 0 = 2/3

(b) Find sin 0 if 0 is acute and tan 0 = 7/4

Sketch the graph

- (a) x sin x
- (b) xs sin x
- 4 Inverse Functions and the Inverse

If a function maps a to b we may wish to switch the point of view and consider the inverse function which sends b to a

the function defined by F = IC + 32 gives the Fahrenheit temperature F as a function

- of the centigrade reading C

If we solve the equation for C to obtain C = 9(F

- - 32) we have the inverse function which produces C,
- given F

the inverse is probably also useful

we discuss inverses in general,

and three inverse trigonometric functions in particular

The inverse function Let f be a one-to-one function

The inverse of f,

- denoted by f

is defined as follows: if f (a) = b then f

-'(b) = a

the inverse maps "backwards" (Fig

Only one-to-one functions have inverses because reversing a non-one-to-one function creates a pairing that is not a function (Fig

Given a table of values for f,

- a table of values for f
- -' can be constructed by interchanging columns

Clearly,

-'(x) = 36

- f and f
- -' are inverses of each other

Figure I shows that if f and f

- -' are applied successively (first f and then f

or vice versa) the result is a "circular" trip which returns to the starting

- 1/Functions
- f-'(f(x)) = x and f(f-'(x)) = x

multiplying a number by 3 and then multiplying that result by 1/3 produces the original number

the centigrade/fahrenheit equations show that if f(x) = 23x + 32 then f-'(x) = y(x

- such as its graph,

often follow easily from the properties of the original function

- -' amounts to com-
- - a(7,2) paring points such as (2,
- 7) and (7,2) (Fig

- the graph off
- -' is the reflection of the graph off in the line y = x,

so that the pair of graphs is symmetric with respect to the line

If f(x) = x2,

and x >_ 0 so that f is one-to-one,

- then f"'(x) = Vx

The symmetry of the two graphs is displayed in Fig

The inverse sine function Unfortunately,

the sine function as a whole doesn't have an inverse because it isn't one-to-one

- sine graph are one-to-one,
- in particular,

any section between a low and a high point passes the horizontal line test and can be inverted

- we use the part between
- - 7r/2 and a/2 and let sin-'x be the inverse of this abbreviated sine function
- that is,
- sin-'x is the angle between
- - 1T12 and 7r/2 whose sine is x

Equivalently,

sin-'a = b if and only if sin b = a and

- -7r/2 s'b < ir/2

The graph of sin-'x is found by reflecting sin x,

- -ir/2 s'x < zr/2,
- in the line y = x (Fig

The domain of sin-'x is [-1,

- 1] and the range is (2)

[-zr/2,

The sin-' function is also denoted by Sin-' and arcsin

the abbreviation ASN of arcsin is often used

- then sin x = 2

- sin(-330°) = 2,
- sin 150° = 2,

We must choose the angle between

- -90° and 90°
- therefore sin-' 2 = 30°,
- in radians,
- sin-' 2 = zr/6

Example 3 Find sin-'(-1)

Of all the angles whose sine is

the one in the interval [- ir/2,

Therefore,

- sin-'(-1) =

- 5 Warning 1

- -wr/2 and 3ir/2 are coterminal angles
- that is,

as rotations from the positive x-axis,

they terminate in the same place

- -ir/2 and 3ir/2 are not the same angle or the same number,
- and aresin(-1) is
- not 3ir/2

- -'(f (x)) = x,
- sin-(sin 200°) is not 200°

- -90° and 90°

The sine function maps 200°,

- along with many other angles,
- such as 560°,

-160°,

- all to the same output

The sin-' function maps in reverse to the particular angle between

- -90° and 90°

Therefore,

- sin-'(sin 200°) =

The inverse cosine function The cosine function,

- like the sine function,
- has no inverse,
- because it is not one-to-one

By convention,

- we consider the

one-to-one piece between 0 and a,

and let cos-'x be the inverse of this abbreviated cosine function (Fig

cos''x is the angle between 0 and it whose cosine is x

- 1/Functions

cos-'a = b if and only if cos b = a and 0 < b _< ir

- 1) and the range is [0,

- arccos and ACN

Example 4 Find cos-'(-2)

