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Methods of Calculus
Summer-2012
Lecture Notes: 1
Exponents Basic Denitions
For any real number base x , we dene powers of x :
x 0 = 1,
x1 = x,
x 2 = x .x ,
x 3 = x .x .x , etc.
(The exception is 00 , which is considered indeterminate)
Powers are also called exponents.
Exponents Basic Denitions
For any real number base x , we dene powers of x :
x 0 = 1,
x1 = x,
x 2 = x .x ,
x 3 = x .x .x , etc.
(The exception is 00 , which is considered indeterminate)
Powers are also called exponents.
Exponents Basic Denitions
For any real number base x , we dene powers of x :
x 0 = 1,
x1 = x,
x 2 = x .x ,
x 3 = x .x .x , etc.
(The exception is 00 , which is considered indeterminate)
Powers are also called exponents.
Exponents Basic Denitions
For any real number base x , we dene powers of x :
x 0 = 1,
x1 = x,
x 2 = x .x ,
x 3 = x .x .x , etc.
(The exception is 00 , which is considered indeterminate)
Powers are also called exponents.
Exponents Basic Denitions
For any real number base x , we dene powers of x :
x 0 = 1,
x1 = x,
x 2 = x .x ,
x 3 = x .x .x , etc.
(The exception is 00 , which is considered indeterminate)
Powers are also called exponents.
Exponents Basic Denitions
For any real number base x , we dene powers of x :
x 0 = 1,
x1 = x,
x 2 = x .x ,
x 3 = x .x .x , etc.
(The exception is 00 , which is considered indeterminate)
Powers are also called exponents.
Cont
Examples
20 = 1,
23 = (2)(2)(2) = 8
(=3)4 = (=3) × (=3) × (=3) × (=3) = 81.Also, we can
dene√fractional exponents in terms of roots, such as
x 21 = x , the square root of x.
√
√
1
3
Similarly, x 3 = 3 x , the cube root of x , x 5 = 5 x 3 , etc.
√
m
In general, we have x n = n x m , i.e., the nth root of x,
raised to the mth power.
√
1
(81) 2 = 81 = 9 , etc.
Finally, we can dene negative exponents: x −n = x1n .
Thus, x −1 = x1 , etc.
Cont
Examples
20 = 1,
23 = (2)(2)(2) = 8
(=3)4 = (=3) × (=3) × (=3) × (=3) = 81.Also, we can
dene√fractional exponents in terms of roots, such as
x 21 = x , the square root of x.
√
√
1
3
Similarly, x 3 = 3 x , the cube root of x , x 5 = 5 x 3 , etc.
√
m
In general, we have x n = n x m , i.e., the nth root of x,
raised to the mth power.
√
1
(81) 2 = 81 = 9 , etc.
Finally, we can dene negative exponents: x −n = x1n .
Thus, x −1 = x1 , etc.
Cont
Examples
20 = 1,
23 = (2)(2)(2) = 8
(=3)4 = (=3) × (=3) × (=3) × (=3) = 81.Also, we can
dene√fractional exponents in terms of roots, such as
x 21 = x , the square root of x.
√
√
1
3
Similarly, x 3 = 3 x , the cube root of x , x 5 = 5 x 3 , etc.
√
m
In general, we have x n = n x m , i.e., the nth root of x,
raised to the mth power.
√
1
(81) 2 = 81 = 9 , etc.
Finally, we can dene negative exponents: x −n = x1n .
Thus, x −1 = x1 , etc.
Cont
Examples
20 = 1,
23 = (2)(2)(2) = 8
(=3)4 = (=3) × (=3) × (=3) × (=3) = 81.Also, we can
dene√fractional exponents in terms of roots, such as
x 21 = x , the square root of x.
√
√
1
3
Similarly, x 3 = 3 x , the cube root of x , x 5 = 5 x 3 , etc.
√
m
In general, we have x n = n x m , i.e., the nth root of x,
raised to the mth power.
√
1
(81) 2 = 81 = 9 , etc.
Finally, we can dene negative exponents: x −n = x1n .
Thus, x −1 = x1 , etc.
Cont
Examples
20 = 1,
23 = (2)(2)(2) = 8
(=3)4 = (=3) × (=3) × (=3) × (=3) = 81.Also, we can
dene√fractional exponents in terms of roots, such as
x 21 = x , the square root of x.
√
√
1
3
Similarly, x 3 = 3 x , the cube root of x , x 5 = 5 x 3 , etc.
√
m
In general, we have x n = n x m , i.e., the nth root of x,
raised to the mth power.
√
1
(81) 2 = 81 = 9 , etc.
Finally, we can dene negative exponents: x −n = x1n .
