HELLO!!!

So right now i'm in my statistics class and Krystal is sleeping right next to me hhaaha. This is a wild class. Random students are in here.
And it's a shortened day so the bell is about to ring.
Well, GOODBYE and I hope I pass the ap statistics test, but i highly doubt it.

Calculus :O

After so much struggling, we've finally SURVIVED AP Calculus.!!!!

Thanks to all the GREAT friends who helped us overcome obstacles,

and ESPECIALLY to a GREAT teacher, Ms. Hwang!! (:

Let's Hope we PASS this Test!!!!8]

2005 FR 5 D:

The tide removes sand from Sandy Point Beach at a rate modeled by the function R, given by

A pumping station adds sand to the beach at a rate modeled by the function S, given by

Both R(t) and S(t) have units of cubic yards per hour and t is measured in hours for . At time t=0, the beach contains 2500 cubic yards of sand.
(a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure.

fnInt(R(t),t,0,6)

fnInt(2+5sin(4pi t/25),t,0,6)=approximately 31.815 or 31.816 cubic yards of sand.

(b) Write an expression for Y(t), the total number of cubic yards of sand on the beach at time t.

Y(t)= fnInt(S(t)-R(t)dt)+2500

Y(t)=fnInt((15t/1+3t)-(2+5sin(4pi t/25)),t,0,x)+2500

(c) Find the rate at which the total amount of sand on the beach is changing at time t=4.

So you need the rate of change of the equation in part b, at t=4, since the equation is the total amount of sand on the beach, and you need to know the rate at which this is changing.

If you dont want to find the derivative, just RECALL that...

When told the given equations, they are already in the units cubic yards per hour, which is the rate at which it is changing., therefore you just need to subtract the rate at which the tide removes sand from the rate at which the tide adds sand. (S(t)-R(t))

But if you want to take the derivative of Y(t)...

Since you know that Y(t) is the antiderivative, that means that the derivative would just equal the original equation "f(x)", which in this case is S(t)-R(t). <--- It is ALREADY the derivative (Rate of Change)

same thing if you do....

Y'(4)=nDeriv(fnInt(S(t)-R(t))+2500... The integral (fnInt cancels out with the nDeriv, because if you take the antiderivative and then the derivative, then you are left with S(t)-R(t), because of the Fundamental Theorem of Calculus Part 1. Taking the derivative of 2500 is just 0, since it is a constant.

so S(4)-R(4)= 4.6154-6.5241= approximately -1.908 or -1.909 cubic yards/hour

(d) For , at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.

Y'(t)=S(t)-R(t)

0=S(t)-R(t)

(The GRAPH of Y(t))

Using the Calculator, you find that the critical point is t=5.118 (the 2 curves intersect at this point twice if looking at both equations, but if you're looking at the graph of S(t)-R(t), the zero is t=5.118) and the endpoints are t=0 and 6.

By looking at the graph of Y'(t)=S(t)-R(t), its is clear that x=0 is a maximum because Y'(t) is negative after that point and it is also obvious that x=6 is a maximum because Y'(t) is positive before that point. By looking at x=5.118, you see that Y'(t) changes from negative to positive at that point, and therefore it is a minimum.

The Minimum Value of sand can be found by plugging in the minimum value to the equation that outputs "the total number of cubic yards of sand on the beach at time t.", also known as Y(t), the equation we found in part b. (Wow this equation really comes in handy! (:).

So Y(5.118)= fnInt(S(5.118)-R(5.118),t,0,5.118) +2500. (You want the area of the graph from 0 to 5.118 in order to get the amount of sand at the time t.)

Y(5.118)= -7.630517431+2500 = approximately 2492.369 cubic yards of sand.

