2) The cognitive complexity and a structure was studied by Scott using a technique in which a person was asked to specify a number of objects and group them into as many groupings as he or she found meaningful. If a person groups 12 objects into three groups in such a way that: 7 objects are in group A 7 group B 8 group C 3 group A&B 5 group B&C 4 group A&C 2 all three groups
Click this for the Venn Diagram: https://plus.google.com/photos/109661488938063739714/albums/5834016694134940929
Questions and answers: 1) P(A) = 2/12 = 1/6 2) P (B) = 1/12 3) P (C) = 1/12 4) P (A&C but B) = 3/12 = 1/4 5) P (N) = 12/12 = 0 6) P (B&C but not A) = 3/12 = 1/4 7) P (exactly 1) = 4/12 = 1/3 8) P (exactly 2) = 1/2 9) P (A or B) = 3/12 = 1/4
The cognitive complexity of a structure was studied by Scott using a technique in which a person was asked to specify a number of objects and group them into as many groupings as he or she found meaningful. If a person groups 12 objects into tree groups in such a way that 7 objects are in group A 7 in group B 8 in group C 3 in group A&B 5 in group B&C 4 in group A&C 2 in all three groups
1/6 1. P (A) 1/12 2. P (B) 1/12 3. P (C) 1/6 4. P (A and C but not B) 0 5. P (N) 1/4 6. P (B and C but not A) 1/6 7. P (exactly 1) 1/12 8. P (exactly 2) 1/12 9. P (A or B)
Mika Lapus 1-16-2013 Math Homework: Venn Diagram Q: twelve objects in three groups arranged in such a way that: 7 in group A 7 group B 8 group C 3 group A and B 5 group B and C 4 group A and C 2 all three groups
sets of questions: 1. P(A) = 2/12= 1/6 2. P(B)= 1/12 3. P(C)= 1/12 4. P(A and C but not b)= 1/6 5. p(NONE)= 0 6. P (B and C but not A)= 1/4 7. P(exactly 1)= 1/3 8.P(exactly 2)= 1/2 9. P(A or B) = 1/4
The cognitive complexity of a structure was studied by Scott using a technique in which a person was asked to specify a number of objects and group them into as many groupings as he or she found meaningful. If a person groups 12 objects into tree groups in such a way that 7 objects are in group A 7 in group B 8 in group C 3 in group A and B 5 in group B and C 4 in group A and C 2 in all three groups
Questions:
1. P(A) = 2/12 = 1/6 2. P(B) = 1/12 3. P(C) = 1/12 4. P(A&C but not B) = 1/12 5. P(N) = 0 6. P(B&C but not A) = 1/4 7. P(exactly 1) =1/3 8. P(exactly 2) = 1/2 9. P(A or B) = 3/12 = 1/4
1. P (A) = 1/6 2. P (B) = 1/12 3. P (C) = 1/12 4. P (A & C but not B) = 1/6 5. P (N) = 0 6. P (B & C but not A) = 1/4 7. P (exactly 1) = 1/3 8. P (exactly 2)= 1/2 9. P (A or B) = 1/4
1.P(a) 2/12 = 1/6 2.P (b) = 1/12 3. P (c) = 1/2 4.P (a & c not b) = 2/12 + 1/6 5. P (n) = 0/12 = 0 6. P (b & c not a) = 3/12 = 1/4 7. P (exactly 1) = 4/12 = 1/3 8. p (exactly 2) = 6/12 = 1/2 9. P (a or b) 3/12 = 1/4
1. P(A) = 2/12= 1/6 2. P(B)= 1/12 3. P(C)= 1/12 4. P(A and C but not b)= 1/6 5. p(NONE)= 0 6. P (B and C but not A)= 1/4 7. P(exactly 1)= 1/3 8.P(exactly 2)= 1/2 9. P(A or B) = 1/4 Nathanfuster
1. P(A) = 2/12= 1/6 2. P(B)= 1/12 3. P(C)= 1/12 4. P(A and C but not b)= 1/6 5. p(NONE)= 0 6. P (B and C but not A)= 1/4 7. P(exactly 1)= 1/3 8.P(exactly 2)= 1/2 9. P(A or B) = 1/4 Nathanfuster
2) The cognitive complexity and a structure was studied by Scott using a technique in which a person was asked to specify a number of objects and group them into as many groupings as he or she found meaningful. If a person groups 12 objects into three groups in such a way that: 7 objects are in group A 7 group B 8 group C 3 group A&B 5 group B&C 4 group A&C 2 all three groups
Homework in Math: Research about Limits and Continuity 01/17/13
Limit of a Function
The concept of the limit of a function is the starting point of calculus. Every notion in calculus can be expressed in some forms of limits.
Limit: f(x)=A if and only if, for any chosen number ε>0, however small, there exists a number 3>0 such that, whenever 0< |x-a|<3, then |f(x)-A| < 3.