Thus, x −1 = x1 , etc.
Cont
Examples
20 = 1,
23 = (2)(2)(2) = 8
(=3)4 = (=3) × (=3) × (=3) × (=3) = 81.Also, we can
dene√fractional exponents in terms of roots, such as
x 21 = x , the square root of x.
√
√
1
3
Similarly, x 3 = 3 x , the cube root of x , x 5 = 5 x 3 , etc.
√
m
In general, we have x n = n x m , i.e., the nth root of x,
raised to the mth power.
√
1
(81) 2 = 81 = 9 , etc.
Finally, we can dene negative exponents: x −n = x1n .
Thus, x −1 = x1 , etc.
Cont
Examples
20 = 1,
23 = (2)(2)(2) = 8
(=3)4 = (=3) × (=3) × (=3) × (=3) = 81.Also, we can
dene√fractional exponents in terms of roots, such as
x 21 = x , the square root of x.
√
√
1
3
Similarly, x 3 = 3 x , the cube root of x , x 5 = 5 x 3 , etc.
√
m
In general, we have x n = n x m , i.e., the nth root of x,
raised to the mth power.
√
1
(81) 2 = 81 = 9 , etc.
Finally, we can dene negative exponents: x −n = x1n .
Thus, x −1 = x1 , etc.
Cont
Examples
20 = 1,
23 = (2)(2)(2) = 8
(=3)4 = (=3) × (=3) × (=3) × (=3) = 81.Also, we can
dene√fractional exponents in terms of roots, such as
x 21 = x , the square root of x.
√
√
1
3
Similarly, x 3 = 3 x , the cube root of x , x 5 = 5 x 3 , etc.
√
m
In general, we have x n = n x m , i.e., the nth root of x,
raised to the mth power.
√
1
(81) 2 = 81 = 9 , etc.
Finally, we can dene negative exponents: x −n = x1n .
Thus, x −1 = x1 , etc.
Evaluating Functions
To evaluate a function is to Replace (substitute)
its variable with a given number or expression.
Evaluate the function f (x ) = 2x + 4 for
Just replace the variable ”x ” with ”5”:
f (5) = 2 × 5 + 4 = 14
Answer: f (5) = 14
x =5
Evaluating Functions
To evaluate a function is to Replace (substitute)
its variable with a given number or expression.
Evaluate the function f (x ) = 2x + 4 for
Just replace the variable ”x ” with ”5”:
f (5) = 2 × 5 + 4 = 14
Answer: f (5) = 14
x =5
Evaluating Functions
To evaluate a function is to Replace (substitute)
its variable with a given number or expression.
Evaluate the function f (x ) = 2x + 4 for
Just replace the variable ”x ” with ”5”:
f (5) = 2 × 5 + 4 = 14
Answer: f (5) = 14
x =5
Evaluating Functions
To evaluate a function is to Replace (substitute)
its variable with a given number or expression.
Evaluate the function f (x ) = 2x + 4 for
Just replace the variable ”x ” with ”5”:
f (5) = 2 × 5 + 4 = 14
Answer: f (5) = 14
x =5
Evaluating Functions
To evaluate a function is to Replace (substitute)
its variable with a given number or expression.
Evaluate the function f (x ) = 2x + 4 for
Just replace the variable ”x ” with ”5”:
f (5) = 2 × 5 + 4 = 14
Answer: f (5) = 14
x =5
Cont
Consider f (x ) = 1 − x + x 2
The "x" is just a place-holder(a dummy variable)! And
”f ” is just a name.
It would be the same function if I wrote:
f (q ) = 1 − q + q 2
w (A) = 1 − A + A2
h(j) = 1 − j + j2
Cont
I) Let us evaluate that function f (x ) = 1 − x + x 2 for
x = 3:
f (3) = 1 − 3 + 32
=1−3+9
=7
Cont
II) Let us evaluate the function f (x ) = 1 − x + x 2 for
x = 1/r :
f (1/r ) = 1 − (1/r ) + (1/r )2
III) Or evaluate the function for x = a − 4:
f (a − 4) = 1 − (a − 4) + (a − 4)2
= 1 − a + 4 + a2 − 8a + 16
= 21 − 9a + a2
Constant Function
Constant Function is a function of the form y = b,
where b is a constant. It is also written as f(x) =
b.
The graph of a Constant Function is a horizontal line.
The function f (x ) = 10 is constant since the values of
f (x ) do not vary .
Constant Function
Constant Function is a function of the form y = b,
where b is a constant. It is also written as f(x) =
b.
The graph of a Constant Function is a horizontal line.
The function f (x ) = 10 is constant since the values of
f (x ) do not vary .