Well what I did for part D to find Y(5.118), I first plotted the 2 equations of S(t) and R(t) in the "y=" in the calculator. I put R(t) into the "Y1" and the S(t) into "Y2". Next, I just pressed "math+9" to add "fnInt(", then i clicked "VARS"-->"Y-VARS" --->Function---> "Y2". Then I added the subtraction sign and added "Y1" using the same method. Now I have Y2-Y1, which is equal to S(t)-R(t). Then you just add the conditions ",x,0,5.118)", NOT up to 6. Then just hit enter and add "2500" to get your answer.! TADA!

Im not so sure if this is correct after seeing that some have the same answers, but many others do not. If I am mistaken, feel free to correct me, since it is best to learn from failed attempts (:

Mean Value Theorem Edits :&

Well I only put the corrections here, and the rest of the post is still on the original! (:

1. ----------------------
f'(c)= [f(b)-f(a)]/(b-a)

• This means that the slope of the secant line at 2 points A and B, on a continuous and differentiable function between the intervals [a,b], is guaranteed to be equal to the slope of the tangent line at one point, c. On the graph, c may be the midpoint of [a,b].
  
• Because the slopes of the tangent and secant lines are equal, the lines will be PARALLEL to each other.


• Lets try another function that ISN'T f(x)=x^2, because Ms. Hwang doesn't like it's simplicity (:


• So this time, how about... f(x)=e^x

As you can see, the slope of the tangent line at c=0 is equal to the slope of the secant line between the interval (-1,1), so the lines are PARALLEL.

**I don't like how that one turned out, since the secant line looks weird...,so I shall use another example.. f(x)= -sin(x+pi)-1 between the interval (pi, 2pi).

The graph shows that between the interval (pi, 2pi), there is a tangent line at point c=3pi/2 with the equation y=-2 and a secant line, y=-1. The function, being both differentiable and continuous, allows the slope of the tangent line to be equal to the slope of the secant line, therefore allowing the Mean Value Theorem to apply. Both slopes are equal to 0.

2. -----------------------

It only works for continuous and differentiable functions because if between the interval [a,b] the function is discontinuous(because of a jump or asymptote,) or not differentiable (because of a jump, asymptote, cusp,corner...) then there may not be a tangent line at point c (or there may not even be a point c on the function if it is discontinuous where point c should be) that is parallel to the secant line going through points A and B.

• Since I had an equation involving abs of x, (f(x)=abs (x +1)+3), and im not sure whether I have to re-do this one, I will anyway.


• So... ANOTHER example of a CONTINUOUS but NOT DIFFERENTIABLE function is f(x)=x^(2/3)+1 on (-1,1).

There should be a tangent line (y=1) parallel to the secant line (y=2) at point c=0, but it is not possible for such a tangent line to exist due to the fact that the function is not differentiable at point c.(the slope of the line before c=-1 and the slope of the line after c=-1 are not equal) despite the continuity. Whenever there is a cusp at point c, the mean value theorem does not apply because f ' (c) does not exist.

Ohhhh AND to answer your question ms Hwang, related to my LAST post, the secant line of the second equation is y=1

A Very Mean Value Theorem! xD

f'(c)= [f(b)-f(a)]/(b-a)

1. Explain what this means GRAPHICALLY by showing a good example!

• This means that the slope of the secant line at 2 points A and B, on a continuous and differentiable function between the intervals [a,b], is guaranteed to be equal to the slope of the tangent line at one point, c. On the graph, c may be the midpoint of [a,b].
  
• Because the slopes of the tangent and secant lines are equal, the lines will be PARALLEL to each other.
  
