Blog Post 6: Derivatives, but wait... there's more?

1. Derivatives do not exist:

1. f(x)=|x| 
2. cusp
3. jump
4. asymptote

(anywhere with a corner or where the limit does not exist)

2. Implicit differentiation:

Take a derivative of a function with regards to a variable that may not be in the function or is not the only variable in the function. Every time you find the derivative of a piece, you multiply by the derivative of the variable over the derivative of the variable you are finding the derivative with regards to. Example: 

3. The most important thing to remember in a related rate problem is to always multiply by the correct d(variable)/dt so that you can correctly plug in given information to find the correct answer. 

Blog Post 5: Chain Rule, mostly

1. Where f'=0 there is a maximum or minimum of f.

2. Where a function increases or decreases can be determined by plugging numbers greater than and less than the critical point (possible maximums or minimums) into f'. Where these values are positive, the function is increasing; where these values are negative, the function is decreasing.

3. The chain rule is a process for finding the derivative of a composite function. This is done by taking the derivative of the outside function, rewriting the inside function, and multiplying by the derivative of the inside function.

Example:

tangent line at x=1

4. h(x)=f(g(x))

g(-4)=5, g'(-4)=2, f'(5)=20 find h'(-4)

h'(x)=f'(g(x))g'(x)

h'(-4)=f'(g(-4))g'(-4)

h'(-4)=f'(5)2

h'(-4)=20(2)

h'(-4)=40

Blog Post 4: Derivatives, etc.

1. A function is continuous at x= a if

1. the limit exists at x = a
2. f(a) exists
3. limit at x=a equals f(a)

This function is not continuous at x=1

The limit of x=1 exists

The function exists at f(1)

The limit of 1 is 2, which does not equal f(1) which equals 0

Since the limit and the output of the function are not the same at x=1, the function is not continuous at x=1.

2. Intermediate Value Theorem:

1. solution of f(x)=3x-1 on interval [0,4]

f(0) = -1

f(4) = 11

     Since f is continuous on [0, 4] and f(0) = -1 < 0 < 11 = f(4), then there exists c in [0, 4] such that f(c) = 0.

     2.  solution of f(x)=|4x-7|+12 on interval [7,10]

f(7) = 33

f(10) = 45

     Since f is continuous on [7, 10] and f(7) = 33 > 0 < 45 = f(10), then it cannot be concluded that there exists c in [7, 10] such that f(c) = 0. 

3. The derivative of a function is the slope of the tangent line at x = a. Derivatives can be found using the limit as h approaches 0 of the difference quotient or the limit as x approaches a of the slope formula. 

derivative of f(x) = 2x-6 at x = 4

Difference quotient:

Slope formula:

The hardest part is remembering to do all of the steps.

4. Instantaneous velocity is the slope of the tangent line at one point while the average velocity is over an interval.

Blog Post 3: Limits

1. A limit is the behavior of a function as it approaches an x-value from either side.

2. You can evaluate a limit by plugging an x-value into a continuous function. To determine whether or not a function is continuous, you can plug in a number greater than and less than the x-value to see if the two sides of the function match up.
     If you plug the x value into the equation and the denominator is zero, then you need to manipulate the function and try again. You can:
 rationalize:

You can get rid of a radical by multiplying the fraction by the conjugate and simplifying to cancel then plugging in the x-value to determine the limit.

 factor:

 

You can factor the numerator and denominator in order to cancel one of the factors and then plug in the x value to find the limit.

 combine fractions:

You need a common denominator to combine the fractions, then you can use that fraction to determine the limit. 

3. If a function is not continuous (such as with a piecewise) then limits will not always exist.

In this piecewise function, the limit of -2 does not exist, even though the function exists at -2. The two sides of the function do not match up at -2, therefore the limit of -2 does not exist.

4. When the denominator is an extremely small positive or negative number, the limit will be infinite. If one side of the function goes to positive infinity while the other goes to negative infinity then the limit does not exist. If both sides go to either positive or negative infinity, then the limit is infinite.

Since both sides of 0 cause the function to be divided by a very small positive number, so the limit of the function at 0 is positive infinity.

Blog Post 2

f(x)= 6x⁴+5x³-65x² +50x+24

The y-intercept is the place where the graph crosses the y-axis, and its ordered pair is (0,24) because the constant term is positive 24.

Descartes is a process used to determine the number of possible solutions (place where the graph crosses the x-axis) of a polynomial. Since the degree (largest exponent) of the polynomial is 4, there are four solutions.

To implement Descartes, you count the number of times the signs in the equation change to determine the number of possible positive rational roots.
     In this example, the sign changes twice, so there are two possible positive rational roots.

In order to determine the number of possible negative rational roots, plug in negative x and then count the number of times the signs change.
     In this example, the equation for negative roots will be: f(x)=6x⁴-5x³-65x²-50x+24. The signs change twice, so there are two possible rational negative roots.

Imaginary roots come in pairs, so it is possible to have no rational roots, no positive rational roots, no negative rational roots, or all rational roots.

The next step to finding the roots is to implement the rational root test, which involves dividing all of the factors of the constant term by the factors of the leading coefficient:

    ±1,2,3,4,6,8,12,24
    _______________      = ±1,2,3,4,6,8,12,24,3/2,1/3,2/3,4/3,8/3,1/6

           ±1,2,3,6

This list represents all of the possible rational roots of this polynomial.

Synthetic division can be implemented to test these roots. Synthetic division involves writing all of the coefficients (including ones that do not appear in the polynomial, such as if there is a 0x not shown), then placing the factor you are testing on the ledge. The leading coefficient gets carried down then multiplied by the ledge number. The product is added to the next coefficient, then that sum is multiplied by the ledge and so on. If the last number ends up as a zero, then the ledge number is a factor.

       2|      6     5     -65     50     24
                      12     34     -62   -24
      -4|      6    17     -31    -12     0
                      -24     28     12
     3/2|     6    -7       -3       0
                       9        3
    -1/3|     6     2        0 
                      -2
                 6    0
The solutions can then be checked by plugging the polynomial into the y= page of a graphing calculator and looking for the ordered pairs in the table. The ordered pairs for these solutions are: (2,0), (-4,0), (3/2,0), and (-1/3,0).



Triangle 1: sin 45°= (sq rt 2)/ 2
cos 45°= (sq rt 2)/2
tan 45°=1

Triangle 2: sin 30°=1/2
cos 30°=(sq rt 3)/2
tan 30°= (sq rt 3)/3

Triangle 3: sin 60°=(sq rt 3)/2
cos 60°=(1/2)
tan 60°= sq rt 3

First Assignment

1. The unit circle unit in math analysis was one of the hardest units in math, purely because of the quantity of memorization required.

2. Proofs were a fun topic in geometry because proofs are easy. 

3. I do not know for sure what I want to do with my life, but I managed to pick a major: biochemistry. With this major I could become a college professor or a researcher for a company. 

4. One of my main school goals is to get a PhD so that I could be a college professor. I can can achieve this by gaining research opportunities as an undergraduate student, earning my bachelor's degree, and earning my master's degree. 

5. My main out of school goal is to make the Southern Boone TSA program more successful. I can achieve this by promoting TSA to underclassmen, working harder on fundraising, and putting articles about TSA activities in the newspaper. 

6. Functions pass the vertical line test and often have y or f(x) on one side of the equation. An example of a function is y=2x²+3x-7