Constant Function
Constant Function is a function of the form y = b,
where b is a constant. It is also written as f(x) =
b.
The graph of a Constant Function is a horizontal line.
The function f (x ) = 10 is constant since the values of
f (x ) do not vary .
Constant Function
Constant Function is a function of the form y = b,
where b is a constant. It is also written as f(x) =
b.
The graph of a Constant Function is a horizontal line.
The function f (x ) = 10 is constant since the values of
f (x ) do not vary .
Limit and Continuity
To say that
limx →p f (x ) = L,
means that Œ(x) can be made as close as desired to L by
making x close enough, but not equal, to p.
of a constant function is the constant itself.
Mathematically lim (c ) = c where c is a constant.
Limit
x →a
Example
2
limx →3 xx −−11 =
32 −1
3−1
=
8
2
=4
Examples
2
nd limx →1 xx −−11 .
Ans: Although the limit in question is the ratio of two
polynomials, x = 1 makes both the numerator and
denominator equal to zero. We need to factor both numerator
and denominator as shown below.
2
x −1)
limx →1 xx −−11 = limx →1 (x +(x1)(
−1)
= limx →1 (x + 1)
=1+1
=2
example
Evaluate lim a 2+a23a+4−3 .
a→∞
Solution: Here substituting a = ∞ in both polynomial makes
∞
form which is meaningless. So we need to factor out a
∞
highest power of indeterminate from both polynomials as
shown below.
a3 (1+ 2 − 3 )
lim a32+a23a+24−3 = lim a3 (2+a 4 a)3
3
a→∞
2
a→∞
a3
2 − 3
1+ ∞
∞
= 2+ 4
∞
= 1+2+0−0 0
= 21
f is said to be continuous at a point
following conditions are satised.
f (a) is dened
lim f (x ) exists
Function
a
if
the
x →a
lim f (x ) = f (a)
x →a
Roughly speaking, a function is said to be
continuous at a point if it does not have any
breaks, gaps or holes point at that point. Graphs
below are the examples of continuous functtion.
What can be infered from the above graphs?
f is said to be continuous at a point
following conditions are satised.
f (a) is dened
lim f (x ) exists
Function
a
if
the
x →a
lim f (x ) = f (a)
x →a
Roughly speaking, a function is said to be
continuous at a point if it does not have any
breaks, gaps or holes point at that point. Graphs
below are the examples of continuous functtion.
What can be infered from the above graphs?
f is said to be continuous at a point
following conditions are satised.
f (a) is dened
lim f (x ) exists
Function
a
if
the
x →a
lim f (x ) = f (a)
x →a
Roughly speaking, a function is said to be
continuous at a point if it does not have any
breaks, gaps or holes point at that point. Graphs
below are the examples of continuous functtion.
What can be infered from the above graphs?
f is said to be continuous at a point
following conditions are satised.
f (a) is dened
lim f (x ) exists
Function
a
if
the
x →a
lim f (x ) = f (a)
x →a
Roughly speaking, a function is said to be
continuous at a point if it does not have any
breaks, gaps or holes point at that point. Graphs
below are the examples of continuous functtion.
What can be infered from the above graphs?
f is said to be continuous at a point
following conditions are satised.
f (a) is dened
lim f (x ) exists
Function
a
if
the
x →a
lim f (x ) = f (a)
x →a
Roughly speaking, a function is said to be
continuous at a point if it does not have any
breaks, gaps or holes point at that point. Graphs
below are the examples of continuous functtion.
What can be infered from the above graphs?
example
Is f (x ) = xx −−11 continuous at x = 1?
3
Solution. No! because f (1) = 11−−11 =
3
0
0
which is meaningless.
Quiz
or False?
1) If a function f is not dened at x = a then it is not
continuous at x = a.
2) If f is a function such that lim f (x ) does not exist then f is
x →a
not continuous.
True
Intercepts of a straight line
The x-intercept of a line is the point at which the line
crosses the x axis. ( i.e. where the y value equals 0 )
x − intercept =(x , 0)
The y-intercept of a line is the point at which the line
crosses the y axis. ( i.e. where the x value equals 0 )
y − intercept =(0, y )
Intercepts of a straight line
The x-intercept of a line is the point at which the line
crosses the x axis. ( i.e. where the y value equals 0 )
x − intercept =(x , 0)
The y-intercept of a line is the point at which the line
crosses the y axis. ( i.e. where the x value equals 0 )
y − intercept =(0, y )
Example:
Find the x and y intercepts of the equation 3x + 4y = 12.
Solution
To nd the x-intercept, set y = 0 and solve for x .