Solution: The angle between 0° and 180° whose cosine is

- -2 is 120°
- -2= Therefore,
- cos-'(-2) = 120°,
- or in radians,
- cos'(-)

Warning The graphs of sin x and cos x wind forever along the x-axis,

but the graphs of sin-'x and cos-'x (reflections of portions of sin x and cos x) do not continue forever up and down the y-axis

5 and 6

(If either curve did continue winding,

the result would be a nonfunction

The inverse tangent function The tan-' function is the inverse of the branch of the tangent function through the origin (Fig

- tan-'x is the angle between
- -ir/2 and a/2 whose tangent is x

tan-'a = b if and only if tan b = a and

- -a/2 < b < 1r/2

- arctan and ATN

- tan-'(-1) _
- -ir/4 because
- -Ir/4 is between
- -1T/2 and 7r/2 and tan(- 7r/4) =

Equivalently,

restrict x so that the function 2 tan 3x is one-to-one,

and then find the inverse function

Solution: To use tan-' as the inverse of tangent,

- the angle,
- which is 3x
- in this problem,

must be restricted to the interval (-111r,

- that is,
- -2ir < 3x < 2Tr

Consequently,

- we choose
- -ir/6 < x < Tr/6

- 2y = tan 3x tan ' 2y = 3x
- tan-' 2y = x

(divide both sides of the original equation by 2)

(take tan-' on both sides) (divide by 3)

- if f(x) = 2 tan 3x and
- -Ir/6 < x < it/6,
- then f''(x) _ I3 tan'2x

Exponential and Logarithm Functions

If f(3) = 4 and f(5) = 2,

- if possible,

- if it exists
- (c) 1 /x
- (b) Int x

If f(x) = 2x

- - 9 find a formula for f

-'(f(17))

Show that an increasing function always has an inverse and then decide if the inverse is decreasing

? If f is continuous and invertible then f

- -' is also continuous

- ? (a) x2 and (b) xs and 8

- (a) cos-'O
- (b) sin''0
- (f) tan-'I
- (c) sin`2
- (g) tan-'(-1)
- (d) cos-'(-§\) 9

True or False

? (a) If sin a = b then sin-'b = a (b) If sin-'c = d'then sin d'= c

Place restrictions on 8 so that the equation has a unique solution for 8,

- and then solve

(a) z = 3 + s'sin ar8 (b) x = 5 cos(28

Odd and even functions were defined in Problem 8,

Section 1

Do odd (resp

- even) functions have inverses
- ? If inverses exist,
- must they also be odd (resp

Exponential and Logarithm Functions

- 1 by considering the exponential functions and their inverses,
- the logarithm functions

they have important physical applications,

- such as exponential growth,
- discussed in Section 4

- (4)' and 7' are called exponential functions,

as opposed to power functions x2,

- x"' and x'

an exponential function has the form b',

- and is said to have base b

- which are not real

- there is no (real) f (f'),
- f (17),

the domain of (-4)' is too riddled with gaps to be useful in calculus

(The power function x' also has a restricted domain,

- namely [0,

but at least the domain is an entire interval

- ) Because of this difficulty,

we do not consider exponential functions with negative bases

we first make a table of values

- (Remember
- that 2',
- for example,
- is defined as 1/2',
- and 2° is 1

we used integer values of x in the table,

but 2' is also defined when x is not an integer

For example,

- 1/Functions 22"3

E = 231110 =

and the graph of 2' also contains the points (2/3,

4) and (3

1,'2 V7)

and when the pattern seems clear,

connect them to obtain the final graph (Fig

The connecting process assumes that 2' is continuous

t Figure 1 also contains the graphs of (11)' and 3' for comparison

- the most popular base is 10,

while computer science often favors base 2

for reasons to be given in Section 3

- calculus uses base e,

a particular irrational number (that is,

an infinite nonrepeating decimal) between 2

71 and 2

the official definition will be given in that section

Because calculus concentrates on base e,

the function e' is often referred to as the exponential function

It is sometimes written as expx

programming languages use EXP(X)

Figure 2 shows the graph of e',

along with 2' and 3' for comparison

- and correspondingly,

the graph of e' lies between the graphs of 2' and 3

- a value of e',
- such as e2,

may be approximated with tables or a calculator

- 9 will indicate one method for evaluating e' directly

A rough estimate of e2 can be obtained by noting that since e is slightly less than 3,

