Calculus Soruları 2

Chapter 1 : EQUALITY, LOGARİTHM and FUNCTION

Lineer Inequalities

Solve that inequality

2 < | 3x – 4 | Þ 2 < style=""> Þ 6 < style=""> Þ 2 < style="position: relative; top: 12pt;">

Solution Set : { x | x < È x > 2 }

Quadratic Inequalities

x2 – x – 6 > 0 Solve that inequality

x² – x – 6 > 0 Þ (x – 3) (x + 2) > 0

Solution Set : { x | x < -2 È x > 3 }

Solve for x if sin2x – 3cos2x + 2 = 0 is given where xÎ [ 0,2p]

sin²x – 3cos²x + 2 = 0 Þ 1-cos²x – 3cos²x + 2 = 0 Þ 4cos²x = 3 Þ cos x =

Solution set : { x = x = }

COORDINATE PLANE

Two points define a uniqe line on two dimensional plane.

a) Two points where A(x₁,y₁) , B(x₂,y₂)

b) Slope and one point y – y₁ = m (x – x₁)

c) Slope and y-intercept y = m x + b

Defines a uniqe line.

Suppose you are given points A(-2,3) and B(3,5) , write the equation of the line.

m = slope = = ; y = m x + b Þ 3 = . 2 + b , b =

y = . x + or we write 5 y – 2 x – 19 = 0 as standart form.

Paralel lines , Perpendicular lines

Parallel lines have equal slopes.

Perpendicular lines have slopes that are negative reciprocals.

Exercises:

I- Given l_{1 //} to the x-axis, write the equation of l₁ if it passes from (-3,2).

y = m x + n y = n is the equation of a horizontal line.

y = 0.-3 + 2 so we find y = 2 line

II- Suppose you are given 2x + 3y = 6 write the equation of the perpendicular line that passes from the origin.

a. Find the slope of the original line

b. Find negative reciprocal.

_c. Write equation of

a) 2x + 3y = 6 Þ 3y = -2x + 6 Þ y = x + 2 m_original =

b) =

c) y = x

III- Write the equation of a linet hat has y intercept 2, and passes from (-2,4)

y = m x + n Þ 4 = -2m + 2 , m = -1 Þ y = - x + 2

Intersection of Lines

Find the intersection of 2x + 3y = 6 and x – 2y = -2

2x + 3y = 6

x – 2y = -2 y = x = Point, (,)

CIRCLE EQUATION

(x-a)² + (y-b)² = r² Circle, with center (a,b) radius r

Circle & Line intersection

1 point of int. 2 pts. of int. No intersect.

Given x2 + y2 + 6x – 2y + 6 = 0 is the eq. of the circle. Find center, find radi.

(x-a)² + (y-b)² = r² is circle eq.

(x² + 6x + 9 ) + (y²– 2y +1) + 6 – 9 – 1 = 0 Þ (x + 3)² + (y - 1)² = 2²

Center : C(-3,1) radius : 2

Definition: A function is a relation such that there exists only 1 “y” for each “x”

Sketch y = x ¹ -2

x > -2 x < -2

y = = 1 y = = -1

Sketch y = | x |

y = x veya y = -x y =

Sketch y = 2| x | + 3

Sketch y = | x - 2|

Sketch y = | x + 1| - 3

Describe the graph of the given equation x2 + y2 + x = 0

( x² + x + ) + y² - = 0 Þ ( x - )² + (y – 0)² =

x = - ; y = 0 Þ Center: C(- , 0 ) radius :

Find an equation of the perpendicular bisector of the line segment joining points. (2,1) and (1,2)

Mid points

Slope of original line : m = = -1

Slope of perpendicular line : = + 1

Point :

Equation of line : y - = 1 ( x - ) Þ y = x - + Þ y = x

Length of segment : d = =

Sketch the graph of y = | x + 1 | + | x – 1 |

Case I x + 1 0 x – 1 0

x -1 x 1

x 1 Þ y = x + 1 + x – 1 Þ y = 2x

Case II x + 1 0 x – 1 0

x -1 x 1

x 1 Þ y = x + 1 + x – 1 Þ y = 2

More About Functions

Domain Polynomial : f(x) = a0 + a1x + a2x2 + ..........+ anxn Domain : R Ratianal Function : R(x) = Domain : Q(x) ¹ 0 Even –root func. : f(x) = Domain: g(x) ³ 0 Odd –root func. : f(x) = Domain: R

Ex: Find the domain of f(x) =

7x + 4 ³ 0 Þ 7x ³ -4 Þ x ³

Ex: Find the domain of f(x) =

x ¹2 Ù ³ 0

D : { x | 1 £ x <>

Ex: Find the domain of f(x) =

Ex: Find the domain of f(x) =

D : x – 1 ³ 0 Þ x ³ 1 D : x > 1

COMPOSITION OF FUNCTIONS

h(x) = , g(x) = are given

goh = hog = ^hoh =

Ex: Consider g(x) = Try to write it is a composition of two functions

1. f(x) = ; h(x) = | x | Þ g(x) = f o h =

2. f(x) = ; h(x) = 1 + | x | Þ g(x) = f o h =

3. f(x) = ; h(x) = x² Þ g(x) = f o h = =

DOMAIN OF COMPOSITION OF FUNCTIONS

If f(x) and g(x) are two function then (fog)(x) = f (g(x))