Note that, the limit of a function gives an idea about how the function behaves close to a particular x value. It does not necessarily give the value of the function at x. That means, f(x) and f(c) may or may not be equal.
Continuity
LIMITS AT INFINITY
In some cases we may need to observe the behavior of the function when the independent variable x increases or decreases without bound, that is, x+1 or x-1.
The limit of a function is concerned with the behaviour of a function near a given point.To be more precise, if “the limit of the function f(x) as x approaches the point a” is the value L then this is denoted as limx→af(x)=L.
Limit and Continuity • Definition: Limit at a real number c (Both side limit).
Given > 0, there exists > 0 such that | f ( x ) - L | < whenever 0 < | x - c | < , then the limit of f ( x ) at x = c is L, and denoted by , otherwise f ( x ) has no limit.
Roughly speaking, means that whenever x approaches to c
from either side of c, the graph point (x, f(x)) approaches to the point (c, L)
on the plane.
• Definition: Continuity for f ( x ) at x = c ( both side continuity ).
Given > 0, there exists > 0, such that | f ( x ) - f ( c ) | < whenever | x - c| < , then f ( x ) is continuous at x = c. ( i.e ), otherwise f ( x ) is not continuous at x = c. ( i.e f ( x ) is discontinuous at x = c ).
Roughly speaking, means that whenever x approaches
to c from either side of c, the graph point (x, f(x)) approaches to the
We write lim x→a f(x) = L or f(x) → L as x → a to mean that f(x) approaches the number L as x approaches (but is not equal to) a from both sides. A more precise way of phrasing the definition is that we can make f(x) be as close to L as we like by making x be sufficiently close to a.
We write
lim x→a+ f(x) = L or f(x) → L as x → a+ and
lim x→a- f(x) = L or f(x) → L as x → a- to mean that f(x) → L as x approaches a from the right or left, respectively. For limx → af(x) to exist, the left and right limits must both exist and must be equal. We write
lim x→+∞ f(x) = L and
lim x→-∞ f(x) = L to mean that f(x) → L as x gets arbitrarily large or becomes a negative number with arbitrarily large magnitude, respectively. Examples 1. As x → 3, the quantity 3x2-4x+2 approaches 17, and hence
lim x→3 (3x2-4x+2) = 27 Notice this is just the value of the function at x = 3 (see "Algebraic Approach" below). 2. Notice that the function
x2 - 9
x - 3 is not defined at x = 3. However, for other values of x, it simplifies to x2 - 9
x - 3 = (x - 3)(x + 3)
x - 3 = x + 3, and, as x → 3, this quantity approaches 6. Therefore x2 - 9
x - 3 → 6 as x → 3,
or
lim x→3 x2 - 9
x - 3 = 6
*Continuous Functions A function f is continuous at a if limx → a f(x) exists, and is equal to f(a).
The function f is said to be continuous on its domain if it is continuous at each point in its domain. The algebraic approach to limits above is based on the fact that any closed form function is continuous on its domain.
Examples The function f(x) = 3x2-4x+2 is a closed form function, and hence continuous at every point in its domain (all real numbers).
The function
g(x) = 4x2+1
x - 3 is also a closed form function, and hence continuous on its domain (all real numbers excluding 3). On the other hand, the function
h(x) = -1 if -4 ≤ x < -1 x if -1 ≤ x ≤ 1 x2-1 if 1 < x ≤ 2 is not a closed-form function, and is in fact discontinuous at x = 1.
The constant rule: This is simple. f (x) = 5 is a horizontal line with a slope of zero, and thus its derivative is also zero.
The power rule: To repeat, bring the power in front, then reduce the power by 1. That’s all there is to it. The power rule works for any power: a positive, a negative, or a fraction.
The constant multiple rule: What if the function you’re differentiating begins with a coefficient? Makes no difference. A coefficient has no effect on the process of differentiation. You just ignore it and differentiate according to the appropriate rule. The coefficient stays where it is until the final step when you simplify your answer by multiplying by the coefficient.
Don’t forget that ð (~3.14) and e (~2.72) are numbers, not variables, so they behave like ordinary numbers. Constants in problems, like c and k, also behave like ordinary numbers.So, for example, the derivative of π^3 is zero, not 3π^2, the derivative of √ k is also zero, not 1/2k^-1/2.
The sum rule: When you want the derivative of a sum of terms, take the derivative of each term separately.
What's f'(x) = x^6 + x^3 + x^2 + x + 10?
Solution : just use the power rule for each of the first four terms and the constant rule for the final term
thus, f'(x) = 6x^5 + 3x^2 + 2x + 1
The difference rule: If you have a difference (that’s subtraction) instead of a sum, it makes no difference. You still differentiate each term separately.
thus, if y = 3x^5 - x^4 - 2x^3 +6X^2 + 5x, then y' = 15x^4 - 4x^3 - 6x^2 + 12x + 5
The addition and subtraction signs are unaffected by the differentiation.
In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's instantaneous velocity.
