Download Calculus

Calculus
Functions
Def. Relation
A relation is a correspondence between two sets.
For us, we are usually interested in a correspondence between a subset of the real numbers
and a second subset of the real numbers ( they may or may not be identical).
Def. 1.1 Functions
A function is a rule that assigns to each element of one set ( the domain ) exactly one element of
another set (the range)
Since there are two sets we normally think of the first set as corresponding to the variable x
and the second set as corresponding to y.
ex. f(x) = x2
The domain:
looking for all values of x which are “permissible” ( values of x which yield a result for y).
When we write f( 2 ), we are looking for the value obtained when x is replaced with 2
in this case f(2 ) = ( 2 ) 2 = 4. This means when x =2, y will have a value of 4.
The domain is the set of all allowed input (x’s) values. For f(x) = x2, any real number will work
the domain of f :
we write Dom (f ) or Dom f =
in this example we get Dom f = all real numbers and in interval notation we write ( - ∞, ∞)
The range of a function: looking for output values
( values of y – that result from using all permissible values of x)
We write Range ( f) = Range f
and in this example Range f = all real numbers ≥ 0 ,in interval notation we write [0, ∞ ]
It is usually easier to find the range if you have a graph in mind.
f(x) = x2 is the same as y = x2
and the graph of y = x2 ( or any function of the form f(x) =az2 + bx + c ) is a parabola that
open upward if a > 0 ( downward if a < 0 )
You can see from the graph that all x’s are being used and the
only y’s generated are those that are ≥ 0
1
ex. Look at x2 + y2 = 4. Functions of this type (x2 + y2 = r2 ) represent circles centered at the
origin (with radius r ).
Let x = 0 you will get y2 = 4 which means y =  2 . So, for x = 0 we get two values of y; 2 and -2
This means we do not have a function. We have a relation between x and y.
Graph of x2 + y2 = 4
Circle with center at the origin (0,0)
and of radius 2.
Cross the x-axis at 2 and -2
these are called x-intercepts
Cross the y-axis at 2 and -2
these are called y-intercepts
You can draw a vertical line in such a way that you will cross the graph in more than one point ( one
value of x produces more than 1 value of y corresponding to the chosen x)
– this means that the graph is not a function. It is just a relation.
Ex. Look at y =
4  x 2 . Is y a function of x ? In other words, if a permissible value of x (domain) is
chosen, will you get one single value of y that corresponds to that x ? If the answer is yes, then you have function.
let x = 2 → y = 4  (0) 2 = 2, let x = 1, then y = 4  (1) 2 = 3
It appears that we may have a function. This is not a proof – just an example.
What is the domain ? you can only use values of x so that 4 – x2 ≥ 0 → -2 ≤ x ≤ 2
What about its range ? May not be so obvious but
4  x 2 ≥ 0 and since y =
4  x 2 ≥0
so, y ≥ 0 ( the range ).
The graph may be a better place to look for the range. You have to understand that y =
represents the top portion of a circle: square both sides →
4  x2
y2 = 4 – x2 → x 2 + y2 = 4 but only the top portion of the circle. Why only the top portion ?
2
Def. Let f(x) represent a function from the set A into the set B. If the range of f (x) is equal to the
set B, we say that f(x) is onto ( onto the set B ).
ex. f(x) = x2 is a function from the set of real numbers ( the domain ) into the set of real
numbers ( not the range ). The range is actually all real numbers ≥ 0.
We can say that f(x) is a function on the set of nonnegative real numbers ( range → [0, ∞)
Def. one-to-one (1-1)
A function is one-to-one iff x1 ≠ x2  (implies) f(x1) ≠ f(x2 )
that is, different values of x always have correspondingly different values of y
Note: horizontal line test
if a horizontal line crosses the curve at more than one point it is not 1-1.
Used in tandem with the vertical line test, you can determine if the you have a function that is 1-1.
f(x) =x2 + 4
You can see this curve represents a function by the vertical line test. When you draw
a horizontal line, it’s possible to intersect more than two points on the curve. f(x) is
a function that is not 1-1.
Notation:
we use iff to mean if and only if
these are equivalent statements: mean the same thing
1. if A , then B; if B, then A
2. A is true if and only if B is true
3. A is true iff B is true
3
Def. 1.2 Even Functions.
A function is f(x) is said to be even iff f(-x ) = f(x) for all x  Dom(f ).
If x is replaced with –x and no change occurs, then f(x) is said to be even. The graph of an even
function is symmetric with respect to the y – axis ( mirror image across the y-axis).
This means if (x,y ) is a point on the graph of f(x), then so is ( -x,y)
ex. f(x) = x2 → f( - x ) means replace x with – x → f( - x ) = ( - x ) 2 = x2,
no change occurred
:. f (x) = x2 must be even
You can see the symmetry across the y-axis.
ex. f(x) = 4  x 2 , find f(- x ) = 4  ( x) 2 = 4  x 2
same thing as what you started with
f(-x ) = f( x ) → f (x) must be even.
You may not be able to see it, but the graph is the top half of a circle.
You have symmetry about the y-axis.
4
Rules from trigonometry.
1. If f(x ) = sin x, then f ( - x ) = sin ( - x ) = - sin x
2. If f(x) = cos x, then f( - x ) = cos ( - x ) = cos x
3. If f(x) = tan x, then f(-x ) = tan ( - x ) = - tan x
You can see that only cos x remains unchanged when x is replaced with –x.
So, f(x) = cos x must be an even function.
ex. What about f(x) = | x |.
x
if x ≥ 0
The actual definition of | x | =
- ( x ) if x < 0
This is a branch function.
So, if f(x) = | x | ,
be even.
Whose graph is a V shape curve with vertex at x = 0
f( - x ) = | - x | = | x | , no change occurred → f(x ) = f( -x ) , f(x) must
We define an odd function in a similar way. It turns out that unlike whole numbers, functions do
not necessarily have to be even or odd, they could be neither.
5
Def. 1.3 Odd Functions
A function f(x) is said to be odd provided f(-x) = -f(x) for all x’s in the domain of f.
Odd functions are symmetric with respect to the origin. This means if (x,y ) is a point on the graph of f(x),
then ( -x, -y) is also a point on the graph.
When we write f(-x) = - f(x) we mean to say that when x is replace with –x
we do not end up with what we started with ( f(x) ) but with the opposite of what we started with ( - f(x))
ex. Let f(x) = x3 – 3x.
Find f(-x ). f( -x ) = ( -x )3 - 3(-x) = -x3 + 3x
This means the function is not even.
is - x3 + 3x = x3 – 3x ( does f(-x) = f(x) ) ? NO!
but if - x3 + 3x = - ( x3 – 3x ) ( does f(-x ) = - f(x) ) ? Yes – we got the opposite of what we
started with.
We have an odd function.
Notice: f( 2 ) = (2)3 – 3(2) = 8 – 6 = 2 what about f( - 2 ) ? Since we have an odd function, we
expect to get the opposite of f(2) → -2.
f( - 2 ) = ( -2 )3 – 3(-2) = - 8 + 6 = - 2.
Notice the mirror image across the origin (symmetry)
ex. Let f(x) = x3 + 4. Is f(x) odd or even ?
Try to see if f( - x ) = f(x) or if f(-x) = - f(x) or maybe it’s neither !
6
Other examples:
1. Find the domain
1
of f(x)=
, the only value of x that is not permissible ( allowed) is x =3. Why ?
x3
1
1
f( 3 ) =
=
which represents an undefined value ( can not divide by zero). x = 3 is the only
33 0
value of x that generates this kind of problem.
( -∞, 3)  ( 3, ∞ )
Domain of f(x) is: the set of all real numbers not equal to 3 or
2. What about f(x) =
2x ?
In the real number system you are not allowed to have the square root of negative numbers →
2 – x must be nonnegative → 2 – x ≥ 0.
So, 2 ≥ x ( or x ≤ 2 ) .
The domain of f(x): all real numbers x, so that x ≤ 2 and in interval notation ( - ∞, 2 ]
3. A third example: Find the domain of f(x) =
1
9x
2
.
As done in the two previous examples; 9 – x2 ≠ 0 and 9 – x2 ≥ 0. Which means 9 –x2 > 0.
Recall from algebra on how to solve 9 – x2 > 0. This can be written as x2 – 9 < 0
Factor and find critical values: x2 – 9 = 0 → ( x – 3)(x + 3) = 0 → critical values: x = 3, - 3
if x > 3: x2 – 9 > 0 that’s not what we want → we can not use values of x > 3
if -3 < x < 3: x2 – 9 < 0 that’s what we want → we will use values of x, -3 < x < 3
if x < -3: x2 – 9 > 0 that’s not what we want → use can not use values of x < - 3
The solution to x2 – 9 < 0 → all x’s so that
-3<x<3
The domain: all x’s so that x is between -3 and 3 ; -3 < x < 3
and in interval notation we write ( -3, 3 )
7
Composition of Functions
Given two functions: f(x) = 3x -2 and g(x) = x2, we can find f(2) = 3 (2) – 2 = 4 and also g(4) = 16.
We write (g o f )(x) to mean (g o f ) (x) = g ( f ( x ) ). This means evaluate f at x and the result is used
to evaluate g.
(g o f ) ( 2 ) = g ( f (2) ) : First find f( 2 ) = 3 (2 ) – 2 = 4, then replace f(2) with 4,
(g o f ) (2 ) = g ( f (2) ) = g ( 4 ). Evaluate g( 4 ) = 42. So, (g o f ) (2 ) = 16.
Notice:
Normally ( g o f ) (x ) ≠ ( f o g ) (x ) .
A special relationship exists between the functions when they are equal.
ex. f(x) =
x  2 , Dom( f ) = ? ______________________, x – 2 ≥ 0 → x ≥ 2 or [ 2, ∞ )
g(x) = x2 + 5 , Dom (g) = ? _____________________, all real numbers; ( - ∞, ∞ )
Find (f o g ) (x ) = f( x2 + 5 ) =
Find (g o f )(x) .
g(
( x 2  5)  2 =
x 2  3 ; domain of f o g: all real numbers
x  2 ) = ( x  2 )2 + 5 = x – 2 + 5 = x + 3 → (g o f) ( x ) = x + 3
domain of (g o f )( x ) is determined by f(x): [2, ∞ )
even though the function y =x + 3 has a domain = set of all real numbers
8
example: page 25: #15
|x| + 1
if x < 1
f(x) =
- x + 1 if x ≥ 1
what is the domain ? _____________________________ Range ? _________________________
Graph:
Domain: all real numbers
Range: y ≥ 1
or
y≤0
9
Inverse Functions:
Def. 1.4 one-to-one (1-1)
A function is one-to-one iff x1 ≠ x2  (implies) f(x1) ≠ f(x2 )
that is, different values of x always have correspondingly different values of y
Ex. Let f(x) = 2x + 4 → if x1 ≠ x2 then 2x1 ≠ 2x2 → 2x1 + 4 ≠ 2x2 + 4 → f (x1 ) ≠ f(x2 )
“ this way of thinking may lead to trouble” – think of it slightly different manner.
The idea of being one-to-one can also be illustrated with a graph. Just as we use vertical lines to
determine if a relation is a function – we can use a horizontal line test on a function to determine if it
is one-to-one.
When we draw f(x) = 2x + 4 we get a slant line similar to the following graph.
No matter what horizontal line we select, the line will always
cross the graph ( line ) at only one point → one-to-one.
The vertical line test would show that we have a function.
You would normally show 1-1 by working with the contrapositive of a statement;
If A  B ( A implies B ) ,
the converse of this statement is
If B  A
the contrapositive of this statement would be; not B  not A
Assume that f(x1 ) = f(x2 ), show that x1 = x2.
For f(x) = 2x + 4
if f(x1) = f(x2 ) → 2x1 + 4 = 2x2 + 4 → 2x1 = 2x2 → x1 = x2.
By using the contrapositive f(x) = 2x + 4 must be 1-1.
10
Def. 1.5 Inverse Function
Let f(x) be a given 1-1 function defined on some interval I. f(x) has an inverse function, written f -1( x)
with the property that f(f -1 (x) ) = x = f -1(f(x) ) .
Notes:
1.
1
,
f ( x)
i.e., the idea of inverse function does not mean the reciprocal of a function.
f -1 (x) does not mean f
-1
(x ) =
2. If f(x) does represent a 1-1 function, then f (x) has an inverse function.
In some cases it is relatively easy to find an inverse relation – which in turn may end up being a function.
ex. f(x) = 2x – 3 → Find the inverse function of f(x), f -1 (x).
Think of this function as y = 2x – 3.
Process:
1. interchange x and y; x  y →
2. Solve for y. →
x = 2y - 3
x + 3 = 2y →
x3
y
2
3. Replace y with f -1 (x). This is the inverse relation ( in this case its more – it is a function)
x3
2
y = f -1 (x ) =
To convince yourself that f(x) and f -1 (x) are inverse functions find
(f o f-1 )(x) or (f-1 o f ) (x). In either case you should get an answer of x , unless they are not
inverse functions.
ex. Try the same method with f(x) = x3 + 2. if it is one-to-one we can find it’s inverse function.
Think of it as y = x3 + 2 → interchange x and y: → x = y3 + 2 →
solve for y: y =
f -1 ( x ) =
3
3
x2
x2
11
Brief Review of Trigonometry.
Angle:
Initial and terminal side of an angle:
550
2200
4000
Measurements:
degree: Divide a circle into 360 equal pie sections with a common point at the center of the circle.
Each of the central angles is said to be of measure 10.
radians: Take any circle of radius r and some central angle θ (vertex at the center of the circle )
If the length of the intercepted arc is also of length r, then the angle is said to be of measure
1 radian.
Relationship between radians and degrees:
1 revolution is equivalent to 3600 . Why ? This is how we defined a degree ( 1 degree ).
Also, 1 revolution is equivalent to 2π radians. Why ? Think of the circumference of a circle.
C = 2πr. How many radians in 1 revolution ( in C ) ? 1 radian intersects a length r
C/r represents the number of radians → 2πr/r = 2π → how many radians.
12
So, we have 1 revolution = 360o = 2π radians → 1800 = π radians.
This can be used to convert from degrees to radians ( or from radians to degrees)
ex. convert 1500 to radians → 1500 = 1500 •
ex. convert