Ex: f(x) = g(x) = - x2 compute domain of fog and gof

f(x) = g(x) = - x² Þ fog = D :

Þ gof = - []² = - x D : x ³ 0

It is because of the domain of

Ex: Find the domain and range of y = 2 +

D : x ³ 1

R : [ 2 , ¥ )

y = y = y = 2 +

EXPONANTIAL FUNCTION

Definition : The functions of the form f(x) = a^x a ÎR⁺ - {1} are called exponential functions.

Ex: f(x) = 3^x Ex: f(x) =

Domain : R Domain : R

Range : (0 , ¥) Range : (0 , ¥)

Some properties

i-) a^x = a^k Û x = k

ii-) (a^x)^t Û a^x.t

iii-) a^x . a^k Û a^x+k

* f(x) = a^x is an exponential function a ÎR⁺ - {1}

* f(x) = e^x is an exponential function which is called “naturel exponential function.”

Function – Inverse Function

In order f to have an inverse function f should be one-to-one and onto

Now let’s consider the function :

f(x) = a^x f ^-1(x) = log_ax

Domain : R Domain : (0,¥ )

Range : (0,¥ ) Range : R

Some Properties of Logarithmic Functions

Logn (A.B) Û Logn A + Logn B

Logn () Û Logn A - Logn B

Log Ap Û Loga A

Loga A Û

Loga A Û “Taban Değiştirme”

Loga b . Logb a = 1

Log b = - Loga b

Loga b . Logb c . Logc d . . . . . . Logy z = Loga z

Ex: Using the logarithmic properties rewrite the expressions below in terms of lna, lnb , lnc a) ln a2 b) ln c) ln d) ln

a) ln a² = ln a² + ln = 2 . ln a + ln b + ln c

b) ln = ln b – ln (a³.c) = ln b – 3.ln a – 3.lnc

c) ln = ln - ln (a.b) = ln c – ln a – ln b

d) ln = ln = [ ln a + 3.lnb -2.lnc ] = lna + ln b – ln c

Ex: Rewrite the expressions below as a single logarithm a) 4 log 2 – log 3 + log 16 b) log x – 3.log 2x + 3.log 2 c) 2.ln (x+1) + ln x – 2.ln

a) 4 log 2 – log 3 + log 16 = log 16 – log 3 + log 16 = log + log 16 = log

b) log x – 3.log 2x + 3.log 2 = log - log 8x³ + log 8 = log x^-3/2

c) 2.ln (x+1) + ln x – 2.ln = ln (x+1)² + ln x^1/3- ln x = ln

Domain of logarithm Function

If f(x) = log_{a X} a ÎR⁺ - {1} Domain : (0 , ¥) Range : R

If f(x) = log_g(x) h(x) Domain of f(x) = { x| h(x) > 0 } Ç { x| g(x) > 0 , g(x) ¹ 1 }

Ex: Find the domain of the following function f(x) = log3

There are two conditions.

1.) > 0 2.) a = 3 > 0 , a ¹ 1

> 0 x + 1 = 0 ; x – 1 = 0

x = -1 ; x = 1

Domain of f(x) : (-¥,-1) È (1, ¥ )

Some Notes About Functions

Definition : A function is a set of ordered pairs (x,y) in which no two distinct pairs have the same first number .

This is not a function.

Definition : The set of all “x” is called domain of a function, set of all “y” is called Range of a function.

Ex: Consider f(x) = , sketch the graph, find the domain and range

y

Domain : R⁺È {0} = [ 0,¥) Range : R⁺È {0} = [ 0,¥)

x

Ex: Consider f(x) = , sketch the graph, find the domain and range

Solution : x² – 9 ³ 0 Þ (x-3) (x+3) ³ 0 x = 3 , x = -3

Domain : R⁺È {0} =(-¥,-3]È[3, ¥)

Range : R⁺È {0} = [ 0,¥)

Ex: Find the domain and range, and draw the graph of; f(x) =

Soln: Domain : R Range : { -3, 1, 4 }

.