• This is proven through the graph of the function f(x)=x^2

The secant line (green line) goes through the points (0,0) and (2,4), where a=0 and b=2. It's slope should be equal to at least one slope of a tangent line(blue line) at a point c. In our case c=1, which is the midpoint between a=0 and b=2.

slope of secant line=[f(b)-f(a)]/(b-a)
=[f(2)-f(0)]/(2-0) ----use(f(x)=x^2)
=(4-0)/2
=4/2
=2

You know that c=1 because

f'(c)= 2 (2 is the slope of the secant line)
2c=2 (slope of tangent is 2x[or 2c],found by taking derivative of f(x)= x^2)
c= 1

f'(c)=2c
f'(1)=2(1)
f'(1)=2 (Slope of tangent line)

2=2
Slope of tangent line=slope of secant line (PARALLEL)


2. Explain why this only works for continuous and differentiable functions.

• It only works for continuous and differentiable functions because if between the interval [a,b] the function is discontinuous(because of a jump or asymptote,) or not differentiable (because of a jump, asymptote, cusp,corner...) then there may not be a tangent line at point c (or there may not even be a point c on the function if it is discontinuous where point c should be) that is parallel to the secant line going through points A and B.
  
• An example of a discontinuous function that fails the Mean Value Theorem is 1/abs(x).
  
  
  
  A point c does not exist between the interval [-2,2] where f'(c)=[f(b)-f(a)]/b-a.
  
  (NOTE: A function may be discontinuous between an interval [a,b] and STILL have a tangent line at a point c that is equal to [f(b)-f(a)]/b-a]!!! Because of the discontinuity, there is just not a GUARANTEED point c, since the Mean Value Theorem fails).
  
• An example of a continuous but not differentiable function that fails the Mean Value Theorem is abs(x+1)+3 on (-2,0).
  
  

There should be a tangent line parallel to the secant line (green line y=4) at point c=-1, as shown in the picture (blue line), but it is not possible for the tangent line to actually exist because the function is not differentiable at the point c (the slope of the line before c=-1 and the slope of the line after c=-1 are not equal)despite the continuity.

The Function f(x) from the Graph f'(x) (:

1. Where is the function, f(x), increasing? Where is it decreasing? How can you tell from this graph? Explain.

• f(x) is increasing between (-2,0)U(0,2). This is due to the fact that when f '(x)>0, the graph of f(x) is increasing.
• f(x) is decreasing from (- infinity, -2)U(2, infinity). I can tell from this graph because whenever f '(x)< 0, it means that the graph of f(x) is decreasing.
• (On the graph f '(x), anything above the x axis is where the graph f(x) is increasing and anything below the x axis is where f(x) is decreasing. **Remember that f '(x) graphs the slope of f(x).)

2. Where is there an extrema? Explain. (There are no endpoints.)

If you're talking about the extrema on the graph f(x)....

• Local Minimum: x=-2 (-2,0) A Local minimum occurs at the point where f ' (x)< 0 and then changes to f ' (x)>0, also known as a critical point ( f ' (x)=0 or undefined ) [the function has to change from negative to positive slope in order for it to be a local minimum).
• Local Maximum: x=2 (2,0) A Local maximum occurs at the point where f ' (x)>0 and then changes to f ' (x)<0, which is also a critical point. (the function has to change from positive slope to negative slope in order for it to be a local maximum).

(Way clearer explanation than on my test, >:\)

3. Where is the function, f(x). concave up? Where is it concave down? How can you tell from this graph?

• Concave up: (-infinity, -1.25)U(1.25, infinity). The function is concave up whenever f "(x)>0
• Concave down: (-1.25, 0)U(0, 1.25). The function is concave down whenever f " (x)<0

4. Sketch the graph f(x) on a sheet of paper. Which power function could it be? Explain your reasoning.

The graph f(x) appears to be an x^5. Since the graph of f ' (x) looks like an x^4, my prediction is reasonable since an x^4 is the derivative of an x^5 graph.