The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. Informally, the derivative is the ratio of the infinitesimal change of the output over the infinitesimal change of the input producing that change of output. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point. In higher dimensions, the derivative of a function at a point is a linear transformation called the linearization.[1] A closely related notion is the differential of a function.
The process of finding a derivative is called differentiation. The reverse process is called antidifferentiation. The fundamental theorem of calculus states that antidifferentiation is the same as integration. Differentiation and integration constitute the two fundamental operations in single-variable calculus.
Rules of Differentiation of Functions in Calculus The basic rules of Differentiation of functions in calculus are presented along with several examples:
1 - Derivative of a constant function. The derivative of f(x) = c where c is a constant is given by f '(x) = 0
Example f(x) = - 10 , then f '(x) = 0
2 - Derivative of a power function (power rule). The derivative of f(x) = x r where r is a constant real number is given by f '(x) = r x r - 1
Example f(x) = x -2 , then f '(x) = -2 x -3 = -2 / x 3
3 - Derivative of a function multiplied by a constant. The derivative of f(x) = c g(x) is given by f '(x) = c g '(x)
Example f(x) = 3x 3 , let c = 3 and g(x) = x 3, then f '(x) = c g '(x) = 3 (3x 2) = 9 x 2
4 - Derivative of the sum of functions (sum rule). The derivative of f(x) = g(x) + h(x) is given by f '(x) = g '(x) + h '(x)
Example f(x) = x 2 + 4 let g(x) = x 2 and h(x) = 4, then f '(x) = g '(x) + h '(x) = 2x + 0 = 2x
5 - Derivative of the difference of functions. The derivative of f(x) = g(x) - h(x) is given by f '(x) = g '(x) - h '(x)
Example f(x) = x 3 - x -2 let g(x) = x 3 and h(x) = x -2, then f '(x) = g '(x) - h '(x) = 3 x 2 - (-2 x -3) = 3 x 2 + 2x -3
6 - Derivative of the product of two functions (product rule). The derivative of f(x) = g(x) h(x) is given by f '(x) = g(x) h '(x) + h(x) g '(x)
Example f(x) = (x 2 - 2x) (x - 2) let g(x) = (x 2 - 2x) and h(x) = (x - 2), then f '(x) = g(x) h '(x) + h(x) g '(x) = (x 2 - 2x) (1) + (x - 2) (2x - 2) = x 2 - 2x + 2 x 2 - 6x + 4 = 3 x 2 - 8x + 4
7 - Derivative of the quotient of two functions (quotient rule). The derivative of f(x) = g(x) / h(x) is given by f '(x) = ( h(x) g '(x) - g(x) h '(x) ) / h(x) 2
Example f(x) = (x - 2) / (x + 1) let g(x) = (x - 2) and h(x) = (x + 1), then f '(x) = ( h(x) g '(x) - g(x) h '(x) ) / h(x) 2 = ( (x + 1)(1) - (x - 2)(1) ) / (x + 1) 2 = 3 / (x + 1) 2
in partucular the many definitions of continuity employ the limit if the condition o/begin matrix a given is to be more precise if the limit of the function f(x) and the f(c) this is as f(x) = l
Definition: Limit at a real number c (Both side limit).
Given > 0, there exists > 0 such that | f ( x ) - L | < whenever 0 < | x - c | < , then the limit of f ( x ) at x = c is L, and denoted by , otherwise f ( x ) has no limit.
Roughly speaking, means that whenever x approaches to c
from either side of c, the graph point (x, f(x)) approaches to the point (c, L)
on the plane.
Definition: Continuity for f ( x ) at x = c ( both side continuity ).
Given > 0, there exists > 0, such that | f ( x ) - f ( c ) | < whenever | x - c| < , then f ( x ) is continuous at x = c. ( i.e ), otherwise f ( x ) is not continuous at x = c. ( i.e f ( x ) is discontinuous at x = c ).
Roughly speaking, means that whenever x approaches
to c from either side of c, the graph point (x, f(x)) approaches to the
[IMG]http://i1352.photobucket.com/albums/q660/carlosmat619/3-circle-venn-diagram-blank_zps90d2d4b0.jpg[/IMG]
ReplyDeletehttp://i1352.photobucket.com/albums/q660/carlosmat619/3-circle-venn-diagram-blank_zps90d2d4b0.jpg
1. P(A)= 1/5
2. P(B)= 1/10
3. P(C)= 1/10
4. P(A and C but B)= 1/5
5. P(N)= 0
6. P(B and C but A)= 3/10
7. P(exactly one)= 1/5
8. P(exactly 2)= 1/10
9. P(A or B)= 3/10
1.P(A) 1/6
ReplyDelete2.P(B) 1/12
3.P(C) 1/12
4.P(A&C but not B) 5/12
5.P(N) 12/12=1
6.P(B&C but not A) 1/3
7.P(Exactly 1) 1/3
8.P(Exactly 2) 2/3
9.P(A or B) 1/4
A=2
B=1
C=1
A&B=1
A&C=2
B&C=3
B&C&A=2
Homework.