180 o
radians =
150 0
5
• π radians =  radians
0
6
180
2(180 0 )
2
2
2
180 0
 radians to degrees →  radians =  radians •
=
= 1200
3
3
3
 radians
3
Def. Let θ be an angle so that it’s initial side is on the positive x-axis with vertex at the origin. We
say θ is in standard position.
θ is in quadrant I if its terminal side is in quadrant one
quadrant II if its terminal side is in quadrant two
quadrant III if its terminal side is in quadrant three
quadrant IV if its terminal side is in quadrant four
Sign of an angle in standard position:
1. If the angle being measured is found by moving the initial side in a counterclockwise direction, then
the angle is said to be a positive angle.
2. If the angle being measured is found by rotating the initial side in a clockwise direction, then the
angle is said to be a negative angle
Draw each of the following angles in standard position and determine its quadrant.
a) 3000
b) - 4000
c) 1400
13
We can define the trigonometric functions in terms of a point as well as in terms of the sides of a right
triangle.
Def. Let θ be an angle in standard position with point P(x,y) a point on the terminal side of θ.
The distance between the point P(x,y) and the origin (vertex of the angle) is called the radius vector
of θ. We will use r to represent this length.
You can find the value of r by the following relationship: x2 + y2 = r2.
comes from ?
Do you know where this
Define the six trig. functions in terms of x, y , and r. You do not need to have a triangle just a point
on the terminal side of θ.
The sine of θ: sin θ =
y
r
the cosine of θ: cos θ =
x
r
the secant of θ: sec θ =
r
x
, and the cotangent of θ: cot θ =
x
y
the tangent of θ: tan θ =
y
x
Also,
the cosecant of θ: csc θ =
r
,
y
P( -3, 4)
For the point P, x= -3, y = 4 → x2 + y2 = r2 . What is the value of r ? r = _______
It is also useful but not necessary to either think of the point P as being on the unit circle centered at the
origin. x2 + y2 = 1. This does is provide a simple value for r (r=1).
14
Special angles:
Axes angles: 00, 900, 1800, 2700, 3600, .... as well as negative angles → - 900, - 1800, - 2700, ...
(90o)
( - 270o)
( 4500 )
Examples.
Find cos 900. Think of a point on the terminal side of 900. How about ( 0, -1 ) ?
Remember r measures the distance from the point to the origin or use x2 + y2 = r2.
If x = 0, y = -1, then r must equal → 1.
→
cos 900 =
x
0
=
r
1
What about csc 1800 ? use a point on the terminal side of 1800 → P ( -1, 0 ) .
Any point will do – but these are small values to work with.
What about r ?
csc 1800 =
r = 1. r > 0 for any point.
r 1
 but what results from a fraction in which the denominator is zero ?
y 0
an undefined value → this means csc 1800 is undefined.
15
Find
tan ( - 3600 ). _______________
cot 900 = __________
sin ( - 900 ) . ________________
cos ( -π ) = _________
Def. coterminal angles
Angles A and B are said to be coterminal if they are written in standard position and they have the
same terminal side.
ex.
ex.
900 and ( -2700 ) are coterminal angles
1000 and 4600 are coterminal
Note: functions of coterminal angles will always be equal
ex. sec 230 = sec ( - 3370 ) = sec ( 3830 ) because all three angles are coterminal.
Def. Related angles.
Let angle A be given in standard position. The acute angle Ar formed with the x-axis and the
terminal side of A, is called the related angle of A.
16
ex. Find the related angle of
1) 1400
2) - 3000
Answer: 400 ( must be positive )
Answer: 600
Other angles that are of special interest:
45o ( in radians:

)
4
600 ( in radians:

)
3
300 ( in radians:

)
6
If an angle has a related angle with a measure equal to the angles above, we will also think of those angles
as special angles.
These two triangles are of special interest ( because of their angles).
900 – 450 – 450 right triangle:
450
2
One angle is equal to 900 and the two acute angles are both equal to 450
Regardless of the value of the hypotenuse, the length of the other two legs
will be identical. For simplicity, we usually use the values shown here.
1
900
450
1
900 – 600 – 300 right triangle:
600
The side opposite 300 is always the shorter
leg and it is always ½ the length of the
hypotenuse.
2
1
900
30o
3
17
Now that we have an idea of triangles we define the trigonometric functions in terms of sides of the right
triangle with respect to an angle; hypotenuse, adjacent, and hypotenuse. The definitions still have to
match the definitions using x, y, and r.
Define:
sin θ =
y
→ now we write
r
cos θ =
hypotenuse
adjacent
x
→ now we write cos θ =
→ Also, sec θ =
r
hypotenuse
adjacent
tan θ =
opposite
y
→ now we write tan θ =
x
adjacent
sin θ =
opposite
hypotenuse
→ Also, csc θ =
hypotenuse
opposite
→ Also, cot θ =
adjacent
opposite
Find
sin 300. ___________ sin 300 = opposite / hypotenuse. Use the triangle in the previous page to get the
answer.
sec 600. ___________
tan 450. ___________
The angles do not have to be in standard position.
Use these two right triangles to find the values.
A
7
D
E
41
40
24
C
9
B
25
F
Find tan B. __________
sec F. ________
cos A. ___________
cot E. ________
sin B. ____________
csc F. _________
18
What happens when the angles are
1) greater than 900 ( not acute angles )
2) negative angles
3) angles that are not special
The answer to the third part requires more study of trigonometry or the use of a calculator.
But in the special cases of 1) and 2) we can use the idea of a related angle.
Property.
Let A be a given angle with related angle Ar
( such as 1500 and 300: 300 is the related angle of 1500 )
The trigonometric function of A will be numerically equal to the same trig. function of Ar.
The sign may be different(it will depend on what quadrant angle A is in), but the numbers will
be the same.
If you use your calculator ( or remember how to do it ), you will find that
cos 150o = -
3
3
while cos 300 =
. Almost the same value except for the sign.
2
2
Because of the way the trig. functions are defined we can easily remember the sign of each trig.
functions in a given quadrant.
All Students Take Calculus ( A, S, T, C )
S(students)
A ( all )
T(take )
C (calculus)
This will remind us that ;
All functions are positive in quadrant I,
the sine and its reciprocal function ( cosecant ) are positive in quadrant II
the tangent and its reciprocal function (cotangent) are positive in quadrant III
the cosine and it reciprocal function ( secant ) are positive in quadrant IV
19
Find
sin 2100.
The related angle of 2100 is 300. Draw it to see it.
So that means that sin 2100 is numerically equal to sin 300
Go back to the special angles (triangle) and find that sin 300 =
opposite
1