Mindsets :O

1. Well I believe that I have components of both mindsets in me, depending on certain subjects. For example, sometimes I don't like taking challenges, which shows the Fixed Mindset. Like at this moment, I hate the challenge of taking AP English, and I want a way out of it, yet at the same time, I know that it will help me improve my analytical writing and I want to stay. The problem is that I don't like writing essays, and since the class is based on essay writing, I find it very stressing.
At the same time, I am enthusiastic about taking the AP Calculus challenge, probably because I love math and I want to be able to know more, which shows the Growth Mindset. If I have trouble, I know that I can always ask for help, and even if I fail, i'll get right back up and keep attempting.
On the other hand, I do believe that intelligence can be developed. Also, i've recently learned to not care about how people see me. I only care about how I see myself, and If I fail, its my problem and no one elses because at least i'll know that I tried. Yes, it may discourage me and may take me a while to recover, but that doesn't mean that life should end there. There is still plenty of time to get back up and keep adding effort in order to succeed. That's another thing: I don't see effort as useless. It is always important to work hard even if it doesn't get you anywhere at that moment. Later on you'll realize that it made you into a stronger person and gave you more confidence to keep trying. Also, I like receiving criticism. It tells me what i'm doing wrong so that I can change and improve on what i'm doing wrong. There always has to be criticism because not everyone is perfect and if they were, then what is the point of going to school? You go to school to learn and fail, and then get back up, fail, and keep trying until you succeed.
2. This mindset has helped me in math because I'm always up for the mathematical challenge. So far, i've reached calculus, and im enjoying every bit of the challenge. I receive criticism, and it just makes me put in more effort to fix my errors and understand the lesson even further. Obstacles help me grow, even though they may make me gloomy at first. Once I master it, i get ecstatic and realize that I am one step closer to overcoming the challenge. I know that if I need help, I can always count on my other classmates to help me. That is the perk of having classmates that may have succeeded more in that certain area of math.
3. Well, by finding out that you can train your brain, I was given more hope to keep going and learn more.This has reasurred me that school does have a purpose and it is okay to fail, because it just makes me stronger as a person. Everyone struggles, and it just takes time to adjust to it.
4. Now, I feel more confident in accepting challenges and overcoming obstacles. It's not going to be easy, but slowly i'll be able to let go and allow myself to take the risk, even if it leads to failure. No one is perfect, and everyone has had an experience that has allowed them to see this. This doesn't mean that I have to give up if I fail. I have to keep trying, no matter how difficult it may be. I know that I feel crazy writing this right now, but I know that it is the right thing to do. Yes, I may not follow my advice all the time, but it takes some time to get used to it and let it soak in. I've done it before, like in AP Biology. If I didn't get something, I would keep trying to understand until I finally did. It was a struggle that I overcame. This will allow me to succeed in college. I've heard about how difficult it is, and if I fail, I have to keep trying. This article will truly affect my future in a positive way.

Algebra vs. Calculus :}

1. What is the DIFFERENCE between finding the limit of a function at x=c and actually plugging in the number x=c? When are the two cases the SAME?

• When finding the limit of a function at x=c, you're finding what y value f(x) gets closer to as it approaches c. There doesn't actually have to be a point at x=c. There can be a hole at x=c and the limit can still exist, as long as f(x) passes through only one point at x=c.
• In turn, when plugging in the number x=c, there cannot be a hole at that point. You are finding the exact value of y when x equals a constant. Even though there cannot be a hole, the function doesn't have to actually pass through the point. There could be 2 open dots, one coming from the left and approaching a value different from f(c) and one coming from the right and approaching an entire new value from the other two. As long as there is a closed dot at x=c, f(c) does exist.
• They are the same when there is continuity at x=c. If the lim f(x) x->c = f(c), then you will get the same value for both and therefore f(x) does pass through the exact point at x=c.

2.What are the SIMILARITIES between finding the derivative and finding the slope of a line? What are the DIFFERENCES between the two?

• The similiarities between finding the derivative and finding the slope of a line is that you use change of y/change of x to find the slope, even though the specific formula for finding them are different.
• For finding the derivative, limits are involved. One of the most commonly used formulas is limf(x)h->0 = f(a+h)-f(a)/h. To find the derivative, you are bringing one point closer and closer to a main point and a tangent line is formed. This is done by making h approach 0, which closes the gap between the 2 points since h is the distance between the first point and the main point. With a derivative, you are finding the slope of a tangent line on a curve at a specific point (depending on where they want you to find the slope). There can be many different slopes on the curve as well.
• When finding the slope of a line, you are finding only the slope of that line. It is a specific slope (unlike the many slopes you can find when finding the derivative of a curve).

ex. 2x+1. The only slope on this line is 2.

x^2. This parabola has many slopes with different ones at each point on the parabola where the tangent lines are formed.