ReplyDelete2) The cognitive complexity and a structure was studied by Scott using a technique in which a person was asked to specify a number of objects and group them into as many groupings as he or she found meaningful. If a person groups 12 objects into three groups in such a way that:
7 objects are in group A
7 group B
8 group C
3 group A&B
5 group B&C
4 group A&C
2 all three groups
Click this for the Venn Diagram:
https://plus.google.com/photos/109661488938063739714/albums/5834016694134940929
Questions and answers:
1) P(A) = 2/12 = 1/6
2) P (B) = 1/12
3) P (C) = 1/12
4) P (A&C but B) = 3/12 = 1/4
5) P (N) = 12/12 = 0
6) P (B&C but not A) = 3/12 = 1/4
7) P (exactly 1) = 4/12 = 1/3
8) P (exactly 2) = 1/2
9) P (A or B) = 3/12 = 1/4
1.) P(A) 2/12 = 1/6
ReplyDelete2.) P(B) 1/12
3.) P(C) 1/12
4.) P(A&C but not B) 2/12 =1/6
5.) P(N) 0/12
6.) P(B&C but not A) 3/12
7.) P(Exactly 1) 1+1+2 / 12 4/12 = 1/3
8.) P( Exactly 2) 3+2+1 / 12 = 6/12 = 1/2
9.) P(A or B) 2+1 / 12 = 3/12
Math Homework for January 16, 2013
ReplyDeleteThe cognitive complexity of a structure was studied by Scott using a technique in which a person was asked to specify a number of objects and group them into as many groupings as he or she found meaningful. If a person groups 12 objects into tree groups in such a way that
7 objects are in group A
7 in group B
8 in group C
3 in group A&B
5 in group B&C
4 in group A&C
2 in all three groups
Questions:
1) P(A) = 2/12 = 1/6
2) P(B) = 1/12
3) P(C) = 1/12
4) P(A&C but not B) = 2/12 = 1/6
5) P(N) = 0
6) P(B&C but not A) = 3/12 = 1/4
7) P(exactly 1) = 4/12 = 1/3
8) P(exactly 2) = 6/12 = 1/2
9) P(A or B) = 3/12 = 1/4
HW
ReplyDelete1. P(A)
2/12
= 1/6
2. P(B)
= 1/12
3. P(C)
= 1/12
4. P(A&C but B)
2/12
=1/6
5. P(N)
= 0
6. P(B&C but not A)
= 3/12
7. P(exactly 1)
1+1+2/12
4/12
= 1/3
8.P(exactly 2)
3+2+1/12
6/12
=1/2
9. P(A or B)
2+1/12
=3/12
1)a=2
ReplyDelete2)b=1
3)c=1
4)a&c=2/3
5)n=0
6)b&c=1
7)exactly 1=3/7
8)exactly 2=3/7
9)a or 5 = 1/4
Venn Diagram: http://www.flickr.com/photos/92267447@N03/
1/6 1. P (A)
ReplyDelete1/12 2. P (B)
1/12 3. P (C)
1/6 4. P (A and C but not B)
0 5. P (N)
1/4 6. P (B and C but not A)
1/6 7. P (exactly 1)
1/12 8. P (exactly 2)
1/12 9. P (A or B)
Mika Lapus
ReplyDelete1-16-2013 Math Homework: Venn Diagram
Q: twelve objects in three groups arranged in such a way that:
7 in group A
7 group B
8 group C
3 group A and B
5 group B and C
4 group A and C
2 all three groups
Diagram: http://break-into-oblivion.tumblr.com/image/40683479069
sets of questions:
1. P(A) = 2/12= 1/6
2. P(B)= 1/12
3. P(C)= 1/12
4. P(A and C but not b)= 1/6
5. p(NONE)= 0
6. P (B and C but not A)= 1/4
7. P(exactly 1)= 1/3
8.P(exactly 2)= 1/2
9. P(A or B) = 1/4
1. P(A) 2/12 = 1/6
ReplyDelete2. P(B) 1/12
3. P(C) 1/12
4. P(A and C but not B) 2/12 =1/6
5. P(N) 0/12
6. P(B&C but not A) 3/12
7. P(1) 1+1+2 / 12 4/12 = 1/3
8. P(2) 3+2+1 / 12 = 6/12 = 1/2
9. P(A or B) 2+1 / 12 = 3/12
Homework
ReplyDelete1. P (A)
2. P (B)
3. P (C)
4. P (A and C but not B)
5. P (N)
6. P (B and C but not A)
7. P (exactly 1)
8. P (exactly 2)
9. P (A or B)
ANSWERS:
1.) 1/6
2.) 1/12
3.) 1/6
4.) 1/6
5.) 0
6.) 1/4
7.) 1/6
8.) 1/12
9.) 1/12
hw in math.. venn diagram jan 17 2013 late hw..