hypotenuse 2
That means sin 2100 = to either ½ or - ½. Which one ? 2100 is in quadrant III and only the tangent
and its reciprocal are positive in quadrant III. This means that sin 2100 is negative.
Answer: sin 2100 = - ½ .
More Examples:
1) Find tan ( - 450 ). _______________
2) csc ( - 150 0 ) . _____________
3) cos ( 4200 ) . _______________
Write the trigonometric functions as
f(x ) = sin x, f(x) = cos x, f(x) = tan x , .....
Now we can graph each one of them. Concentrate on the first three
The first time you see these functions you would create a table of values for x and y and plot the points
(x,y). The only restriction we make is that x must be in radians.
20
Graph of f(x) = sin x
Notice that it repeats every 2π units→ period is 2π
Graph f(x) = cos x
Graph of f(x) = tanx
The dark vertical lines represent vertical asymptotes – they are not part of the graph – but they help us draw the graph
unlike the sine and cosine ( and their reciprocal functions) – the tangent and cotangent do not repeat every 2π units but every π
units. We say the have a period of π.
21
Using the idea of reciprocals we construct the graph of the other three functions.
Graph of f(x) = sin x
and g(x) = csc x
Graph of f(x) = cos x
and g(x) = sec x
Graph of f(x) = tanx and g(x) = cot x
In all three cases above – ignore the vertical lines – they just represent asymptotes.
22
Identities and Properties
There are times that certain forms of a trig. equation are preferable or necessary over other forms.
We use identities to convert from one to the other.
Reciprocal Identities:
1. csc x =
1
sin x
1
→
csc x
csc x • sin x = 1
2. sec x =
1
1
or if needed cos x =
→
cos x
sec x
sec x • cos x = 1
3. cot x =
1
1
or if needed tan x =
→
tan x
cot x
cot x • tan x = 1
or if needed sin x =
Thus we can separate the six trig. functions in pairs in terms of being reciprocals of each other.
sin x, csc x;
csc 32o =
ex.
cos x, sec x;
1
sin 32 0
cos 230o =
tan x, cot x
1
sec 230 0
cot ( -24o ) =
1
tan( 24 0 )
We can also separate them into pairs in terms of what we call cofunctions;
sinx , cos x are cofunctions of each other
sec x and csc x are cofunctions
tan x and cot x are cofunction
Two acute angles are said to be complementary if their sum equals 900 ( 600, 300; 70o, 200 ... )
Property:
Given two complementary angles and a pair of cofunctions;
the function of one angle will be equal to the cofunction of the second angle.
ex.
sin 24o = cos 66o,
tan 120 = cot 780,
sec 500 = csc 40o
Make sure not to get the pair of reciprocal functions mixed up with cofunctions – they are different
pairs, except for tangent and cotangent functions.
23
More on Identities
sin x
cos x
4. tan x =
5. cot x =
cos x
sin x
We defined Trig. functions in terms of x, y, and r ( initially) and we invoked Pythagorean Thm (theorem).
→ x2 + y2 = r2
What happens if you divide both sides by
(we use θ instead of x so as not to confuse x the angle with x the x-coordinate)
y2
x2
+
r2
r2
6) r2: →
7) y2: →
y2
x2
+
y2
y2
y
x
r
r2
→ ( )2 + ( )2 = ( )2 → cos2 θ + sin2 θ = 1
2
r
r
r
r
=
=
y
x
r
r2
→ ( )2 + ( )2 = (
) → cot2 θ + 1 = csc2 θ
2
y
y
y
y
8) x2 → .................................................................................... → tan2 θ + 1 = sec2 θ
These three identities are generally referred to as Pythagorean Identities
A couple more useful identities:
9. Double angle identity (for the sine function )
sin 2x = 2sin x cos x
ex. sin 1200 = sin 2(600 ) =
2 sin 600 cos 600
ex. sin 6θ = sin 2(3θ ) = 2 sin 3θ cos 3θ
10. Double angle identity (for the cosine function) – there are three different forms.
a) cos 2x = cos2 x - sin2 x
c) cos 2x = 1 – 2sin2 x
b) cos 2x = 2cos2 x - 1
ex. cos2 150 - sin 2 152 = cos 2(150) = cos 300 =
ex. 2cos2 ( 22.50 ) - 1 = cos 2(22.50 ) = cos 450 =
3
2
1
2
=
2
2
24
Review of Algebra.
Factoring: writing a polynomial as a product of other polynomials.
GCF:
greatest common factor.
2x + 4y = 2 ( x + 2y )
What about
2sinA + 4cosA ? 2sinA + 4cos A = __________________
Difference of Squares:
x2 – 9y2 = (x )2 - ( 3y )2 = ( x – 3y ) (x + 3y )
4x2 – 49 = (2x)2 - ( 7)2 = __________________
what about sin2 A - cos2 A ?
sin2 A - cos 2 A = ( sin A - cos A ) ( sin A + cos A )
What about 9 – tan2 B ? 9 – tan2 B = ______________________________
Multiply: (2x – 3y)(2x+3y). _____________
Sum (Difference ) of Cubes :
8 – x3 = (2)3 - (x)3 = ( 2 – x ) ( 22 + 2x + x2 )
1st factor
2nd factor
1st factor: obvious by rewriting the original polynomial
2nd factor: sopps → square the first (2)2, opposite sign, product of terms (2) • (x), positive
sign, and square of the last term (x)2
8x3 + 27y3 = ( 2x)3 + (3y)3 = ( 2x + 3y) ( (2x)2 - (2x)(3y) + (3y)2 ) = (2x + 3y)( 4x2 – 6xy + 9y2)
Factor:
27 – x3 = ______________________
27 + 64x3 = _____________________
25
Exponential Functions.
Let f(x) = 3x
Find f( 2).
f(2) = 32 = 9; f(0 ) = 30 = 1, f( -2 ) = 3-2 =
1
32
What does the graph of f(x) = 3x look like ?
Graph is asymptotic to the x-axis ( gets close to the axis ( closer an closer) but never
crosses it)
Logrithmic functions
f(x) = log 3 x
Find f(9).
f(9) = log 3 x use y instead of f(9).
y = log 3 x
The logarithm represents an exponent with the following pattern.
if y = log 3 x, then 3 y
=
x.
in our notation x = 9 → 3 y = 9, → y = 2
So, log 3 9 = 2.
Find log 4 64. ______________
log 2 16 = ____________
Graph of f(x) = log 3 x.
curve is asymptotic to the y-axis.
26
Special bases.
1) if b = 10 we write log 10 x = log x and call it the common logarithm
2) if b = e, we write log e x = ln x and call it the natural logarithm.
Note:
as n → ∞, ( 1 + 1/m)m → e, e  2.718281228
Three Basic Properties of logarithms
1) log xy = log x + log y for any base
2) log (
x
) = log x - log y for any base
y
3) log xk = k log x for any base
There are other properties(more than what’s listed here).
4) logb b = 1
5) log b 1 = 0
6) log b bk = k , .....
Back to inverse functions.
We now have one more example of inverse functions.
f( x) = 4x and g(x) = log4 x are inverse functions.
27
Three forms of the equation of a line:
1) slope intercept form: y = mx + b
2) point-slope form: (y- y1) = m ( x – x1)
3) intercept form:
x y
 1
a b
examples:
1. find the equation of the line that has slope 3 and passes through the point ( 4, 1)
Since m = 3 → equation: y = 3x + b we can substitute for x and y : ( 4, 1)
→ (1) = 3(4) + b → 1 – 12 = b
→ - 11 = b , now we know the value of b
→ your equation: y = 3x – 11
2) Find the equation of the line that passes through the point ( 2, -4 ) and has slope m = -3
You can use the process from above or
y – (-4) = - 3 ( x – 2 ) and simplify → y + 4 = - 3x + 2 → y = - 3x – 2
3) Given the equation:
x y
 1
3 2
where does the line cross the x-axis (x-intercept) → 3
where does the line cross the y-axis ( y-intercept) → 2
28
Parabolas:
The graph of a function of the form f(x) = ax2 + bx + c is called a parabola
1) it opens upward if a > 0
2) opens downward if a < 0
3) it has a vertex at V(x, y ) where
x=
b
2a
and y = f( x ) = f (
b
)
2a
ex. Sketch the graph of f(x) = 2x2 – 4x + 5
parabola that opens upward with vertex at V(x,y)
x=
 (4)
b
b
=
= 1 , y = f(
) = f( 1) = 2(1)2 – 4(1) + 5 = 2 -4+5 = 3
2a
2a
2( 2)
Vertex at V(1, 3), you can also find the y-intercept by setting x = 0 and solving for y.
f(0) = 2(0)2 – 4(0) + 5 = 0 – 0 + 5 = 5.
Graph:
ex. What about the graph of f(x) = -x2 + 2x + 1
29
Limits
Examples:
1) To what value(s) does y (f(x) )approach as x gets “real close” to c.
lim f ( x ) = ?
x c
if f (x) = 2x + 3, and c = 5
lim f ( x ) = lim (2 x  3) = _________________
x 5
x c
the closer x gets to 5, the closer 2x gets to 10, the closer 2x + 3 gets to 13.
answer: 13.
2. We can use tables to find out the answer.
When we say x gets close to 5 we mean
a) x gets close to 5 from the left
lim f ( x ) = lim (2 x  3)
written
x
f(x)
4
11
4.5
12
x c 
x 5
4.9
12.8
4.99
12.98
4.999
12.998
it appears the answer is 13, i.e., lim (2 x  3) =13
x 5
b) x gets closet to five from the right
written
x
f(x)
6
15
5.5
14
lim f ( x ) = lim (2 x  3)
x c 
5.1
13.2
x 5
5.01
13.02
5.001
13.002
it appears the answer is 13, i.e.,
lim (2 x  3) =13
x 5 
Whenever these two limits are equal we can say that
lim (2 x  3) =13
x 5
i.e., if lim f ( x) = lim f ( x ) = L, then lim f ( x) exists and it equal L
x c
x c
x c
30
3. Other examples
x2
. You are being asked the question “ as approaches 2,
x 2 x  x  2
x2
what happens to the value of y,
( 2
)?
x x2
a) Find lim
2
Look at it using the same approach as before. We can also see that
1
x2
x2
=
=
as long as x ≠ 2
2
x 1
x  x  2 ( x  2)( x  1)
1
x2
we work with lim
.
x