- e2 is somewhat less than 9

tThe connecting process also provides a definition of 2' for irrational x,

- that is,

when x is an infinite nonrepeating decimal,

- such as ir

and by connecting the points to make a continuous curve,

we are defining 2' by the following sequence of inequalities: 234 5

- (o) ex = e-x

-in x = 4

- (p) x In x = 0
- (g) In(-x) = 4
- (q) xex + 2ex = 0
- (h) esx+3 = e2x
- (r) ex In x = 0 25

- (j) aresin ex = it/6 6
- 2 +1n 3x = 5

Show that In sV simplifies to

-s In 2

A scientist observes the temperature T and the volume V in an experiment and finds that In T always equals

-1 In V

Show that TV2" must therefore be constant

The equation 4 In x + 2(In x)2 = 0 can be considered as a quadratic equation in the variable In x

- and then solve for x itself

- ? (a) If a = b,
- then e° = e'
- (b) If a + b = c,
- then e° + e" = e`

Find the mistake in the following "proof"that 2 < 1

- so ln(Y)2 < In 4

Thus 2 In s'< In 1

Cancel In s'to get 2 < 1

A simple inequality such as 2x + 3 > 11 is solved with the same maneuvers as the

equation 2x + 3 = 11 (the solution is x > 4),

- in general,
- inequalx s'X 2x 1 > 0,

we ities are trickier than equations

- to solve
- 1/Functions

want to multiply on both sides by x

- - 5 to eliminate fractions

- - 5 is negative and multiplication by x
- - 5 reverses the inequality
- if x > 5,
- - 5 is positive,

and the inequality is not reversed

- (For equations,

this type of difficulty doesn't arise

- ) This section

offers a straightforward method for solving inequalities of the form f (x) > 0,

- f (x) < 0,

or equivalently for deciding where a function is positive and where it is negative

In order for a function f to change from positive to negative,

- or vice

its graph must either cross or jump over the x-axis

a nonzero continuous f cannot change signs

its graph must lie entirely on one side of the x-axis

- -3 and x = 2,

and is discontinuous only at x = 5,

so that within the open intervals (-w,

- (-3,2),(2,5) and (5,-),f is nonzero and continuous

we have the following method for determining the sign of a function f,

- that is,
- for solving the inequalities
- f(x)>0,f(x)0

x2-2x+1 5,

and negative for x < 1 and for 1 < x < 5

the solution to the first inequality in (1) is x > 5,

and the solution to the second inequality is x < 1 or 1 < x < 5

These are places where f might (but doesn't have to) change sign by crossing or jumping over the x-axis

Indeed,

- in this example,

f changes sign at x = 5 but not at x = 1

The graph in Fig

- 2 shows what is happening

At x = 1,

f touches the x-axis but does not cross,

- so there is no sign change

f happens to jump over the axis,

- so there is a sign change
- z Problems for Section 1

- 10x2 9(x

- - 3)2 (b)x+i
- e` (d) x
- (e)x2+x-6
- (c) x2-x+2 2

- (b) 2x+
- 1/Functions

Graphs of Translations,

Reflections,

Expansions and Sums

Considerable time is spent in mathematics finding graphs of functions because graphs can be extremely useful

It is possible to see from a graph where a function is positive,

- negative,
- increasing,
- decreasing,
- one-to-one,
- discontinuous,
- and so on,

when it may be very hard to do this from a formula

Suppose that the graph of y = f(x) is known

in trigonometry it is shown that the graph of sin 2x can be obtained easily from the graph of sin x by changing the period to or

Similarly,

the graph of 2 sin x can be derived from the graph of sin x by changing the amplitude to 2

the problem will be to find the graph of a variation off,

assuming that we have the graph off

Perhaps it was found by plotting many points,

possibly it was generated by a computer,

it may be a standard curve such asy = e' or it may have been drawn using techniques of calculus,