ReplyDelete1. P(A)= 1/5
2. P(B)= 1/10
3. P(C)= 1/10
4. P(A and C but B)= 1/5
5. P(N)= 0
6. P(B and C but A)= 3/10
7. P(exactly one)= 1/5
8. P(exactly 2)= 1/10
9. P(A or B)= 3/10
Homework in Math:
ReplyDeleteThe cognitive complexity of a structure was studied by Scott using a technique in which a person was asked to specify a number of objects and group them into as many groupings as he or she found meaningful. If a person groups 12 objects into tree groups in such a way that
7 objects are in group A
7 in group B
8 in group C
3 in group A and B
5 in group B and C
4 in group A and C
2 in all three groups
Questions:
1. P(A) = 2/12 = 1/6
2. P(B) = 1/12
3. P(C) = 1/12
4. P(A&C but not B) = 1/12
5. P(N) = 0
6. P(B&C but not A) = 1/4
7. P(exactly 1) =1/3
8. P(exactly 2) = 1/2
9. P(A or B) = 3/12 = 1/4
1. P (A) = 1/6
ReplyDelete2. P (B) = 1/12
3. P (C) = 1/12
4. P (A & C but not B) = 1/6
5. P (N) = 0
6. P (B & C but not A) = 1/4
7. P (exactly 1) = 1/3
8. P (exactly 2)= 1/2
9. P (A or B) = 1/4
http://instagram.com/p/UkHVV7RdTh/
1.) P(A) 2/12 = 1/6
ReplyDelete2.) P(B) 1/12
3.) P(C) 1/12
4.)a&c=2/3
5.)n=0
6.) P (B&C but not A) = 3/12 = 1/4
7.) P (exactly 1) = 4/12 = 1/3
8.) P( Exactly 2) 3+2+1 / 12 = 6/12 = 1/2
9.) P(A or B) 2+1 / 12 = 3/12
1. P(A) 2/12 = 1/6
ReplyDelete2. P(B) = 1/12
3. p(C) = 1/12
4. P(A and C but B) 2/12 = 1/6
5. P(N) = 0/12 = 0
6. P(B and C but A) = 3/12 = 1/4
7. P(exactly 1) = 4/12 = 1/3
8. P(exactly 2) = 6/12 = 1/2
9. P(A or B) = 3/12 = 1/4
Venn Diagram here: https://pbs.twimg.com/media/BAxTH3mCUAAU4jQ.jpg:large
1.P(a) 2/12 = 1/6
ReplyDelete2.P (b) = 1/12
3. P (c) = 1/2
4.P (a & c not b) = 2/12 + 1/6
5. P (n) = 0/12 = 0
6. P (b & c not a) = 3/12 = 1/4
7. P (exactly 1) = 4/12 = 1/3
8. p (exactly 2) = 6/12 = 1/2
9. P (a or b) 3/12 = 1/4
Image: https://twitter.com/yasmeencoo/status/291699051706073088/photo/1
1/6
ReplyDelete1/12
1/12
1/6
0
1/12
1/3
1/2
1/4
Karl Claricia
ReplyDelete10-Pascal
http://instagr.am/p/UkIFDLRdUc/
Nile Galingan(Using Karl's account)
10-Pascal
http://instagr.am/p/UkH0nyRdUM/
1)1/6 1. P (A)
ReplyDelete2)1/12 2. P (B)
3)1/12 3. P (C)
4)1/6 4. P (A and C but not B)
5)0 5. P (N)
6)1/4 6. P (B and C but not A)
7)1/6 7. P (exactly 1)
8)1/12 8. P (exactly 2)
9)1/12 9. P (A or B)
Wackie De Guzman
ReplyDelete10-Pascal
http://instagr.am/p/UkI3-RRdVj/
http://instagram.com/p/UkJKs4xdV8/
ReplyDeletediego domagas
http://instagr.am/p/UkHVV7RdTh/
ReplyDelete1. P(A) = 2/12= 1/6
2. P(B)= 1/12
3. P(C)= 1/12
4. P(A and C but not b)= 1/6
5. p(NONE)= 0
6. P (B and C but not A)= 1/4
7. P(exactly 1)= 1/3
8.P(exactly 2)= 1/2
9. P(A or B) = 1/4
Nathanfuster
http://instagr.am/p/UkHVV7RdTh/
ReplyDelete1. P(A) = 2/12= 1/6
2. P(B)= 1/12
3. P(C)= 1/12
4. P(A and C but not b)= 1/6
5. p(NONE)= 0
6. P (B and C but not A)= 1/4
7. P(exactly 1)= 1/3
8.P(exactly 2)= 1/2
9. P(A or B) = 1/4
Nathanfuster
2) The cognitive complexity and a structure was studied by Scott using a technique in which a person was asked to specify a number of objects and group them into as many groupings as he or she found meaningful. If a person groups 12 objects into three groups in such a way that:
ReplyDelete7 objects are in group A
7 group B
8 group C
3 group A&B
5 group B&C
4 group A&C
2 all three groups
Answers:
1)A = 2
2)B = 1
3)C = 1
4)A&C = 2/12 = 1/6
5)none = 0
6)B&C = 3/12 = 1/4
7)exactly 1 = 2/12 = 1/6
8)exactly 2 = 6/12 = 1/2
9)A or B = ?