2
x 1
x x2
1
1
1

as x gets close to 2,
gets close to to
2 1 3
x 1
Instead of working with lim
x 2
Answer: lim
x 2
2
x2
= 1/3
x x2
2
sin x
. In the previous example it did not take long to recognize the fact that you can get the
x
sin 0
answer by substituting for x. That does not work here because you get
undefined
0
(indeterminate) value
b) Find lim
x 0
You can use the table value to arrive at generally good guess.
x
f(x) =
-0.1
-0.01
0.1
0.01
sin x
x
sin x
= _________
x 0
x
What is your guess ? lim
31
4. If you have a graph, you can also find limits by looking at the “functional values” – not always.
x2  4
Graph f(x) =
, may look complex, but with a little bit of algebra →
x2
x2
x 2  4 ( x  2)( x  2)
 x  2 if x ≠ 2
=
=
1
( x  2)
x2
when you look at f(x) = x +2, y=x +2, it’s a lot easier to graph.
this is almost the graph of
x2  4
f(x) =
, what is the difference ?
x2
Answer: they differ at one value @x = 2. Why ?
x 2  4 lim
Find lim
= x  20
x2 x  2
( x  2)( x  2)
x2
= lim
x2
x2
= 4. , this works as long as x ≠2, and when
1
we say x→2, that means x gets close to 2, but never
reaches it.
Note: f(2) does not exist – in this case x =2 which means you can not reduce by canceling.
Find lim
x 2
x2
. ( x ≠2)
x x2
2
lim
x 2
x2
1
1
x2
= lim
= lim
= .
3
x  x  2 x  2 ( x  2)( x  1) x  2 x  1
2
32
Note: it is possible for a function to
1. have a limit at x=c → at most 1point of discontinuity, no breaks or gaps
but not a y value → i.e., lim f ( x) exists but f(c) does not exist
x c
2. have a limit at x=c but have a y value different from the limit → i.e., lim f ( x) ≠ f(c).
x c
3. have a y-value at c ( f(c) exists ) but no limit
ex. 1 f(x) =
ex. 2 f(x) =
x2  4
has a limit at x =2 but f(2) does not exist.
x2
x+ 1
5
2
x -1
if x > 2
if x = 2
if x < 2 ,
Why is this true ?
f(2) = 5 it exists, lim f ( x) =3, but they are not equal
x 2
Note: to find this limit you need to use the left-hand limit and the right-hand limit
You need to decide which of the three branches has to be used when calculating the limit.
x2 -1 if x ≥ 2
ex. 3 f(x) =
x
if x < 2,
f(2) = 3 it exists, lim f ( x) does not exist because
lim f ( x) = 2
x2
x 2
and
lim f ( x) =3 → different values → lim f ( x) does not exist
x2
x 2
Another example. (see notes: branch function with missing points – jumps)
33
Definition of a limit
When we write
lim f ( x ) = L,
x c
we mean to say that “ the closer that x gets to c, the closer that f(x) gets to L. The following is the
mathematical definition.
Definition: For every  > 0 (epsilon), there is a   0 (delta: which usually depends on  )
such that
0 < | x – a | <  → | f(x) – L | <  .
i.e., if x is within  of c but is not equal to c, then f(x) is within  of L.
Look at the graph below: from your class notes
Basic Limit Theorems
1. lim k = k, : the limit of a constant function is that constant
x a
ex. lim 3 = 5
x 5
1
=½
x 5 2
ex. if f(x) = ½ find lim f ( x) . lim f ( x) = lim
x 5
x 5
2. lim x  x , the limit of x as x → a will always equal a.
xa
3. lim x p  x p
x a
4. lim [ f ( x)  g ( x)]  lim f ( x) + lim g ( x ) ( same idea for subtraction)
xa
x a
x a
if these individual limits exist
34
examples:
page 83 #17
lim x  1 =
x 3
lim ( x  1) =
x 3
lim ( x)  lim (1) =
x 3
x 3
3 1  2
page 83 #19
lim sin x = sin (π/2) = 1
x  / 2
Sometimes when we try to find the limit by using f(c) – substituting for x – we end up with zero in the
denominator – in that case we try something similar to the following problems.
page 84 #50.
(e x ) 2  1
(e x  1)(e x  1)
e 2x  1
ex 1
lim
lim x
= lim
=
=
= e0 + 1 = 1 + 1 = 2
lim
x
x
x

0
x 0 e  1
x 0
x

0
1
e 1
e 1
page 84 #54.
lim
x 3
( x  3)( x  2)
x2  x  6
x2
3 2
= lim
= lim
=
=5
2
x 3 x  2
x 3 ( x  3)( x  2)
32
x  5x  6
page 84 #56.
lim
x 0
3 x  3
, multiply by the “conjugate” of the numerator
x
3 x  3
•
x
3 x  3
3 x  3
=
3 x 3
=
x
x( 3  x  3 )
x( 3  x  3 )
1
=
if x ≠0
( 3  x  3)
So,
lim
x 0
3 x  3
3
1
1
= lim
=
=
x 0
x
6
3 x  3
2 3
35
Another Example:
#62
( x  x) 2  x 2
, x replaced by 0 generates 0/0
x 0
x
lim
2 x  x
( x  x) 2  x 2
x 2  2 xx  x 2  x 2
2 xx  x 2
=
=
=
= 2x+ x
1
x
x
x
→
So,
( x  x) 2  x 2
= lim (2 x  x) = 2x + 0 = 2x
x 0
x 0
x
lim
The “Squeeze Theorem”
Let f(x), g(x) and h(x) be given functions with the following conditions;
a) f(x) ≤ g(x) ≤ h(x) and
b) lim f ( x) = L , lim h( x) = L,
x a
x a
then
lim g ( x ) = L .
x a
The proof of the following theorem uses the theorem above.
Theorem:
lim
 0

sin 
= 1, θ is measured in radians.
Proof.(see notes)
36
As a consequence of the theorem above:
lim
 0
sin 
1
= lim

 0
(
1
=

sin 
)
lim (
 0
=

sin 
)
1
=1
1
Fundamental Trigonometric Limits
1.
lim
sin 
 0
2. lim

= lim

Pf.

=

lim

•

cos   1

sin   sin 
1  cos 
1  cos 2 
sin 2 
•
=
=
=
1  cos 
 (1  cos  )  (1  cos  )  (1  cos  )
sin 
(1  cos  )
1  cos 
 0
= 0

1  cos 
=
sin 
Now,
Since
cos   1
 0
1  cos 
=1
sin 
 0
1  cos 
 0

= lim
= lim
 0
=-
sin 

(1  cos  )

sin 
0
= (1) • (
)=0
 0 1  cos 
11
• lim
, then lim
 0
cos   1

= lim
 0
 (1  cos  )

= - lim
 0
1  cos 

= -(0)=0
Note:
sin kθ ≠ k sin θ, in other words, you can not factor out a coefficient of θ inside a trig. function
sin 3θ ≠ 3 sin θ
One more example:
Find
sin 3 x
sin 3x
sin 3 x
)] = 3 lim
= lim [3(
if you think of 3x as θ ( replace 3x with θ )
x 0
x 0
x 0
x
3x
3x
sin 
= 3 lim
= 3( 1 ) = 3
lim
 0

37
More on Basic Limit Properties:
--- you have already seen the first four properties:
1. lim k = k, : the limit of a constant function is that constant
x a
2. lim x  x , the limit of x as x → a will always equal a.
xa
3. lim x p  x p
x a
4. lim [ f ( x)  g ( x)]  lim f ( x) + lim g ( x ) ( same idea for subtraction)
xa
x a
x a
if these individual limits exist
-------------------------------------------5. lim [ f ( x)  g ( x)] = lim f ( x) • lim g ( x )
xa
x a
x a
in words: the limit of a product is equal to the product of the limits
6. lim
x a
lim f ( x)
f ( x)
= xa
as long as the individual limits exist
lim g ( x)
g ( x)
xa
in words: the limit of a quotient is equal to the quotient of the limits
7. If we use the fifth property, we get
lim cf ( x ) = c lim f ( x )
x a
x a
Def. (recall definition of limit)
38
Graph of y =
Graph of y =
sin x
, the graph below is over a larger interval ( see more of the graph)
x
cos x  1
, the graph below is for a larger interval (see more of the graph)
x
39
Homework examples:
page 84 #60
lim [(
x 0
1(4)  1( x  4) 1
4  x  4) 1
1
1 1
 )  ] = lim [(
) ]  lim[ (
] =
x