I've Reached My Limit! >=/

What do you still not understand about limits? Choose 3 problems or types of problems anywhere from chapter 2 that were the most difficult and that you would like to get more help on. OR you can simply explain 3 ideas / concepts that still elude you.

1. I still find it sort of confusing at times to find the limit of something as it approaches infinity and negative infinity. It is not difficult with a graph in front of me, or if it is a simple equation. But when the equation gets really difficult or confusing, I do understand that you have to find the end behavior model and go from there. Sometimes, I just get confused and I would like more practice on this in order to understand it even better.
2. When it comes to finding vertical asymptotes, I understand it most of the time. What I do have trouble with sometimes is finding the horizontal asmyptotes. I would understand this concept and forget it continuously, and I think that the only way to engrave this in my head is by practicing it more. On friday by 4th period after various attempts to understand horizontal asymptotes, I recalled that the limit of f(x) as x approaches + and - infinity=c, then c is the horizontal asymptote. I also remember that if you find the end behavior model, you can just imagine the graph in your head and see where the horizontal asymptotes are, if any.
3. At times I get confused on how to "Describe the behavior of f(x) to the left and right of each vertical asymptote". I know how to do this, but when I cannot imagine the graph in my head, I don't know whether it approaches negative or positive infinity from the left or from the right of the vertical asymptote. It is difficult to determine this on my own, like on:

f(x)={x^3-4x, x<1

{x^2-2x-2, >or= to 1

This problem also greatly affects me with piecewise equations when it asks to find the limit of f(x) as x approaches +or-c from the +or- side of a equation that has a difficult graph to imagine, and I have to say whether it is negative or positive infinity. Sometimes end behavior isnt enough for me to figure it out, so if it is, I would like to have that concept explained to me better.

Majors and Colleges ")

3 Majors

1. Biomedical Engineering: This major interested me since I love Biology/Anatomy as well as Math and it has both of them integrated! Well in this major, people try to find ways to "fix" people whenever they need to be "fixed". They use engineering to help solve problems related to health (Yay). Their job is to invent new vaccines, invent new technology, etc. that will benefit those who need it in order to be "fixed". With tons of math and science, this major really caught my attention.
2. Biochemistry: As I was reading the description, I knew exactly what they were talking about! I love learning about proteins and enzymes as well as about the chemical reactions that go on inside organisms. In AP Biology, even though many people hated learning about the detailed process of photosynthesis which included the light reactions and about what occurs in mitochondria to create energy, I really enjoyed it and got into it. In the major, I will also learn about how living things function. The creation of DNA is also very intriguing to me, and I hope to study something that I will never get tired of since I will be doing it for the rest of my life.
3. Accounting: I will learn to help businesses so that they can make the best financial decisions. By looking at financial records, I will be able to predicit what the company should do financially and what they should not do. I will "advise"in order to try to guarantee the success of a business. I'll also learn how to prepare financial statements. OK I LOVE MATH! There are many careers that can be obtaind by this major such as becoming a budget analyst (ive liked this since 9th grade!). I think all of this is fun and I would love to be able to do this!