- coming later

We will first consider three variations in which an operation is performed on the variable x in the equation y = f(x),

resulting in horizontal changes in the graph

Then we examine three variations obtained by operating on the entire right-hand side of the equation y = f(x),

resulting in vertical changes in the graph

Results are summarized in Table 1

Finally we consider the graph of a sum of functions,

- given the individual graphs

- - I is given in Fig

- - 7)' + 3(x

look for a connection between the two tables of values

Y = x' + 3x2

2' + 3(22)

- 1 = 19

53 + 3(52)

- 1 = 199

- y=(x-7)'+3(x-7)2-1
- 2'+3(22)-1=19

53 + 3(52)

- 1 = 199

Substituting x = 9 into the new equation involves the same arithmetic (because 7 is immediately subtracted away) as substituting x = 2 in the

Graphs of Translations,

Expansions and Sums

- original equation

x = 12 in the new equation produces the same

calculation as x = 5 in the old equation

In general,

b) is in the old table then (a + 7,

- b) is in the new table

Now that we have a connection between the tables,

- how are the graphs related
- ? The new point (9,
- 19) is 7 units to the right of the old point (2,

- given the (old) graph
- of y = f (x),
- the (new) graph of y = f (x
- - 7) is obtained by translating (i

shifting) the old graph to the right by 7 units (Fig

- - 7)2 + y2 = r2 is a circle centered at the point (7,0),
- that is,
- translated to the right by 7
- ? Similarly,

the graph of y = f (x + 3) is found by translating y = f (x) to the left by 3 units

Horizontal expansion/contraction Consider the following two equations with their respective tables of values

23 + 3(22)

- - 1 = 19 5' + 3(52)

- 1 = 199

- (5x)3 + 3(5x)2

23 + 3(22)

- - 1 = 19 53 + 3(52)

- 1 = 199

Substituting x = 2/5 in the new equation produces the same calculation as x = 2 in the old equation (because each occurrence of 2/5 in the new equation is immediately multiplied by 5)

b) is in the old table then (a/5,

- b) is in the new table

given the graph of y = f (x) (Fig

the graph of y = f (5x) is obtained by dividing x-coordinates by 5 so as to contract the graph horizontally (Fig

the graph of y = f (qx) is found by tripling x-coordinates so as to expand the graph off horizontally (Fig

Note that in the expansion (resp

- contraction),

points on the y-axis do not move,

but all other points move away from (resp

toward) the y-axis so as to triple widths (resp

- divide widths by 5)

The expansion/contraction rule says that the graph of y = sin 2x is drawn by halving x-coordinates and contracting the graph of y = sin x horizontally

This agrees with the standard result from trigonometry that y = sin 2x is drawn by changing the period on the sine curve from 21r to ir,

- a horizontal contraction
- - 1 /Functions

OeiZpNfAL HogizDn-' A L'coN W& 1,10N

Horizontal reflection Consider the following two equations and their respective tables of values

- y = X' + 3x2

23 + 3(22)

- 1 = 19

5' + 3(52)

- I = 199

- y=(-x)'+3(-x)2- 1 2' + 3(22)
- - I = 19 5' + 3(52)

- I = 119

- -2 into the new equation results in the same calculation as x = 2 in the original

b) is in the old table then (-a,

- b) is in the
- new table

In general,

given the graph of y = f(x) (Fig

the graph of y = f (-x) is obtained by reflecting the old graph in the y-axis (Fig

- 3d) so as to change the sign of each x-coordinate

y =x'+3x2- 1 and y =(x'+3x2- 1) + 10

they value for the second equation is 10 more than the first

- given the graph of y = f (x),

the graph of y = f (x) + 10 is obtained by translating the original graph up by 10

- t Similarly,
- the graph of y = f (x)
- - 4 is found by translating the graph of y = f (x) down by 4

y =x'+3x2- 1 and y =2(x'+3x2- 1)

For any fixed x,

the y value for the second equation is twice the first y

- given the graph of y = f (x),

the graph of y = 2f (x) is obtained by doubling the y-coordinates so as to expand the original graph vertically

the graph of y = 3f(x) is found by multiplying heights by 2/3,

so as to contract the graph of f (x) vertically

tThe conclusion that y = f (x) + 10 is obtained by translating up by 10 may be compared with a corresponding result for circles,

provided that we rewrite the equation as (y

- 10) f(x)