0
x4 4 x
4( x  4)
x
x 0
4( x  4) x
= lim [(
x 0
x
1
1
1
1
)  ] = lim [(
)] =
= 
x

0
16
16
4( x  4) x
4( x  4)
page 84 #79
1  ex
1
, since e-x = x
x
x 0 e  1
e
lim
ex 1
1
x
x
1  ex
→ lim x
= lim x e = lim xe
x 0 e  1
x 0 e  1
x 0 e  1
1
ex 1
1
1
1
 x
= lim x = =1
x
x 0
x 0 e
1
e
e 1
= lim
page 84 #71
lim
x 0
sin x(1  cos x)
, notice that
2x 2
1 sin x (1  cos x)
sin x(1  cos x)
= •
•
2
2
x
2x
x
sin x(1  cos x)
=
x 0
2x 2
So, lim
= lim ( (
x0
lim (
x0
1 sin x (1  cos x)
•
•
)
2
x
x
(1  cos x)
1
1
sin x
) • lim (
) • lim (
)= (
)(1)(0)= 0
x0
2 x0
x
2
x
Recall:
x if x≥ 0
|x | =
-(x) if x < 0
40
| x|
,
x 0 x
this is one of those cases in which you actually have to take a left and right limit individually
Find lim
a) lim
x 0
 ( x)
|x|
= lim
, why is |x| = - (x) or - x ?
x0
x
x
= lim
x 0
b)
lim
x0
1
1
1
( x)
|x|
= lim
, why is |x| = x ?
x0
x
x
1
= lim  1
x 0 1
Two horizontal rays (ignore the vertical line segment)
Conclusion:
Since the limit from the right does not agree with the limit from the left –
| x|
does not exist
x 0 x
lim
Continuity and limits (see notes)
a) continuous at x=a
b) discontinuous but limit exists
c) discontinuous and limit does not exist
d) discontinuous, f(a) does not exist,
but limit exists
41
Def. (continuous)
We say f(x) is continuous at x =a iff
1) f(a) is defined ( you can actually find a y-value at x=a )
2) lim f ( x) exists ( you can actually find a limit )
x a
3) lim f ( x) = f(a) , ( the values from the first two parts are equal )
x a
ex.
Is f(x) = | x | continuous at x = 0 ?
1) f(0) = | 0 | = 0 , it exits
2) lim | x| = | 0 | = 0 , it exists
x→0
3) Does lim |x | = f( 0 ) ? If so, f(x) = |x | is continuous at x = 0.
x→0
Is f(x) =
sin x
continuous at x = 0 ?
x
1) f(0 ) =
sin 0
, an undefined value → f(0) does not exist !
0
2) We know that the limit as x → 0 exists , but it will not matter because
3) lim f(x) ≠ f(0), since f(0) does not exist
x→0
:. f(x) is discontinuous at x = 0, and hence discontinuous.
x2  9
is it continuous at x=3 ? No ! f(3) does not exist. Are there any other
x3
points of discontinuity ? No ! This is called a removable discontinuity.
ex. What about f(x) =
The limit lim f ( x) exits. What is it ? lim f ( x) = ________
x 3
x 3
42
Another Example:
The Greatest Integer Function
let f (x) = | x |,
| 0 |= 0
| x | represents the greatest integer that is less than or equal to “x”
| 0.2 | = 0
| 1 | =
1
| 0.99 | =
| 1.5 |
=1
0
| 2.1 | = - 3
| x |
is discontinuous at all integer values of x.
| x |
is continuous on the intervale [0,1) and over similar intervals
Example from notes:
f is continuous on ( - ∞, -1], (-1, 1], and (1, ∞ )
43
page 94 #10.
Find lim
x4
x 2
0
= ?, a simple substitution will not do since we have an indeterminate form;
0
x4
x 2
x 2
x 2
=
•
x4
x4
x 2
(rationalize the numerator, this is sometimes an effective way or removing the indeterminate form)
change the fraction:
x 2
x 2
=
•
x4
x4
x 2
=
x 2
x4
( x  4)( x  2)
=
1
x 2
, this is only true if x ≠4
Now,
lim
x4
x 2
= lim
x4
x4
1
x 2
=
1
4
An alternate way of looking at this limit is by rewriting the denominator as a difference of squares
→
x 2
=
x4
( x )  ( 2)
2
and as before: lim
x4
page 94 #25:
lim ln (x-3) = ?
x 3
x  2
x  2
2
=
( x  2)( x  2)
x 2
= lim
x4
x4
1
x 2
=
=
1
x 2
1
4
as x → 3-, (x-3) → 0-, values are negative
you can not take the natural log. (ln ) of negative values – ln (x-3 ) is undefined for these values – no
limit ! ( actually -∞)
44
page 94 # 44:
If f(x) =
x3
if given, find values of x at which f is not continuous
x2  9
find the points of discontinuity
answer: since you have a polynomial over a polynomial ( a rational function)
the only problem will be when the denominator = 0.
x2 – 9 = 0 → x =
 3 → It will discontinuous at x = 3 or -3
page 94 #64:
Look at the following branch function;
4 sin x
, if x < 0
x
g(x) =
a – 2x, if x ≥ 0 , a is a constant ( but unknown)
Find the value of a so that g(x) is continuous on ( -∞, ∞).
The only problem ( possible point of discontinuity) is at x = 0.
1) f(0) = a – 2(0) = a , y-value exists and it equals a.
2) To find the limit , you need to consider the
a) limit from the left
lim g(x) = lim
x 0
x 0
4 sin x
sin x
= 4 ( lim 
)= 4(1)=4
x 0
x
x
b) limit from the right
lim g(x) = lim ( a – 2x ) = a – 2(0) = a
x 0 
x 0
c) the limit will exist if
lim g(x) = lim g(x) → 4 = a
x 0 
x 0
Continuous at x =0 if a = 4.
45
page 95 #51
½ x + 1 if x ≤ 2
f(x) =
3 - x, if x > 2
For x < 2 ( f(x) represents a polynomial – polynomials are continuous everywhere
Similarly, for x > 2 ....
The only question is at x=2.
1) f(2) = ½ ( 2) + 1 = 2
2) lim f(x) = ?
x→2
lim f(x) = lim (3  x) = 3 - (2) = 1
x 2 
x2
1
1
lim f(x) = lim ( x  1) = ( 2)  1 =2 , they are different
x 2
x2 2
2
Hence,
lim f(x) = does not exist.
x→2
f is not continuous at x=2.
Def. (removable discontinuity)
Let f discontinuous at x=a. We say that x=a is a removable discontinuity if
lim f(x) exists, but f(a) does not.
x→a
a
example in which f is discontinuous at
x=a and limit does not exist
46
#65
2, if x ≤ -1
f(x) = ax + b, of -1 < x < 3
-2, if x ≥ 3
Want it to be continuous at all x’s – in particular at x=-1 and x=3 ( those are the only values of x that
are in question.
x=-1: to be continuous here you want f(-1) = 2 = lim f(x)
x→-1
lim f ( x) = lim  2 = 2
x  1
x  1
and lim  f ( x) = lim  ax  b = a(-1) + b = -a + b
x  1
x  1
→ -a + b = 2
Also,
x= 3: to be continuous at x=3, you want f(3) = - 2 = lim f ( x)
x 3
lim f ( x) = lim (ax  b) = 3a+b and lim f ( x) = lim ( 2) = -2
x  3 
x 3
x 3
x 3
→ 3a+b = - 2
Now we solve the following system of equations.
-a + b = 2
3a + b = - 2
→
multiply top equation by 3
-3a + 3b = 6
3a + b = - 2 →
Add: 4b = 4 → b = 1 → a = -1.
47
Infinite Limits
ex.
1
1
= + ∞, if you allow x to decrease to zero from both sides x2 → +0 → 2 →+∞
2
x
x
lim
x0
Try a few values of x to convince yourself: x = 0.1 = 1/10 , 0.01= 1/100, ....
ex.
1
=?
x 0 x
1
= + ∞ ( just as you did above
x 0 x
1
lim = - ∞ ( look at values of x to convince yourself; x= -0.1, -0.01, ... )
x 0 x
lim
lim
1
= does not exist – since the left and the right limit do not agree.
x 0 x
→ lim
ex. lim
x 3
x
, just as above the limit will not exist because
x 9
2
x
=+∞
x 9
lim
x 3
#9
lim 
x 3
x
x 9
2
x 3
ex. lim [1  ( x  1) 2 
x 3
x
= -∞
x 9
2
3
(think of x  3.1 ) , x2 → 9 + →