Homework in Math: Research about Limits and Continuity 01/17/13
ReplyDeleteLimit of a Function
The concept of the limit of a function is the starting point of calculus. Every notion in calculus can be expressed in some forms of limits.
Limit: f(x)=A if and only if, for any chosen number ε>0, however small, there exists a number 3>0 such that, whenever 0< |x-a|<3, then |f(x)-A| < 3.
Note that, the limit of a function gives an idea about how the function behaves close to a particular x value. It does not necessarily give the value of the function at x. That means, f(x) and f(c) may or may not be equal.
Continuity
LIMITS AT INFINITY
In some cases we may need to observe the behavior of the function when the independent variable x increases or decreases without bound, that is, x+1 or x-1.
HW. Research about limits and continuity.
ReplyDeletehttp://www.youtube.com/watch?v=HB8CzZEd4xw
http://www.goiit.com/posts/show/0/content-limits-and-continuity-804300.htm
Limits and Continuity :D
ReplyDeleteThe limit of a function is concerned with the behaviour of a function near a given point.To be more precise, if “the limit of the function f(x) as x approaches the point a” is the value L then this is denoted as limx→af(x)=L.
Limit and Continuity
ReplyDelete• Definition: Limit at a real number c (Both side limit).
Given > 0, there exists > 0 such that | f ( x ) - L | < whenever 0 < | x - c | < ,
then the limit of f ( x ) at x = c is L, and denoted by ,
otherwise f ( x ) has no limit.
Roughly speaking, means that whenever x approaches to c
from either side of c, the graph point (x, f(x)) approaches to the point (c, L)
on the plane.
• Definition: Continuity for f ( x ) at x = c ( both side continuity ).
Given > 0, there exists > 0, such that | f ( x ) - f ( c ) | < whenever | x - c| < ,
then f ( x ) is continuous at x = c. ( i.e ),
otherwise f ( x ) is not continuous at x = c. ( i.e f ( x ) is discontinuous at x = c ).
Roughly speaking, means that whenever x approaches
to c from either side of c, the graph point (x, f(x)) approaches to the
point (c, f(c)) on the plane.
My research about limits and continuity.
ReplyDelete*Limits
We write
lim
x→a
f(x) = L
or
f(x) → L as x → a
to mean that f(x) approaches the number L as x approaches (but is not equal to) a from both sides.
A more precise way of phrasing the definition is that we can make f(x) be as close to L as we like by making x be sufficiently close to a.
We write
lim
x→a+
f(x) = L
or
f(x) → L as x → a+
and
lim
x→a-
f(x) = L
or
f(x) → L as x → a-
to mean that f(x) → L as x approaches a from the right or left, respectively. For limx → af(x) to exist, the left and right limits must both exist and must be equal.
We write
lim
x→+∞
f(x) = L
and
lim
x→-∞
f(x) = L
to mean that f(x) → L as x gets arbitrarily large or becomes a negative number with arbitrarily large magnitude, respectively. Examples
1. As x → 3, the quantity 3x2-4x+2 approaches 17, and hence
lim
x→3
(3x2-4x+2) = 27
Notice this is just the value of the function at x = 3 (see "Algebraic Approach" below).
2. Notice that the function
x2 - 9
x - 3
is not defined at x = 3. However, for other values of x, it simplifies to
x2 - 9
x - 3
=
(x - 3)(x + 3)
x - 3
= x + 3,
and, as x → 3, this quantity approaches 6. Therefore
x2 - 9
x - 3
→ 6 as x → 3,
or
lim
x→3
x2 - 9
x - 3
= 6
*Continuous Functions
A function f is continuous at a if limx → a f(x) exists, and is equal to f(a).
The function f is said to be continuous on its domain if it is continuous at each point in its domain. The algebraic approach to limits above is based on the fact that any closed form function is continuous on its domain.
Examples
The function f(x) = 3x2-4x+2 is a closed form function, and hence continuous at every point in its domain (all real numbers).
The function
g(x) =
4x2+1
x - 3
is also a closed form function, and hence continuous on its domain (all real numbers excluding 3).
On the other hand, the function
h(x) =
-1 if -4 ≤ x < -1
x if -1 ≤ x ≤ 1
x2-1 if 1 < x ≤ 2
is not a closed-form function, and is in fact discontinuous at x = 1.
Homework today, January 18, 2013.
ReplyDeleteResearch about Rules of Differentiation.
1 - Derivative of a constant function.
The derivative of f(x) = c where c is a constant is given by
f '(x) = 0
Example
f(x) = - 10 , then f '(x) = 0
2 - Derivative of a power function (power rule).