3x + 4(0) = 12
=⇒ 3x + 0 = 12
=⇒ 3x = 12
=⇒ x = 12/3
=⇒ x = 4
cont
To nd the y-intercept, set x = 0 and solve for y .
3(0) + 4y = 12
=⇒ 0 + 4y = 12
=⇒ 4y = 12
=⇒ y = 12/4
=⇒ y = 3
Therefore, the x-intercept is (4, 0) and the y-intercept is (0, 3).
Slope of a st. line
The slope m of the straight line
(x1 , y1 ) and (x2 , y2 ) is given by
m = yx22 −−yx11
through the points
= vertical rise (x1 6= x2 ) (Slope Formula)
horizontal run
As you go from left to right,
Uphill = Positive Slope
Downhill = Negative Slope
And of course, no change in height means that the line
has zero slope.
Slope of a st. line
The slope m of the straight line
(x1 , y1 ) and (x2 , y2 ) is given by
m = yx22 −−yx11
through the points
= vertical rise (x1 6= x2 ) (Slope Formula)
horizontal run
As you go from left to right,
Uphill = Positive Slope
Downhill = Negative Slope
And of course, no change in height means that the line
has zero slope.
Slope of a st. line
The slope m of the straight line
(x1 , y1 ) and (x2 , y2 ) is given by
m = yx22 −−yx11
through the points
= vertical rise (x1 6= x2 ) (Slope Formula)
horizontal run
As you go from left to right,
Uphill = Positive Slope
Downhill = Negative Slope
And of course, no change in height means that the line
has zero slope.
graph of slopes
Examples
Find the slope of the line segment joining the points (1, −4)
and (−4, 2).
Solution: Label the points as x1 = 1, y1 = −4, x2 = −4, and
y2 = 2.
To nd the slope m of the line segment joining the points, use
the slope formula :
4)
m = 2−−(−
4−1
= 2−+54
= − 65
So, m = − 65 .
Graph with positive and negative slopes
Parallel Lines
Parallel lines have the same slope and will never
intersect.
Parallel lines continue, literally, forever without touching
(assuming that these lines are on the same plane).
If two lines l1 and l2 with slope m1 and m2 respectively are
parallel then m1 = m2 .
Parallel Lines
Parallel lines have the same slope and will never
intersect.
Parallel lines continue, literally, forever without touching
(assuming that these lines are on the same plane).
If two lines l1 and l2 with slope m1 and m2 respectively are
parallel then m1 = m2 .
Parallel Lines
Parallel lines have the same slope and will never
intersect.
Parallel lines continue, literally, forever without touching
(assuming that these lines are on the same plane).
If two lines l1 and l2 with slope m1 and m2 respectively are
parallel then m1 = m2 .
Perpendicular lines
Perpendicular lines intersect each other at 90 degree.
slope of perpendicular lines are the negative
reciprocals of each other
If two lines l1 and l2 with slope m1 and m2 respectively are
perpendicular to each other then m1 = − m12 .
the
Perpendicular lines
Perpendicular lines intersect each other at 90 degree.
slope of perpendicular lines are the negative
reciprocals of each other
If two lines l1 and l2 with slope m1 and m2 respectively are
perpendicular to each other then m1 = − m12 .
the
Equation of a straight line
If the equation of the line is given in the form
y = mx + b (Slope intercept form)
m is the slope and b is y-intercept
This form of a line's equation is called the slope-intercept
form, because b can be interpreted as the y -intercept of
the line, the y-coordinate where the line intersects the
y-axis.
The slope of the line dened by the linear equation
ax + by + c = 0 is: − ba .
Equation of a straight line
If the equation of the line is given in the form
y = mx + b (Slope intercept form)
m is the slope and b is y-intercept
This form of a line's equation is called the slope-intercept
form, because b can be interpreted as the y -intercept of
the line, the y-coordinate where the line intersects the
y-axis.
The slope of the line dened by the linear equation
ax + by + c = 0 is: − ba .
Equation of a straight line
If the equation of the line is given in the form
y = mx + b (Slope intercept form)
m is the slope and b is y-intercept
This form of a line's equation is called the slope-intercept
form, because b can be interpreted as the y -intercept of
the line, the y-coordinate where the line intersects the
y-axis.
The slope of the line dened by the linear equation
ax + by + c = 0 is: − ba .
Equation of a straight line
If the equation of the line is given in the form
y = mx + b (Slope intercept form)
m is the slope and b is y-intercept
This form of a line's equation is called the slope-intercept
form, because b can be interpreted as the y -intercept of
the line, the y-coordinate where the line intersects the
y-axis.
The slope of the line dened by the linear equation
ax + by + c = 0 is: − ba .