0
=
lim 
x 1
lim
2
x
x 9
2
1
1000 x  1  1
7
x 2  9 → 0+
=-∞
] = ? _____________
48
Asymptotes -vs- Removable discontinuities
ex.
x
, has a vertical asymptote at x = 2
x2
Notice that
ex. lim
x 2
lim
x2
x
=+∞
x2
lim
x2
x
= -∞
x2
( x  2)( x  1)
( x  1)
x2  x  2
= lim
= lim
= 3,
x 2
x2
x2
1
x2
we say f has a removable discontinuity at x = 2 ( missing point – “hole” )
-- no asymptote
#12
g(x) =
2 x
, has vertical asymptotes at x= 0 and x = 1
x (1  x)
2
#24
4( x  3)( x  2)
4( x 2  x  6)
4 x 2  4 x  24
f(x) = 4
=
=
3
2
2
3
2
x  2 x  9 x  18 x
x[ x ( x  2)  9( x  2)]
x( x  2 x  9 x  18)
=
4( x  3)( x  2)
=
x[( x 2  9)( x  2)]
4( x  3)( x  2)
,
x[( x  2)( x  3)( x  3)]
V.A at x = 0, x = 3 with removable discontinuities at x = 2, x = -3
#27
h(t) =
V.A. at t = - 2
49
#32
g(θ) =
tan 

=
sin 
, a problem will occur when denominator is equal to zero
 cos 
→ θ = 0 or cos θ = 0
V.A. or whenever cos θ = 0 ( this occurs at (2n+1)π/2 or at π/2 + nπ )
Why not at θ = 0 ?
When you find the limit as θ → 0 you get
sin 
sin 
1
lim
= lim
= 1(1) = 1
  0  cos 
 0 
  0 cos 
lim
the limit exists
→ θ = 0 is a removable discontinuity.
50
Tangent Lines to a curve
You know that you can find the slope of a line if
a) you are given two points; P(x1, y1) and Q(x2, y2) → slope = m =
b) an equation; ax + by = c → y = -
y 2  y1
, or
x 2  x1
a
a
x c, m=
b
b
We will try to describe slope with respect to a general curve. The difference is that
for a line the slope is the same throughout the line – that may not be true for a general curve.
Different points may generate different slopes.
Note: physical example → if x = time and y = f(x) represents displacement (distance), then the slope
will represent the velocity.
51
We need the idea of secant and tangent lines(and their slopes).
3
2
f(x) = x x 6 x3
A secant line requires two points on f(x). The moment you have two points – you can find the slope of
the secant line. Let’s say the points are
are at (x, f(x) ) and ( x + h, f(x+h) ), h: represents the distance from the first point to the second point
msec =
y
f ( x  h)  f ( x )
=
, the difference of the y’s over the difference of the x’s ( x+h – x = h )
x
h
if x represents time and y represents distance, then msec = average velocity
Tangent line requires only one point on the curve, say (x, f(x) ) .
f ( x  h)  f ( x )
h 0
h
If x = time and y= distance, then mtan = instantaneous velocity ( not over a time interval –
but at an exact time)
We define the slope of the tangent line by m tan = lim
If lim
h 0
f ( x  h)  f ( x )
exists, then we call it derivative of f at x, denoted by several notations.
h
Here are two of the most common;
f / (x ) = lim
h 0
f ( x  h)  f ( x )
( read f prime of x )
h
dy
f ( x  h)  f ( x )
= lim
,
h 0
dx
h
if this value exists, we say f is differentiable at x.
52
ex. Find the slope of the tangent line to the graph of y = x2 at the point (1, 1)
find the slope of the tangent line.
mtan = lim
h 0
f ( x  h)  f ( x )
, except here x =1. → (make the replacement at the end)
h
f ( x  h)  f ( x )
( x  h) 2  x 2
( x 2  2 xh  h 2 )  x 2
mtan = lim
= lim
= lim
h 0
h 0
h 0
h
h
h
h( 2 x  h)
( 2 x  h)
2 xh  h 2
= lim
= lim
= lim
=2x
x

0
h

0
h 0
h
1
h
The fact that your answer for the slope has a variable – means that the slope will vary
depending on the value of x. In this case x = 1 → mtan = 2(1) = 2.
That’s the derivative of f(x) at x =1 → f / ( 1 ) = 2 ( the slope of the tangent line)
Once you know the slope of the line m = 2 and a point that it passes through P(1, 1) - you can find
the equation of the tangent line.
by using the slope intercept form:
or the point-slope form:
y = mx + b
(y – y1 ) = m (x –x1 )
y = 2x + b
y - (1) = 2 ( x – ( 1) )
Use (1, 1) to solve for b
Simplify
(1) = 2(1) + b
y -1 = 2x – 2
-1 = b
→
y = 2x -1
→ y = 2x – 1
53
Ex. Find the equation of the tangent line to y = 1/x at the point where x = -2.
Graph of y = 1/x
Need the slope of f(x) at x = -2, need f / ( -2 ).
1
1
1
1
(
 )
(
 )
f
(
x

h
)

f
(
x
)
1
f / (x) = mtan = lim
= lim x  h h = lim [ x  h h  ]
h0
h 0
h 0
h
h
1
h
= lim [(
h 0
= lim [
h 0
xxh 1
1
1 1
x
xh 1
 )  ] = lim [(

)  ] = lim [
 ]
h

0
h

0
xh x h
xh
x
h
x ( x  h) h
1
1
1
h
1
1
 ] = lim
 ] = lim [
= 2 , we wanted f / (x) at x = -2
h

0
h

0
x ( x  h) 1
x ( x  h)
x ( x  h) h
x
→ f / (-2) =
1
1
=
= mtan at x = -2
2
4
( 2)
To find the equation of the tangent line → need slope = m =
at x = -2, find y. f(-2) =
1
and a points (x,y)
4
1
1
1
=
, since f(x) =
2
x
( 2 )
Now find the equation as in the previous example:
y – (- ½ ) = - ¼ (x – ( - 2 ) → ......
Answer: y =
1
x 1
4
HW: page 120: #1, 2, 5-42 / 9, 21, (27, 29: (a) part )
54
Notes from Thursday – June 9, 2005
Recall the definition of the derivative ( the slope of tangent line at x )
f / (x) = mtan = lim
h 0
f ( x  h)  f ( x )
h
notice that we can re-write as lim
h 0
f ( x  h)  f ( x )
f ( x  h)  f ( x )
1
= lim
•
h 0
h
1
h
In some cases – the algebra looks simpler in the latter form.
#21.
f(x) =
1
- find f / (x).
x 1
f / (x) = mtan = lim
h 0
f ( x  h)  f ( x )
f ( x  h)  f ( x )
1
= lim
•
h