3 Colleges

1. University of California Los Angeles: UCLA is a public university with only 23% of applicants admitted. It is located in a very large city in an Urban setting. 92% of students had a GPA of 3.75 percent or higher, which makes acceptance a lot more difficult. The SAT Reasoning Test or ACT (plus writing) are required as well as the SAT Subject Tests. 70% of Scholarships/Grants are awarded to students as well.
2. Massachusetts Institute of Technology: MIT is located in Cambridge Massachusetts and only 12% of applicants are admitted into the school. This makes it difficult to get in because there is a lot of competition to attend. It offers a Bachelor's, Master's, and Doctoral degree and it is a Private University. It is located in a small city with an urban setting and 97% of 1st year students are in the top 10th of graduating class.
3. University of Southern California: USC is a private university in a very large city in an Urban setting. It costs about $39,274 to attend but 77% of Financial Aid/ Scholarships are awarded to undergraduates and 23% as loans/jobs. Only 22% of applicants are admitted.

Im still looking around and investigating,but these 3 colleges caught my attention. Im hoping to find other choices as well. As for the majors, these are just some I have liked so far, but there are THOUSANDS out there that I still need to look into. Everyday I have new likes and dislikes, and my choices in majors and colleges may even be different by tomorrow :]

Tips and Hints 8]

1. Share how you remember transformations. Do you have any tips or hints that help you to remember?

In order to remember transformations, I first have to memorize what occurs when i change the parent function.

• If I add or subtract a number to the x, then i know that the graph should go either left or right. For example, f(x)=sin(x+1)...Moves ONE unit to the LEFT. I would instinctly believe that the graph would move one unit to the right because of the addition sign, but in reality, it has to move to the left. If you have f(x)=sin(x-1)... then the graph moves ONE unit to the RIGHT (NOT TO THE LEFT!). In order to remember it, I just remember that it moves OPPOSITE to the addition/subtraction sign.


• If I multiply x by a number, then I know that the graph has to shrink horizontally. For example, if I have f(x)=sin2x... then it looks like the graph goes (2x) "faster", compared to its parent function (sinx). It will also compress horizontally because the PERIOD changes to pi instead of the original 2pi. (TO FIND THE PERIOD, REMEMBER THAT YOU DIVIDE THE PERIOD OF THE GRAPH BY THE NUMBER IN FRONT OF THE X). If the number in front of the x is a fraction, then the graph will stretch horizontally instead, since the PERIOD will increase.
  

• If the output has been multiplied by a number, f(x)=2sinx, then the graph stretches vertically. The period still stays the same (if the function has a period). The graph will not be any "faster". If the number in front of the equation is a fraction, then the graph will compress vertically.
• If the number in front of the equation is negative, you just flip the graph. The output has been made negative. You reflect the parent graph across the x axis.

EX. f(x)= -Sinx

• If a number is added to the equation, such as f(x)=sinx+1 (without parenthesis), then the graph will shift up(in this case, up one unit). If a number is subtracted, f(x)=sinx-1, then the graph will shift down (in this case, down one unit).

The only tip I can give you is to MEMORIZE all this information and remembering how the graph will shift, shrink, or compress depending on how the parent function is manipulated.

2. Share how you remember or understand trigonometry. Do you have any tips or hints that help you remember/memorize all those facts?

To understand trigonometry, you have to know that the coordinates on the unit circle are not just random numbers that were made up. They actually MEAN something. Take for example, pi/6. The angle is 60 degrees, since 180 degrees (half of the unit circle) divided by 3 is 30 degrees.

The coordinates are (sqrt of 3/2, 1/2). The x axis is sqrt of 3, divided by the hypotenuse 2, and you get your x coordinate. Same thing for the y coordinate.

You can also remember this by keeping in mind that the longer side of the triangle is sqrt of 3/2 and that the shorter side is 1/2.

For pi/4, just remember that the sides of the triangle are the same (excluding the hypotenuse) and therefore the x and y coordinates are the same. (sqrt 2/2, sqrt 2/2).

The way I remember which coordinate is which for pi/6 and pi/3, (and any other coordinate at "?/6"&"?/3")...

FOR pi/3, I take into consideration that the "/3" comes second, and therefore the 2nd coordinate (y coordinate) has to be sqrt 3/2... (1/2, sqrt 3/2).They will both come SECOND because they include the number "3". Since pi/6 does not have the 3 at the bottom of the fraction like "/3"(the 3 doesn't come second), then the sqrt of 3/2 cannot be the 2nd coordinate (it cannot come second). So the coordinates for pi/6 is (sqrt 3/2, 1/2). The "3" comes FIRST.

