Calculus

IMPORTANT REVIEW TOPICS ON CALCULUS FUNCTIONS There are two fundamental processes of Calculus, Differentiation and Integration. These processes are applied to functions. To carry them out, one has to be familiar with functions. A function is a Rule. Usually` we use small` letters f, g, h` ...... to denote different functions a a a or different rules, such as f x = x @ 1, g x = 10x 2 , h x = x 2 + 3x @ 2 etc.

Definition: Let A and B be two non-empty sets. A function f from A to B is a rule that associates, with each value of x in set A, exactly one value f(x) in set B. The function is indicated by the notation f :A Q B. f(x) is read “f of x”. We usually consider functions for which sets A and B are sets of real numbers. f(a) is` the a value of f(x) when ` a x=a. If f x = 2x @ 3 then f 2 = 2.2 @ 3 = @ 1

The set A, which contains all possible values of the variable x is called the Domain, and the set B which contains the corresponding values of f(x) or y is called the Range of the function (Range consists only of those elements of B which are actually paired with elements of A). We call x the independent variable because we choose it first, and from that we calculate y, which is called dependent variable. Example: Find the Domain & Range of the`function. a (a) f(x) = Sin x. (b) f x = 2x 2 @ 8x + 2 Solution : xQ can be any real number, so the Domain is the set of real numbers R or a) Here P x 2 R . All values of Sin x lie between -1 and 1. So, Range is @ 1 ≤ y ≤ 1 or R

y 2 R :@ 1 ≤ y ≤ 1 . S

b) It can be rewritten as

y = f x = 2x 2 @ 8x + 2 = 2 x 2 @ 4x + 2 ` a

b

c

= 2 x 2 @ 4x + 4 @ 8 + 2 = 2 x @ 2 @ 6 b

c

a2

`

Here x can be any real number, so the Domain is the set of real numbers R or As the value of square of any real number cannot be negative therefore, B c R S Thus y ≥ @ 6. So, Range is the interval @ 6, 1 or y 2 R : y ≥ @ 6 .

`

P

Q

x 2R .

x @ 2 ≥ 0. a2

g b 1f +f hf @ ff 1f w w w w w w wc ff 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Example : Let f x = 2x @ 5x + 3. Evaluate f 2 , f @ f , f pa , f 2 h 2

` a

`

f

` a

a

Solution : Substituting the values of x in f(x) we get ` a 2 f 2 = 2 A2 @ 5.2 + 3 = 8 @ 10 + 3 = 1 g2

1f 1f 1f 1f f f f f 5f f f @ f = 2A @ f @5 @ f +3 = f + f + 3=6 2 2 2 2 2 f

g

f

f

g

bw bw bw w w w w w wc w w w w w wc2 w w w w w wc w w w w w w w f p a = 2 p a @ 5 p a + 3 = 2a @ 5 p a + 3

B ` C B C a2 ` a 2 ` a ` a @ 5 1 + h + 3 @ 2 A1 @ 5.1 + 3 2 1 + h ff 1f + hf @ ff 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f

=

h

h

D b

E

2 1 + 2h + h @ 5 1 + h + 3 @ 2 @ 5 + 3 c

2

`

a

@

A

f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = f h B C 2 @ 5 @ 5h + 3 @ 0f 2 + 4h + 2h f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = f h 2 @ hf +f 2h f f f f f f f f f f f f f f f f f f f f f f f f f f f f = = @ 1 + 2h h

A function can also be defined differently for different sets of values of x. `

a

` a

Example : Find f @ 1 and f 2 where the function f(x) is defined as follows : X for x<0 ` a \2x + 1 f x =Z 2 for x ≥ 0 1 @ 2x Solution : ` a Since @ 1<0, using f x = 2x + 1

we get f @ 1 = 2 @ 1 + 1 = @ 1

Since 2 ≥ 0, using f x = 1 @ 2x 2

we get f 2 = 1 @2.2 = 1 @ 8 = @ 7

` a

`

` a

a

`

a

2

` a

GRAPHS OF FUNCTIONS The graph of f is the set of all points (x,y) such that y = f(x) Or The graph of f is the graph of the equation y =f(x) To draw the graph of y = f x we make a table consisting two columns : one for x and one for y [or f(x)]. We take different values of x and calculate the corresponding values of y [or f(x)]. Then we plot the points on a graph paper and join them with a smooth curve to get the graph of the function . ` a

Example : Draw the the graph of f x = x 3 ` a

Solution : We first make a table. We take different values of x and calculate the corresponding values of y [or f(x)]. Then we plot the points on a graph paper and join them with a smooth curve to get the graph x:

-2

y or f(x) :

-8

3f f @ f 2 27 f f f f f f @ f 8

-1 -1

1f f @ f 2 1f f @ f 8

0 0

1f f f 2 1f f f 8

1 2

3f f f 2 27 f f f f f f f 8

2 8

ABSOLUTE VALUE The absolute value of a real number x, written Therefore,

| 5 | = 5,

LM as LxM=

V

x @x

if x ≥ 0 if x < 0 A

| -5 | = -(-5) = 5.

INTERVAL NOTATIONS The concept of intervals is very useful in calculus. An uninterrupted portion of a number line is called an interval. The interval represents the collection of all the infinite number of points in that portion. We may write an interval like [-3,2] which means all the real numbers on the number line from -3 to 2 or all real numbers x where @ 3 ≤ x ≤ 2 A The interval (-3,2) means @ 3<x<2 .

