Unit U Big Questions

Big Question #1. What is continuity? What is discontinuity?

Continuity is when a graph has no breaks, holes, and jumps. The graph goes through from point to point with no disturbance as seen in the images shown below.

      

http://upload.wikimedia.org/wikipedia/commons/thumb/c/c2/Bisection_method.png/250px-Bisection_method.png  http://www.analyzemath.com/calculus/continuity/continuous_1.gif

Discontinuity is a distinct break in the graph. There are two groups of discontinuity removable discontinuities and non-removable discontinuities.  

Removable discontinuity consists of point discontinuity which is known as a hole. The image below shows an example of point discontinuity, as seen there is a hole in the graph.

http://www.wyzant.com/Images/Help/disc1.gif

There are three types of non- removable discontinuities: jump discontinuity, oscillating behavior, and infinite discontinuity. These are discontinuities because they involve holes in there graphs, wiggly lines and  they have breaks as seen in the pictures below.

http://thetwomeatmeal.files.wordpress.com/2010/11/discontinuities.jpg

Big Question #2 What is a limit? When does a limit exist? When does a limit not exist? What is the difference between a limit and a value?

A limit is the intended height of the function. It is where 2 points meet at the same time on the graph. The way we write the limit is “the limit x approaches a # of f(x) is a # (limit).

http://komplexify.com/math/images/haiku/LimitHaiku1.png

A limit exists when the point reaches the same height of both the right and left side. The limit still exist in a hole because is still has the exact height from the left and right side.

http://www.wyzant.com/Images/Help/limitspic.gif

A limit does not exist when there is a break in the graph, this occurs when the left and right sides of the graph don’t meet at the same height. As seen below there is a jump discontinuity and shows that there are two different heights therefore the limit does not exist. For this we need to write a one- sided limit statement. As seen in the picture you place a negative and positive sign beside the number where x approaches.

http://www.mathsisfun.com/calculus/images/discontinuous-function.gif

A limit is the intended height of the graph whereas the value is the actual point of the graph. A limit and value can either be different points for example in this picture we see the limit and value are different. The limit is 5 and the value is 2.

http://00.edu-cdn.com/files/static/mcgrawhillprof/9780071624756/LIMITS_AND_CONTINUITY_PRACTICE_PROBLEMS_15.GIF

In this picture we see that the limit and value occur at the same time. Tracing the graph from the left and right side we can see that the limit and value will be the same. 

Big Question #3. How do we evaluate  limits numerically, graphically and algebraically?

Evaluating the limit numerically we need to use a table. We put in the number that the limit approaches in the middle and on the left (negative) and right (positive) side we put numbers that reach the closest to number. Then we simply plug it into the calculator to find our limit.

Solving it graphically all we need to do is plug in the function to our calculator and graph it. Then we trace the graph from the left and right side to see where the limit falls. The limit can be a whole number or  Does Not Exist.

There are four ways to solve the limit algebraically:

• The first method is direct substitution. For this method we just plug in the number that x approaches in the equation.  There are four answers can get: a numerical answer, 0/#, #/0 and 0/0 which is an indeterminate form. All the answers are correct to get except 0/0 which is indeterminate.  With getting indeterminate form we have to try a different method. The following methods are what we can use.

•  Another algebraic method we use is dividing out/factoring method. We use this when we get indeterminate form. We factor out the numerator and denominator then we cancel out the like terms.  After we do this we use direct substitution to solve for the limit.

• When dealing with problems with rationales and when they are indeterminate we use the rationalizing/conjugate method.  In this method we multiply the numerator and the denominator by the conjugate.  Then we plug in the number as the x approaches  to get our limit.

• Our last algebraic method is limit at infinity. In this method we divide by the denominator highest power.  The next step is to use direct substitution to get our limit. 

    

Cited:

http://apcalculusbc.com/wp-content/uploads/2012/08/onesided-limits-2.png

https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcRQwZw2zv-i_fxGu8_H8FuW_mXd7XsZxO3kG5WBGDDgL-L7IFnY

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjDL1cEra4hLWWyun8CrRy57-FE7lAgx0wkWVFVFGu0jamAQinwTiTS3CbERmX7PS9ccebcI32xbCW6fkHwPzOLMkrLHw3-yXKZTdXARx8B_pvmqkXDzZBaCyTG9s9GZsHf3ZzjNqb5Ibpe/s1600/Function+y+%253D+10x.019.png

Math Mistakes Add.& Sub. Complex Numbers; Link: http://mathmistakes.org/?p=1070

What went wrong?
The mistake that occurred in the first box is that they did not distribute the negative properly. They were also subtracting numbers but then kept the numbers they had subtracted. Another mistake that they did was that in the second box the student foiled the equation when it should have been subtracted.

What they needed to do?
They needed to distribute the negative sign properly and then combine like terms, as shown on the picture below. This problem should not have involved multiplying it involved adding and subtracting only. 

Math Mistakes Sum and Difference;Link: http://mathmistakes.org/?cat=374

   

What went wrong?
The mistake for all the three boxes are that the student did not use the sum and difference formula to solve the equations. They simply look at the Y or X on the degree ordered pair and add or subtract them to get there answer. This is incorrect...

...What they should have done:
What they needed to do was solve the equations with the sum  and difference formulas. This means each degree being used  is labeled as the U or the V and the ordered pairs are separated into  sin,cos,and tan to make it simpler to imput into the equations, as seen in the pictures below. The highlighted problems are the ones that are shown on the Math Mistakes( each colors corresponds to each equation).

Answer to Question

These responses reveal that the student doesn't understanding trig because he/she is just using the X and Y parts of the degree ordered pair instead of the sum and difference formula. He/She needs to study the formulas and separate each ordered pair in order to successfully solve these problems.

Math Mistakes Special Right Triangles; Link: http://mathmistakes.org/?p=1003

What went wrong?
The mistake was that he rationalized the 1 instead of equaling it to N√2 and solving it out that way. He used the Pythagorean identity  which is a big NO NO when solving this equation.

Correction: What he had to do was equal the 1 to N√2 because that is the corresponding formula for that side. As seen below each side has a corresponding formula, therefore when the correct way was to equal the 1 to N√2 which gives you 1/√2. Next you rationalize and you get the answer of  √2/2.

Comment: 

Using the Pythagorean identity is the wrong formula to use, this type of problem does not use this type of formula. This is not used because each side has a formula. If I were in a situation if I forgot the formulas for the sides, I would probably would have used the Pythagorean identity as well.

Math Mistakes: Combining Like Terms; Link: http://mathmistakes.org/?p=953

What does the picture show?

This picture shows an equation of combining like terms. The equation seen above has been solved wrong  but the words in red go through the correct process.

What the viewer needs to pay close attention to in order to understand the concept?

The viewer needs to pay close attention to what the mistake of the student was. The mistake he made was combining all the terms together even though they weren't alike. They combined the numbers and placed the exponent along with the X. 

Correction: What the student had to do was combine  like terms of the problem which were the numbers with the exponent of x^4(10x^4 & 8x^4). Then they just bring down the 12x^5 to what we had combined and the correct answer would be 12x^5+18x^4.

Comment:

When I looked at the picture I could instantly see the mistake. This problem could be easily solved, but he made a cheeses bucket mistake!!