This makes sense in my mind, and I tried my best to explain it. Usually people do not understand what i'm talking about.. haha

I think that you should just have the unit circle MEMORIZED and you should only use the tips if you FORGET some of the coordinates.

TO GRAPH: All I do is remember the main coordinates of the Sin, Cos, Tan graphs (What y is at 0, pi/2, pi, 3pi/2, & 2pi). Then I just continue the graph over again since its a new period and looks the same (For a parent graph). For Tan though, I have to memorize where the asymptotes are at as well. For Csc and Sec, I just graph the sin or cos graphs (sin for csc and cos for sec) and at the top or bottom of the curve is where i draw the parabolas. The asymptotes are where the sin or cos graphs cross the x axis. To remember Cot graphs, I just memorized that the asymptotes are at pi and that the "middle" point is at pi/2. Also i remember that the Cot graphs are the tan graphs flipped horizontally. When transformations are involved, I just follow the rules of transformations that I mentioned above.

3. What still confuses you or worries you about trigonometry?

Well when you shift a graph up, it is difficult to find the zeros of the graph. For example, if the regular graph would have the x intercept at pi/4, etc, and you shift the graph up, I just know that the period will be BETWEEN pi/4 and pi/3. I don't know how to find the exact x intercept without a calculator.

Logs and Inverses :D

1. Recap 4 major concepts that you have understood about either/both of these two topics.

1. Relating to inverses, I believe that I understood the concept of one-to-one rather clearly. I learned that in order for a function and its inverse to be "one-to-one", the graph of the function must pass the horizontal line test, which means that if a horizontal line is passed over the graph of a function, the horizontal line intersects the graph at one point at a time. This horizontal line test determines if the inverse exists as a function. If the graph of the function passes the horizontal line test, then the function is "one-to-one".
2. I understand that the inverses of a parent function are symmetrical to the parent function itself. If you fold the paper between both lines, such as folding the paper of the function f(x)=3 and f^-1(x) at the line y=x, the lines will lay right on top of each other.
3. I also understand how to find the inverse (f^-1(x)) of a function (f(x)) and how to verify that (f o f^-1)(x)=(f^-1 o f)(x) = x. When you multiply the function by its inverse, you should get "x", and vise versa. In order to demonstrate how to find the inverse of a function, I will use the function f(x)=x^2 (x< (or equal to) 0). This function is also equal to y=x^2. First you change x to y, and y to x in the equation. You get x=y^2. The goal is to leave y alone, so you take the square root of both sides of the equation and end up getting sq root of x=y, also known as f^-1(x)=sq root of x. --- To verify that it is the inverse, you plug in sq root of x into the original function and get (f(x)=sq root of x^2). When you solve it by cancelling out the sq root and the ^2, you get f(x)=x. Also if you input x^2 into the inverse function, you get (f^-1(x)=sq root of x^2) also, which is equal to f^-1(x)=x. This proves that f^-1(x)=sq root of x.
4. Today in class (Thursday), as we were solving logarithm equations for x, I believe that I understood how to solve most of them easily. For example, when solving the equation "7-3e^-x=2, when I got to the point of "e^-x=5/3", I knew that I was supposed to take the natural log of both sides so "e" could cancel out and so I could be left with "-x=ln5/3"--> "x= -ln5/3". Even though sometimes I may get confused or forget how to do a step, I know that with a little more practice, I will grasp the concept more fully.