0
h
1
h
1
1

( x  1)  ( x  h  1) 1
1

= lim x  h  1 x  1 
= lim
h 0
h

0
1
h
( x  h  1)( x  1)
h
= lim
h 0
1
1
1
 =
( x  h  1)( x  1) h
( x  1) 2
ex. For any line of the form f(x) = mx + b – we know that the slope is m – does it agree with our idea of
slopes ( f / (x) ) ?
f / (x) = mtan = lim
h 0
f ( x  h)  f ( x )
m( x  h)  b  (mx  b)
mh
m
= lim
= lim
= lim
=m
h

0
h

0
h
0
h
h
h
1
This matches our idea of slopes with respect to lines –
Notice that f / (x ) = m → what is the graph of this function ?
a horizontal line
ex. Graph of y = x2, has one tangent with slope 0 → one horizontal tangent
Graph of y =
x → What is the connection between these two functions ?
55
x if x ≥ 0
ex. f(x) = | x | =
- (x) if x < 0
There is no problem when you look at f(x) = x ( if x ≥ 0 ) we can find that f / (x ) = 1
f(x) = -x ( if x < 0 ) we can find that f / (x) = - 1
The only real question is what happens at x= 0 ? What is f / ( 0 ) ?
Find the derivative by using the definition of the derivative.
f / (x) = mtan = lim
h 0
f ( x  h)  f ( x )
h
and since we know we want x=0 →
f ( 0  h )  f ( 0)
| h | 0
= lim
this can only be evaluated by using the
h

0
h
h
left and right limits separately
f / (0) = mtan = lim
h 0
lim
h0
| h | 0
h0
= lim
=1,
h 0
h
h
lim
h0
 ( h)  0
| h | 0
= lim
= -1
h 0
h
h
the limit from the right is different from the limit from the left → lim
h0
| h | 0
does not exist
h
Alternative Definition of the derivative at a point.
f / ( c ) = lim
h 0
f (c  h )  f (c )
and if we let x = c + h, then as h → 0, x → c
h
→ So, f / ( c ) = lim
x c
f ( x )  f (c )
xc
56
Theorem.
If f(x) is differentiable ( the derivative exists) at x =c, then f(x) is continuous at x = c.
(Differentiability Implies Continuity )
The converse is not true → Continuity does not imply differentiability
Proof.
lim [ f ( x)  f (c)] = lim [
x c
x c
f ( x )  f (c )
f ( x )  f (c )
 ( x  c)] = lim [
]  lim [ x  c]
x c
x

c
xc
xc
= f / (c ) • 0 = 0
So, lim [ f ( x)  f (c)] = 0 → lim f ( x ) - lim f (c ) = 0 →
x c
x c
x c
lim f ( x ) = lim f (c ) = f(c)
x c
x c
This is the definition of continuity → f is continuous at x = c.
Theorem.
If f(x) = c, the f / ( c ) = 0
(The derivative of a constant is always zero )
Proof.
f ( x  h)  f ( x )
cc
= lim
h
0
h
h
0
0
= lim = lim 0 = 0 ( remember that =0 if h≠0 )
h0
h0 h
h
f / (x) = lim
h 0
57
Theorem
If f(x) = xn, then f -1(x) = nxn-1 , for any positive integer n.
f ( x  h)  f ( x )
( x  h) n  x n
= lim
h 0
h 0
h
h
n
n 1
n2 2
n 3 3
x  nx h  _ x h  _ x h  ...  h n  x
= lim
h 0
h
n 1
n2 2
nx h  _ x h  _ x n 3 h 3  ...  h n
= lim
h 0
h
Pf. f / (x) = lim
h(nx n 1  _ x n  2 h 1  _ x n 3 h 2  ...  h n 1 )
h 0
h
n 1
n2 1
(nx  _ x h  _ x n 3 h 2  ...  h n 1 )
= lim
= nx n 1
h 0
1
= lim
Example:
f(x) = x2 → f / (x ) = 2x2-1 = 2x1 = 2x
Notation: remember that
dy
d
y ) is an alternate way of saying f / (x )
( or
dx
dx
so,
d 2
x = 2x
dx
Example:
y = x4 → y / = 4x3 or
d 4
x = 4x3
dx
Basic Derivative Rules:
d
c=0
dx
ex.
d r
x = rx r-1
dx
d
x =?,
dx
d
d 1/ 2
x =
x = ½ x1/2 – 1 = ½ x
dx
dx
-1/2
=
1
2 x
58
Theorem
d
d
d
[ f ( x)  g ( x)] =
f (x) +
g (x )
dx
dx
dx
This can be proved from the fact that the limit of a sum is equal to the sum of the limits
[ f(x) + g(x) ] / = f / (x ) + g / (x) or
Pf.
[ f ( x  h)  g ( x  h)]  [ f ( x)  g ( x)]
d
[ f ( x)  g ( x)] = lim
h 0
dx
h
[ f ( x  h)  f ( x)]
[ g ( x  h)  g ( x)]
lim
+ lim
= f /(x) + g / (x).
h 0
h

0
h
h
Theorem
If f is differentiable and c is any constant, then so is (cf)(x) = c(f(x) ) and
d
d
cf (x) = c
f (x)
→ if y = cf(x), then y / = c f /(x).
dx
dx
Proof.
[cf ( x  h)  cf ( x)]
c[ f ( x  h)  f ( x)]
[ f ( x  h)  f ( x)]
d
cf (x) = lim
= lim
= c lim
h

0
h

0
h

0
dx
h
h
h
= c f / (x).
d x
e = ex proof ?
dx
Also,
d
sin x = cos x (proof below )
dx
Theorem.
If y = sin x , then y / = cos x
Proof.
( i.e. ( sin x ) / = cos x )
[sin( x  h)  sin( x)]
d
sin x = lim
h 0
dx
h
since you get a zero in the denominator – simple substitution will not work
[sin( x  h)  sin( x)]
[sin( x) cos( h)  cos( x) sin( h)  sin( x)]
= lim
h 0
h 0
h
h
(because of a trig. identity
sin( x)[cos( h)  1]
sin( x)[cos( h)  1]  cos( x) sin( h)
cos( x) sin( h)
= lim
= lim
+ lim
h 0
h

0
h

0
h
h
h
→ lim
[cos( h)  1]
sin( h)
+ cos x lim
= sin(x) • (0) + cos(x) • 1 = cos x.
h 0
h 0
h
h
= sin x lim
59
Homework Problem.
d
cos x = sin x
dx
More examples:
#29 y =
Note:
x
x
→ y=
d
x = 1,
dx
d 1 / 2
x1 / 2
1
1
x
= x -1/2 →
= - ½ ( x )-1/2 – 1 = - ½ x -3/2 =
=
3/ 2
dx
x
2x
2x x
d
ax = a
dx
#42 Find the derivative
if h(x) =
1
1
2 x 2  3x  1
2x 2 3x
, then h(x) =
+
= 2x – 3 +
= 2x – 3 + x -1
x
x
x
x
x
→ h / (x) = 2 – 0 – 1(x )-2 = 2 – x-2 = 2 -
#49
2x 2  1
1
or
x2
x2
f(x) = 6 x + 5cosx
f(x) = 6 x + 5cosx = 6x1/2 + 5cosx → f / (x) = 6 ( ½ )(x)-1/2 + 5 (sin x )
f/ (x ) = 3x-1/2 + 5sinx =
3
+5sinx
x
60
#59 y = x + sin x on the interval 0 ≤ x ≤ 2π
→ Find all points at which the graph has a horizontal tangent line. ( slope of tangent line = 0 )
Want to find all x’s for which y/ = 0.
Find y / and set it equal to zero.
y / = 1 + cos x setting y / = 0 → 1 + cos x = 0
→ cos x = -1 this happens whenever x = π
If x= π, then y = x + sinx = π + sin π = π + 0 = π → the point is (π, π )
Homework:
61