- while x4 + (y
- - 10)4 = r2 is centered at the point (0,
- that is,
- translated up by 10

Similarly,

- the graph of (y
- - 10) = f (x) is obtained by translating y = f(x) up by 10

Graphs of Translations,

The familiar method for graphing y = 2 sin x (change the amplitude from 1 to 2) is a special case of the general method for y = 2f (x) (double all heights)

Vertical reflection Consider y = f (x) versus y =

- f (x)

- the graph of y =
- -f(x) is obtained from the graph of y = f (x) by reflecting in the x-axis

- 14 of Section 1
- 3 which showed the graphs of y = 2 sin x and y =
- -2 sin x as reflections of one another

Variation of y = f (x)

How to obtain the graph from the original y = f (x)

An operation is performed on the variable x

- y =f(-x)

- y = f (2x)
- y=f(x+2)

Halve the x-coordinates of the graph of y = f (x) so as to contract horizontally Multiply the x-coordinates of the graph of y = f (x) by 3 so as to expand horizontally Translate the graph of y = f (x) to the

- y = f(x

Translate the graph of y = f (x) to the

- y = B36)
- left by 2
- right by 3

An operation is performed on f (x),

- on the entire righthand side

- y = 2f(x)

Double the y-coordinates of the graph of y = f (x) so as to expand

- y = SOX)

Multiply the y-coordinates of the graph of y = f (x) by 3 so as to contract vertically Translate the graph of y = f (x) up

- vertically
- y =f(x)+2
- y = f(x)

Translate the graph of y = f (x) down by 3

Warning The graph off (x

- - 1) (note the minus sign) is obtained by translating f (x) to the right (in the positive direction)

- - 1 (note the minus sign) is found by translating f (x) down (in the negative direction)
- 1/Functions
- :k T-i rRANS%A1E

--- DOWN

- 4zr fi coy-1 "
- x WAND Vj:RrJ cALLY

CONTRA( TN

- - cog REFlkC1

YERToay

-ir FIG

10 FIG M

Graphs of Translations,

- to sketch y = f(x) + g(x),

add the heights from the separate graphs off and g,

- as shown in Fig

For example,

the new point D'is found by adding height

ff to height AC to obtain the new height AD

On the other hand,

since point P has a negative y-coordinate,

the newpooint R is found by subtracting

length PQ from 0 to get the new height

- f(x)+y(x)

draw y = cos x and y = sin x on the same set of axes,

- and then add heights (Fig

For example,

- add height AB to

height X to obtain the new height AD

- at x = ir,
- when the sine height is 0,

the corresponding point on the sum graph is point E,

- lying on the cosine curve

- in each case,

include the graph of In x for comparison

- (a) In(-x) (d) In 2x (b)
- (e) In(x + 2)
- (c) 2 In x
- (f) 2 + In x

Figure 13 shows the graph of a function,

- which we denote by star x

- 1/Functions
- 13 (d) star x
- - 2 (e) star(-x) star(x

- 2) (f)

-star x

- (a) star
- (b) s'star x (c)

Find the new equation of the curve y = 2x7 + (2x + 3)6 if the curve is (a) translated left by 2 (b) translated down by 5

Sketch the graph

(a) y = Isin xl (b) y = Iln xl (c) y = le'I

- (d) y = el'I (e) y = Inlxl

Sketch each trio of functions on the same set of axes

- sin x,x + sin x 6

The variations sin2x,

sin3x and s's n x were not discussed in the section

cubing heights and cuberooting heights on the sine graph

- (a) Find f(-4)

(b) For which values of x is f defined

? With these values as the domain,

- the range off
- (c) Find f(a2) and (f (a))2

(d) Sketch the graph off by plotting points

- if it exists

we need the idea of the remainder in a division problem

we say that the quotient is 2 and the remainder is 2

- 8 is divided by 3,

the quotient is 8 and the remainder is 2

the quotient is 9 and the remainder is 0

If x > 0,

let f(x) be the remainder when x is divided by 3

- (a) Sketch the graph off

(b) Find the range of f (c) Find f

- -'(x) if it exists
- (d) Find f (f (x))