2. Write about what you did NOT understand completely.

• Relating to Logarithms, I can honestly say that I am not so sure how to graph them. I know that their inverses are functions, but I still cannot grasp the concept of graphing them without using a graphing calculator, unless I input the values for x to get the y output. I do not know how to get the inverse function of a logarithm function that is complicated. Sometimes the log functions are kind of complex and.. yeah I need help. :D
• Even though I understand, in general, how to solve the logarithm functions for x, most of the problems on the homework C2 were a little more tricky and I did not understand them. For example, number 35 (e^x+e^-x=3) may look simple at first, but once I tried to solve it, my answer looked nothing like the one in the back of the book. This was "mind-boggling"! I tried to take the ln of both sides to cancel out the e's but I ended up getting "x-x=ln3" --> "0=ln3". Im not sure how to solve this or anything in relation to that...

Even and Odd Functions :]

EVEN Functions

An even function can be defined as f(-x)=f(x) algebraically.

• This means that for every input value of x and its opposite value (x and -x), the output will be the same (y).



• Therefore (x,y) and (-x,y), or (x,-y) and (-x,-y) will be reflections of each other about the y axis if plotted on the same graph.



• Because of this, the graph of an even function will always be symmetrical about about the y axis. (Quadrant 1&2, or Quadrant 3&4).



• Any function with an even power (x^2, x^4, x^6, etc.) or with an absolute value will have these characteristics.

Example: All 3 graphs are symmetrical about the y axis.

In the first graph, x^2, If 1 and -1 (x and -x) are inputted into the equation, the outputs (y) will both be 1.

If 5 and -5 are inputted, the outputs will both be 25.

f(-x)=f(x)

f(-5)=f(5); x= -5, 5

25=25; y=y

ODD Functions

An odd function can be defined as f(-x)= -f(x) algebraically.

• If the point (x,y) or (-x,y) is plotted on a graph, (-x,-y) or (x,-y) must also be a point on the graph, respectively, in order for it to be an odd function.



• The graph of an odd function will always be symmetrical about the origin.



• Symmetry may be on Quadrant 1&3, or Quadrant 2&4 in most, but not all, cases.



• Any function with an odd power (x, x^3, x^5, x^7, etc.) will have these characterisitcs.

Example: All 3 graphs are symmetrical about the origin.

In the first graph, x, if 1 and -1 (x and -x) are inputted, the outputs will be 1 and -1 (y and -y) respectively.

If 5 and -5 are inputted, the outputs will be 5 and -5 respectively.

About Me.

Hello! My Name is Dianna. I am 16 years old, and i'm a Junior at Polytechnic High School. I have 2 sisters, with whom I share good memories, as well as bad. My youngest sister, Sandy, is a Sophomore at Polytechnic H.S. The oldest, Karen, is a freshman at Cal Poly Pomona and is studying to be a Computer Engineer. Right now, I miss my sister Karen, since she has left us for the "college life"... :( I love biology and mathematics and I am looking forward to pursuing a career in both. Yes I am a math GEEK/NERD. =] AP Biology has made me love biology as well since the class was so interesting in my point of view. I strive to go to Stanford University by the time I finish high school and hope to become an engineer.

Furthermore, I enjoy eating food for fun. Lets go to COSTCO and get some free samples! :D

I am called a FATTY by my friend! haha

I am also on the PHS Soccer and Tennis teams which are coached by Ms. Miller.

Im in a variety of clubs, including: 11th Grade Council, Soccer Club-Treasurer, California Scholarship Federation (3 years), National Honor's Society - 11th Grade President, and I enjoy them all very MUCH.

I am also in the Young Senators Program with Richard Alarcon and in the Kaiser Permanente Medical Explorers [:

I'm part of the Early Academic Outreach Program (EAOP), Educational Talent Search (ETS), & the HEART Program:]

I read for fun when I get the chance and I also watch television and use the computer. [: I rarely have time to do this, either because of homework, or because im too lazy to do it.. hehe

I have very INTERESTING friends like Rocio, Denise, Alma, STEPHANIE, and Jessica :D.. and ONE of them wants to sell me.. and I am not her dog.. ALMA. haha

Stanley! <3

On my way back from Arkansas :]

Fishing Trip - Sunrise :)