For example : f(-5) = -3 ; f(-1000000000000) = -3

; f(1000) = 4

Ex: Find the domain and range of f(x) = | x | then draw it’s graph

Soln. f(x) = | x | Domain : R

Range : R⁺È {0} = [ 0,¥) “All nonnegative real numbers”

Ex: f(x) = , find domain , range and draw it’s graph

Soln.

f(x) = Domain : R - { 2 } Range : R – { 1 }

f ^-1(x) = Domain : R – { 1 } Range : R – { 2 }

Range of f(x) = Domain of f ^-1(x) = R – { 1 }

Ex: f(x) = Find the domain, range and draw graph

Solution:

Domain : R Range : R – { 6 }

y = x + 3

Ex: Given f(x) = x2 + 3x – 4 Find a) f(x+h) b) f(x) + f(h)

Soln:

a) f(x+h) = (x+h)² + 3(x+h) – 4

b) f(x) + f(h) = x² + 3x – 4 + h² + 3h – 4

Ex: Given f(x) = , g(x) = find the domain of; a) (f+g)(x) b) (f-g)(x) c) (f.g)(x) d) ()(x)

a) (f+g)(x) = f(x) + g(x) = + x + 1 ³ 0 , x ³ -1 ; -x + 4 ³ 0 , x £ 4 Domain of f(x) : [-1,¥) ; Domain of g(x) : (-¥,4] then, Doman of (f+g)(x) : [-1,¥) Ç (-¥,4] = [-1 , 4 ] b) (f-g)(x) = f(x) -g(x) = - x + 1 ³ 0 , x ³ -1 ; -x + 4 ³ 0 , x £ 4 Domain of (f-g)(x) : [-1 , 4 ]

c) (f.g)(x) = f(x) . g(x) = . ; Domain of (f . g)(x) : [-1 , 4 ] d) ()(x) = = ; Domain of ()(x) : [-1 , 4 )

Definition: Given two function f and g then the composite function defined by

(fog)(x) = f (g(x) )

And domain of (fog)(x) is the set of all numbers x in the domain of such that g(x) is in the domain of f.

Ex: f(x) = 2x + 7 g(x) = x2 – 8 a) (fog)(x) b) (fof)(x)

a) (fog)(x) = f( g(x) ) = 2 (x² – 8 ) + 7 = 2x² – 9

b) (fof)(x) = f( f(x) ) = 2 ( 2x + 7 ) + 7 = 4x + 21

Ex: Find the domain of (fog)(x) if f(x) = , g(x) 2x - 3

Soln: Domain of f(x) : [0,¥) ; Domain of g(x) :R domain of (fog)(x) : [,¥)

Chapter 2 : LIMITS and CONTINUITY

Before giving the definition of a limit, let’s give an example.

Ex: Consider a function f(x) = Domain of f(x) = R \ { 1 } ,

For x ¹ 1 we can write f(x) = = = x + 1 so for x ¹ 1 , f(x) = x + 1

Let’s draw the graph of f(x) ;

f(x) is not defined at x = 1, but when “x” approaches to x = 1 , the value of f(x) approaches to 2. In the limit the value of f(x) at x = 1 is not important. The important thing is where the value of f(x) approaches as x approaches x = 1.

At x = 1 f(x) is not defined by f(x) = 2

Informal Definition of limit:

Let f(x) be defined on an open interval about x = x₀ except possible does not contain x = x₀ ,

İf f(x) gets arbitrarly close to L for all x sufficiently close to x₀ we say approaches the limit L as x approaches to x₀ and we write Formal definition of limit;

f(x) = L means, >0 such that 0 < | x - x₀ | < Þ | f(x) – L | <

Ex: Consider the following functions; f(x) = g(x) = x + 1 h(x) =

f(x) = 2 g(x) = 2 h(x) = 2

LIMIT RULES

If L, M, a and K are real numbers and,

f(x) = L and g(x) = M

1) ( f(x) + g(x) ) = f(x) + g(x) = L + M

2) ( f(x) - g(x) ) = f(x) - g(x) = L - M

3) ( f(x) . g(x) ) = f(x) . g(x) = L . M

4) () = f(x) : g(x) = L : M

5) ( k. f(x) ) = k f(x) = k . L

Theorem: (Limits of Polynomials)

If f(x) = a_n xⁿ + a_n-1 x^n-1 + a_n-2 x^n-2 + . + . + . . . . . . . . .+ a_o

f(x) = f(C) = a_n Cⁿ + a_n-1 C^n-1 + a_n-2 C^n-2 + . + . + . . . . . . . . .+ a_o

Theorem: If f(x) and g(x) are polynomials then;

If g( C ) ¹ 0

Ex: Find the limits of the following functions. a) (x3 + 4x2 – 3 ) b) c)

a) (x³ + 4x² – 3 ) = 1³ + 4 . 1² – 3 = 2

b) = =

c) = =

Ex: Find the limits of the following functions. a) b) c)

a) = then = = 3

b) = then = = =

c) = then = =

=

Sandwich Theorem

Let f(x) £ g(x) £ h(x) for all x in open interval containing x = c , except possibly at x = c

f(x) = L , h(x) = L then g(x) = L

Ex: Given that 1 - £ U(x) £ 1 + for all x ( x ¹ 0 ) Find U(x)

Solution: 1 - £ U(x) £ 1 + ( x ¹ 0 ) Take the limit of both sides

(1 - ) £ U(x) £ (1 + )

1 £ U(x) £ 1

By the SANDWICH ttheorem: U(x) = 1