(a) f(a) = a for all a (b) f(a) # f(b) if a 0 b (c) f (a + 7) = f (a) for all a 4

sketch the graphs of log2x and In x on the same set of axes

Solve for x

(a) y = 2 ln(3x + 4) (b) y = 4 + e3 7

(a) e- sin x (e) sin-' 2x (b) sin-'(x + 2) (f) sin 3irx (c) sin-x + 12or

- (g) 2 cos(4x
- (d) 2 sin-x 8

- - e"`) and cosh x = 2te" + e hyperbolic sine and hyperbolic cosine,
- respectively

(a) Sketch their graphs by first drawing 2e' and 32e (b) Show that cosh2x

- - sinh2x = I for all x

Solve the equation or inequality

- (a) In x

- ln(2x

- - 3) = 4 ( b)
- (c) (d )
- 2e' + 8 < 0 x

- - In 5x simplifies to
- are called the

2/LIMITS

As you read them,

you will become accustomed to the new language and,

- in particular,

see how limit statements about a function correlate with the graph of the function

The examples will show how limits are used to describe discontinuities,

the "ends" of the graph where x

-- x or x

- and asymptotes
- (An asymptote is a line,
- more generally,
- a curve,

that is approached by the graph of f

) Limits will further be used in Sections 3

2 and 5

- 2 where they are fundamental for the definitions of the derivative and the integral,

the two major concepts of calculus

- but not equal to 2,
- f(x) gets closer to 5

- 2 f(x) = 5 and say that as x approaches 2,
- f(x) approaches 5

- -> 2 then f(x)

- the limit of f (x) as x
- -> 2 remains 5

In other words,

if the value off at x = 2 is changed from 3 to anything else ,

- including 5 ,

or if no value is assigned at all to f(2) ,

- we still have limx
- 2 f (x) = 5

In general,

- we write
- lim f (x) = L'x+n
- for all x sufficiently close,
- but not equal,

f (x) is forced to stay as close as we like,

- and possibly equal,

- there is no f(3),
- but we write (1)
- lim f(x) = 4,
- x approaches 3 from the left,
- that is,

through values less than 3 such as 2

- then f (x) approaches 4
- lim f W = 5,

meaning that if x approaches 3 from the right,

- that is,

through values greater than 3 such as 3

- then f (x) approaches 5

The symbols

- 3- and 3+ are not new numbers

they are symbols that are used only in the context of a limit statement to indicate from which direction 3 is approached

if we are asked simply to find limas f(x),

we have to conclude that the limit does not exist

Since the left-hand and right-hand limits disagree,

there is no single limit to settle on

2/Limits

- x-3 A table of values and the graph are given in Fig

3 we write

- lim f (x) = x

meaning that as x approaches 3 from the right,

f(x) becomes unboundedly large

- and we write
- lim Ax) =
- -x to convey that as x approaches 3 from the left,f(x) gets unboundedly large and negative

since the left-hand and right-hand limits do not agree

- 3 f(x) _ ±x

lim_ f (x) = x means that for all x sufficiently close,

- but not equal,

to a,f(x) can be forced to stay as large as we like

- a limit of
- -x means thatf(x) can be made to stay arbitrarily large and negative

- -x For the function in Fig
- we write

- lim f (x) = 4 x

to indicate that as x becomes unboundedly large,

far out to the right on the graph,

the values of y get closer to 4

More precisely,

- lim f (x) = 4x

because the values of y are always less than 4 as they approach 4

Both (3) and (3') are correct,

but (3') supplies more information since it indicates that the graph off(x) approaches its asymptote,

- the line y = 4,
- from below

_x f(x) = x because the graph rises unboundedly to the left

- voltage,
- at time t,
- then lim,

x f(t) is called the steady state height,

- voltage,

and is sometimes denoted by f (x)

- voltage,

speed reached after some transient disturbances have died out

Example 1 There is no limit of sin x as x

- -+ no because as x increases without bound,

sin x just bounces up and down between

-1 and 1

- 2) rises unboundedly to the right,
- lim ex = x

- consider the values e ",
- to see that the limit

- lim ex = 0

consider e" " = 1 /e e I /e to see that the limit is 0 (more precisely,

The limit must be approached,

- but not necessarily attained

_x ex = 0 although ex never reaches 0

- for the function f in Fig
- 2 f(x) = 5 although f(x) never attains 5

- 3) rises unboundedly to the right,
- lim In x = x

- lim In x =

- then limx
- f (x) is simply f (a)