(a,b) is called an open interval

where

[a,b] is called an closed interval

where

b

B

a,b = x | x 2 R and a<x
S

a,b = x | x 2 R and a ≤ x ≤ b C R

[a,b) is an interval open at one end and closed B atctheRother, where S a,b = x | x 2 R and a ≤ x
S

Some moreRcommonly used intervals are : S @ a a, + 1 = x | x 2 R and x ≥ a `

a, + 1 = x | x 2 R and x>a a R

S

@1 ,a = x | x 2 R and x ≤ a

`

A R

@1 ,a = x | x 2 R and x
`

a R

Example :

S

S

Interval @ 2,1 = x | x 2 R and @ 2<x<1 b

c

R

S

Interval @ 2, + 1 = x | x 2 R and @ 2 ≤ x B

c

R

S

STRAIGHT LINES SLOPE OF A LINE

The slope of a line is a number which indicates the direction or the slant of the line. It is generally represented by “m”. For a non-vertical line passing through ` a ` a P x1 , y1 and Q x 2 , y 2 the slope of the line segment PQ is yf @ yf yf f f f f f f f f f f f f f f f f f f Δ f f f f f f f 2f 1f Slope = m = f = f = x 2 @ x1 Δx

Change in yf @ coordinate f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Change in x @ coordinate

The slope of a horizontal line is 0, because as we move along the line the x-coordinate Δy 0f f f f f f f f f f f f f f f f = f = 0A changes, but the y coordinate does not change. Δy = 0 here A So, m = f Δx Δx The slope of a vertical line is not defined, because as we move along the line the xΔy f f f f f f f f Δy f f f f f f f f coordinate does not change, but the y coordinate changes. Δx = 0 here A So, m = f = f Δx 0 which is not defined (a symbol with denominator 0 does not signify a number).

For other lines, if the slope is positive, then it slopes up i.e. with increase in xcoordinate, the y-coordinate increases. If the slope is negative, then it slopes down i.e. with increase in x-coordinate, the y-coordinate decreases. Parallel & Perpendicular lines:

If two lines l 1 and l 2 have slopes m1 and m2 : The lines are parallel or l1 || l 2 if and only if m1 = m2 The lines are perpendicular or l1 ? l 2 if and only if m1 Bm2 = @ 1 In the above picture the lines l 3 ? l 4 as their slopes are 1 and -1 and ` a m1 B m2 = 1 A @ 1 = @ 1

Example : Find the slope of the line joining points A (1,-3) and B (-2,1). Solution : Taking x1 = 1, y1 = @ 3, x 2 = @ 2, y 2 = 1 ` a yf @ yf @ @ 3f 4f 4f f f f f f f f f f f f f f f f f f f f 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 2f 1f =` a = =@ Slope of AB = m = 3 x 2 @ x1 @2 @1 @3 Example : Find the slope of a line perpendicular to the line joining A(-3,1) and B(2,3).

Solution : Slope of AB = m1 =

3f @ 1f f f f f f f f f f f f f f f f f f f f f f f f f f f 2f f f ` a=

2@ @3 5 If the slope of the line perpendicular to it is m2 then m1 Bm2 = @ 1 5f 2f f f f f multiply both sides with [ Bm2 = @ 1 2 5 5f 5f f f f f [ m2 = @ A # Slope of a line perpendicular to AB is @ A 2 2

DIFFERENT FORMS OF EQUATIONS FOR STRAIGHT LINES General or Standard form:

A linear equation of x and y represents a straight line and is called the general form or the standard form of the equation of the line. This is written as Ax + By + C = 0 where A,B,C are real numbers and both A and B are not 0 at the same time w w w w w w

Thus 2x @ 3y @ 5 = 0, p3 x + y = 0, 3x @ 2 = 0, y @ 4 = 0 are equations of straight lines. In 2x @ 3y @ 5 = 0, A = 2, B = @ 3, C = @ 5 A w w w w w w w w w w w w In p3 x + y = 0, A = p3 , B = 1, C = 0 A Many a times we are required to find the general form of equation of a line from its slope, a few known points on the line or its intercepts on the axes. There we take help of the following forms : Point-Slope form : ` a If the line passes through the point x1 , y1 and its slope is m , then the equation of the line can be written in the form ` a y @ y1 = m x @ x1 Intercept Form : If a line intersects the x-axis at A(a,0) and the y-axis at B(0,b) then a is called the xintercept and b the y-intercept. Then the equation of the line can be written in the form

xf f f f yf f f f + f = 1 where a ≠ 0, b ≠ 0 a b

Slope-Intercept form : If the slope of a line is m and it intersects the y-axis at B(0,b) ( i.e. the y-intercept is b ), then the equation is y = mx + b

Example : Find the equation of the straight line which (a) passes through the point (3,4) and has a slope -2. 2f f and crosses y–axis at -3. (b) has a slope f 3 Solution : ` a ` a a Using the point @ slope form y @ y1 = m x @ x1 we get ` a y @4 =@2 x @3 y @ 4 = @ 2x + 6 2x + y @ 10 = 0 this is the required equation A ` a b Using the slope @ intercept form y = mx + b we get 2f f y= f x @3 3 3y = 2x @ 9 2x @ 3y @ 9 = 0 this is the required equation A