The derivative of f(x) = x r where r is a constant real number is given by
f '(x) = r x r - 1
Example
f(x) = x -2 , then f '(x) = -2 x -3 = -2 / x 3
3 - Derivative of a function multiplied by a constant.
The derivative of f(x) = c g(x) is given by
f '(x) = c g '(x)
Example
f(x) = 3x 3 , let c = 3 and g(x) = x 3, then f '(x) = c g '(x)
= 3 (3x 2) = 9 x 2
4 - Derivative of the sum of functions (sum rule).
The derivative of f(x) = g(x) + h(x) is given by
f '(x) = g '(x) + h '(x)
Example
f(x) = x 2 + 4 let g(x) = x 2 and h(x) = 4, then f '(x) = g '(x) + h '(x) = 2x + 0 = 2x
5 - Derivative of the difference of functions.
The derivative of f(x) = g(x) - h(x) is given by
f '(x) = g '(x) - h '(x)
Example
f(x) = x 3 - x -2 let g(x) = x 3 and h(x) = x -2, then
f '(x) = g '(x) - h '(x) = 3 x 2 - (-2 x -3) = 3 x 2 + 2x -3
6 - Derivative of the product of two functions (product rule).
The derivative of f(x) = g(x) h(x) is given by
f '(x) = g(x) h '(x) + h(x) g '(x)
Example
f(x) = (x 2 - 2x) (x - 2) let g(x) = (x 2 - 2x) and h(x) = (x - 2), then
f '(x) = g(x) h '(x) + h(x) g '(x) = (x 2 - 2x) (1) + (x - 2) (2x - 2)
= x 2 - 2x + 2 x 2 - 6x + 4 = 3 x 2 - 8x + 4
7 - Derivative of the quotient of two functions (quotient rule).
The derivative of f(x) = g(x) / h(x) is given by
f '(x) = ( h(x) g '(x) - g(x) h '(x) ) / h(x) 2
Example f(x) = (x - 2) / (x + 1) let g(x) = (x - 2) and h(x) = (x + 1), then
f '(x) = ( h(x) g '(x) - g(x) h '(x) ) / h(x) 2
= ( (x + 1)(1) - (x - 2)(1) ) / (x + 1) 2
= 3 / (x + 1) 2
Homework: Differentials
ReplyDeleteBasic Differentiation Rules
The constant rule: This is simple. f (x) = 5 is a horizontal line with a slope of zero, and thus its derivative is also zero.
The power rule:
To repeat, bring the power in front, then reduce the power by 1. That’s all there is to it.
The power rule works for any power: a positive, a negative, or a fraction.
The constant multiple rule:
What if the function you’re differentiating begins with a coefficient? Makes no difference. A coefficient has no effect on the process of differentiation. You just ignore it and differentiate according to the appropriate rule. The coefficient stays where it is until the final step when you simplify your answer by multiplying by the coefficient.
Don’t forget that ð (~3.14) and e (~2.72) are numbers, not variables, so they behave like ordinary numbers. Constants in problems, like c and k, also behave like ordinary numbers.So, for example, the derivative of π^3 is zero, not 3π^2, the derivative of √ k is also zero, not 1/2k^-1/2.
The sum rule: When you want the derivative of a sum of terms, take the derivative of each term separately.
What's f'(x) = x^6 + x^3 + x^2 + x + 10?
Solution : just use the power rule for each of the first four terms and the constant rule for the final term
thus, f'(x) = 6x^5 + 3x^2 + 2x + 1
The difference rule: If you have a difference (that’s subtraction) instead of a sum, it makes no difference. You still differentiate each term separately.
thus, if y = 3x^5 - x^4 - 2x^3 +6X^2 + 5x, then y' = 15x^4 - 4x^3 - 6x^2 + 12x + 5
The addition and subtraction signs are unaffected by the differentiation.
Homework in Math: Research in Rules of Differentiation
ReplyDelete1. Derivative of a constant function
-The derivative of f(x) = c where c is a constant is given by f '(x) = 0
Example
f(x) = - 10 , then f '(x) = 0
2.Derivative of a power function (power rule)
-The derivative of f(x) = x^r where r is a constant real number is given by f'(x) = rx^r - 1
Example
f(x) = x^-2 , then f'(x) = -2x^-3 = -2/x^3
3. Derivative of a function multiplied by a constant
-The derivative of f(x) = cg(x) is given by f'(x) = cg'(x)
4. Derivative of the sum of functions (sum rule)
-The derivative of f(x) = g(x) + h(x) is given by f'(x) = g'(x)+ h'(x)
5. Derivative of the difference of functions
-The derivative of f(x) = g(x) - h(x) is given by f'(x) = g'(x)- h'(x)
6. Derivative of the product of two functions (product rule).
- The derivative of f(x) = g(x) h(x) is given by f'(x) = g(x)h'(x) + h(x)g'(x)
7. Derivative of the quotient of two functions (quotient rule)
- The derivative of f(x) = g(x) / h(x) is given by f'(x)= ( h(x) g'(x) - g(x)h'(x) ) / h(x)^2
January 17, 2013 Assignment
ReplyDeleteResearch about Limits and Continuity
http://mymathworks.blogspot.com
01.18.13
ReplyDeleteAssignment
Research about Differentiation
http://mymathworks.blogspot.com
Source:
http://www.bbc.co.uk/bitesize/higher/maths/calculus/differentiation/revision/7/
HW. Research about Differentiation
ReplyDeleteIn calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's instantaneous velocity.