- _1 f(x) =f(-1) = 2

- then either limx
- f (x) and f (a) disagree,
- or one or both will not exist

Example 4 The function x3

- - 2x is continuous (the elementary functions are continuous except where they are not defined) so to find

2/Limits

- the limit as x approaches 2,

we can merely substitute x = 2 to get limx

- 2x) = 8

- 4 = 4

Some types of discontinuities Figure 1 shows a point discontinuity at x = 2,

- 2 shows a jump discontinuity at x = 3 and Fig
- 3 shows an infinite
- discontinuity at x = 3

a function f has a point discontinuity at x = a if lim,

f(x) is finite but not equal to f(a),

either because the two values are different or becausef(a) is not defined

The function has a jump discontinuity at x = a if the left-hand and right-hand limits are finite but unequal

Finally,

f has an infinite discontinuity at x = a if at least one of the left-hand and right-hand limits is oc or

Problems for Section 2

- (a) lim,
- (b) lim,
- (f) lim,
- ,,2 tan x
- (c) lim

0 cos x

- (g) lim,

2(x2 + 3x

- (d) lim,,

Find lim Int x as (a) x

- -- 3- (b) x

--* 3+ 3

- -- 0- (b) x

-- 0+ 4

Find urn tan x as (a) x

- --> 1- (b) x

(a) Draw the graph of a function f such that f is increasing,

- but lim,
- f (x) is not x

(b) Draw the graph of a function f such that lim,

- f (x) = x,

but f is not an increasing function

- (a) lim,

f(x) = 2 and f(3) = 6 (b) lira,-,

- f(x) (c) lim
- f(x) = 4 and lim,
- 2_ f(x) = 7 (d) lim,
- -- and lim,
- ,_ f(x) = 5 7

Does lim,,

o f(2 + a) necessarily equal j(2)

- and letf(x) = I otherwise

- f(100) = 0,((1000) = 0,
- f(983) = 1

Finding Limits of Combinations of Functions

- (a) limx
- es f (x) (b) lima
- (c) lim,

Use the graph of f(x) to find limx

- f(x) if (a) f(x) = x sin x
- (b) f(x) =

The preceding section considered problems involving individual basic functions,

- such as e',
- sin x and In x

- of basic functions,
- that is,

limits of elementary functions in general,

and continue to apply limits to curve sketching

Limits of combinations To find the limit of a combination of functions we find all the "sublimits" and put the results together sensibly,

as illustrated by the following example

Consider

We can't conveniently find the limit simply by looking at the graph of the function because we don't have the graph on hand

finding the limit will help get the graph

- and finding the limit as x
- -+ 0+ will give information about how the graph "begins

" We find the limit by combining sublimits

- 0+ then

- 5 remains 5 and In x

- the first near 0,

the second 5 and the third large and negative,

- is itself large and negative

Therefore,

- the numerator approaches

In the denominator,

- --> 1 so 2e'

- the final answer is

We abbreviate all this by writing lim

- x2+5+lnx 0+5+ 2e ' = 2

find the individual limits and then put them together

The last section emphasized the former so now we concentrate on the latter,

especially for the more interesting and challenging cases where the individual limits to be combined involve the number 0 and/or the symbol x

an abbreviation for a limit problem where the numerator grows unboundedly large and the denominator approaches 0 from the left

To put the pieces together,

- examine say 100
- which leads to the answer

In abbreviated notation,

Consider 2/x,

an abbreviation for a limit problem in which the numerator approaches 2 and the denominator grows unboundedly large

Compute fractions like

- - 2/Limits 1

002001,

- to see that the limit is 0

- 2/x = 0 or,
- more precisely,

2/x = 0+

we list more limit results in abbreviated form

If you understood the preceding examples you will be able to do the following similar problems when they occur (without resorting to memorizing the list)

- 0x0=0 0+0=0

-2xx=-x

- x + x = x
- - xx (6-)xx=x
- xxx=x x

4°= 1 0+

- (0+)' = 0 cI = x

x1/2 = x

- (0+)' = 0 Example 1
- ex In x = x x x = x,
- ,,+ e' In x = l'x

The graph of a + be" Consider the function f(x) = 2

- 7 we know that the graph can be obtained from the graph of e' by reflection,
- contraction and translation