The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. Informally, the derivative is the ratio of the infinitesimal change of the output over the infinitesimal change of the input producing that change of output. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point. In higher dimensions, the derivative of a function at a point is a linear transformation called the linearization.[1] A closely related notion is the differential of a function.
The process of finding a derivative is called differentiation. The reverse process is called antidifferentiation. The fundamental theorem of calculus states that antidifferentiation is the same as integration. Differentiation and integration constitute the two fundamental operations in single-variable calculus.
From: Wikipedia
Rules of Differentiation of Functions in Calculus
ReplyDeleteThe basic rules of Differentiation of functions in calculus are presented along with several examples:
1 - Derivative of a constant function.
The derivative of f(x) = c where c is a constant is given by
f '(x) = 0
Example
f(x) = - 10 , then f '(x) = 0
2 - Derivative of a power function (power rule).
The derivative of f(x) = x r where r is a constant real number is given by
f '(x) = r x r - 1
Example
f(x) = x -2 , then f '(x) = -2 x -3 = -2 / x 3
3 - Derivative of a function multiplied by a constant.
The derivative of f(x) = c g(x) is given by
f '(x) = c g '(x)
Example
f(x) = 3x 3 ,
let c = 3 and g(x) = x 3, then f '(x) = c g '(x)
= 3 (3x 2) = 9 x 2
4 - Derivative of the sum of functions (sum rule).
The derivative of f(x) = g(x) + h(x) is given by
f '(x) = g '(x) + h '(x)
Example
f(x) = x 2 + 4
let g(x) = x 2 and h(x) = 4, then f '(x) = g '(x) + h '(x) = 2x + 0 = 2x
5 - Derivative of the difference of functions.
The derivative of f(x) = g(x) - h(x) is given by
f '(x) = g '(x) - h '(x)
Example
f(x) = x 3 - x -2
let g(x) = x 3 and h(x) = x -2, then
f '(x) = g '(x) - h '(x) = 3 x 2 - (-2 x -3) = 3 x 2 + 2x -3
6 - Derivative of the product of two functions (product rule).
The derivative of f(x) = g(x) h(x) is given by
f '(x) = g(x) h '(x) + h(x) g '(x)
Example
f(x) = (x 2 - 2x) (x - 2)
let g(x) = (x 2 - 2x) and h(x) = (x - 2), then
f '(x) = g(x) h '(x) + h(x) g '(x) = (x 2 - 2x) (1) + (x - 2) (2x - 2)
= x 2 - 2x + 2 x 2 - 6x + 4 = 3 x 2 - 8x + 4
7 - Derivative of the quotient of two functions (quotient rule).
The derivative of f(x) = g(x) / h(x) is given by
f '(x) = ( h(x) g '(x) - g(x) h '(x) ) / h(x) 2
Example f(x) = (x - 2) / (x + 1)
let g(x) = (x - 2) and h(x) = (x + 1), then
f '(x) = ( h(x) g '(x) - g(x) h '(x) ) / h(x) 2
= ( (x + 1)(1) - (x - 2)(1) ) / (x + 1) 2
= 3 / (x + 1) 2
hw in math research about limit and continuity
ReplyDeletein partucular the many definitions of continuity employ the limit if the condition o/begin
matrix a given is to be more precise if the limit of the
function f(x) and the f(c) this is as f(x) = l
Research in Math //
ReplyDeleteLimit and Continuity
Definition: Limit at a real number c (Both side limit).
Given > 0, there exists > 0 such that | f ( x ) - L | < whenever 0 < | x - c | < ,
then the limit of f ( x ) at x = c is L, and denoted by ,
otherwise f ( x ) has no limit.
Roughly speaking, means that whenever x approaches to c
from either side of c, the graph point (x, f(x)) approaches to the point (c, L)
on the plane.
Definition: Continuity for f ( x ) at x = c ( both side continuity ).
Given > 0, there exists > 0, such that | f ( x ) - f ( c ) | < whenever | x - c| < ,
then f ( x ) is continuous at x = c. ( i.e ),
otherwise f ( x ) is not continuous at x = c. ( i.e f ( x ) is discontinuous at x = c ).
Roughly speaking, means that whenever x approaches
to c from either side of c, the graph point (x, f(x)) approaches to the
point (c, f(c)) on the plane.
source(s):http://people.usd.edu/~ylio/Calcul/Limit_and_Continuity.html