- but in a different location

The fastest way to determine the new location is to take limits as x

-- x and x

and perhaps plot one convenient additional point as a check: f(x)=2-ex=2-x=-x

- f(-x)=2-e-x2-02
- as a check,

f(x) and sketch the graph off in the vicinity of x = 5

- if x re-

mains larger than 5 as it approaches 5,

- - x remains less than 0 as it approaches 0

- -x and (similarly)

20+ = x

- 5 f (x) does not exist

the one-sided limits are valuable for revealing that f has an infinite discontinuity at x = 5 with the asymptotic behavior indicated I/

Rather,

- 2/0+ = x while 2/0- =

in a problem which is of the form (non 0)/0,

it is important to examine the denominator carefully

Example 3 Letf(x) = e where f is not defined

1&2 (Fig

lim e- = e-""+ = e-' = 0+ "o Therefore f has a point discontinuity at x = 0

we can remove the discontinuity and make f continuous

In other words,

- for all practical purposes,
- is 0 when x = 0

In general,

if a function g has a point discontinuity at x = a,

the discontinuity is called removable in the sense that we can define or redefine

g(a) to make the function continuous

jump discontinuities and infinite discontinuities are not removable

- 3) so as to remove the infinite discontinuity and make f continuous

Problems for Section 2

- (f) e"= (g) 1/e"
- (b) (c)
- l \4/ 3
- (t) (J)

(-'k') '

- (d) lim e'-'
- (b) lim
- (e) lim ln(3x

- In x)

- (f) Inn

2/Limits

- (g) Iim x(x + 4)
- lim x cos
- (h) Jim e'
- lim 3 '
- /s sin x
- (k) h m
- e' In x

- (a) lim
- 2 (b) Iim I (c) Iim sin x

Use limits to sketch the graph:

- (a) e-'
- (b) 3 + 2e5=

Decide if the discontinuity is removable and,

remove it with an appropriate definition of f(0)

(a) Try to find the limit as x

In this case,

f has a discontinuity which is neither point nor jump nor infinite

- (b) Find the limit as x

(c) Use (a) and (b) to help sketch the graph off for x > 0

but deliberately avoided the forms 0/0,

- - x and a few others

- an abbreviation for lsm

function f(x) which approaches 0 as x

- -> a function g(s) which approaches 0 as x

Unlike problems say of the form 0/3,

- which all have the answer 0,
- 0/0 problems can produce a variety of answers

we have the following table of values:

- numerator denominator

But consider a second possible table of values:

- numerator 2/3
- denominator

In this case the quotient approaches 2

- numerator denominator

the limit form 0/0 is called indeterminate

a limit form is indeterminate when

Indeterminate Limits

different problems of that form can have different answers

an indeterminate form is a conflict between one function pulling one way and a second function pulling another way

the small numerator is pulling the quotient toward 0,

while the small denominator is trying to make the quotient x or

result depends on how "fast" the numerator and denominator each approach 0

In a problem of the form x/x,

the large numerator is pulling the quotient toward x,

while the large denominator is pulling the quotient toward 0

- the base,

which is positive and nearing 0,

is pulling the answer toward 0,

- while the exponent,
- which is nearing 0,

is pulling the answer toward 1

- and on how "hard" they pull

In a problem of the form 0 x x,

the factor approaching 0 is trying to make the product small,

while the factor growing unboundedly large is trying to make the product unbounded

the base tugs the answer toward x while the exponent,

- which is nearing 0,
- pulls toward 1

- the base,
- which is nearing 1,
- pulls the answer toward I,

while the exponent wants the answer to be x if the base is larger than 1,

or 0 if the base is less than 1

In a problem of the form x

the first term pulls toward x while the second term pulls toward

V and x are also indeterminate

- -xx=x0 x x x,0 x

- x,(-x)

- - (-x'),(0+)° 1" x° xa Every indeterminate limit problem can be done

we do not accept "indeterminate" as a final answer

if a problem is of the form 0/0,

there is an answer (perhaps 0,

- or "no limit"),