Blog Post #6

1. Graphs/Functions where derivatives do not exist: Absolute Value, Asymptotes, Cusps and Jumps

2. Implicit Differentiation

•     Step One: Determine the variable you are using for your differentiation.
•     Step Two: Find the derivative in relation to the variable chosen. (If you find the derivative of a term with another variable, you must multiply that derivative by the change in that variable over the change of the original variable that you chose.) 

3. The most important thing to remember in a rate related problem is to know which variable you are regarding to.

Blog Post #5

1. f'=0 represents a point in the graph when the slope is zero. This could be a maximum or minimum or a critical point. (Possible maximum or minimum.)

2. You can determine where a function increases or decreases by using the derivative to determine the slope at different points. By doing this, you can also determine the location of maximums or minimums if they are in the graph.

3.The chain rule is the process of finding the derivative of a composite function. To apply the chain rule, first take the derivative of the outside and rewrite the interior function. Then multiply the derivative of the inside.

Ex: y = (4-2x)^3 Find the Tangent Line at x=3

3(4-2x)^2 (-2)

-6(4-2x)^2

-6(4-2(3))^2

-6(4-6)^2

=-24

The slope at x=3 is -24.

(4-2(3))^3

(-2)^3

y=-8 when x=3

Plug in to point slope.
y+8=-24(x-3)

y=-24x+64

4. h(x)=f(g(x))
g(-4)=5, g'(-4)=2, f'(g)=20, Solve for h'(x)

f'(g(x)) * g'(x)

f'(5) * g(-4) * g'(-4)

20 * 5 * 2

200

Calculus Blog 4

1. Continuity

The limit exists at x=a, f(a) exists and there are no holes or asymptotes, and the limit at x=a is equivalent to f(a).

Example where continuity does NOT work:

             x^2+5, x<0

f(x)=      10,     x=0

              3x+5, x>0

Step 1: Limit as x approaches zero from theft and right is 5.

Step 2: f(0)=10. This means that there is no hole or asymptote.

Step 3: 5≠10 so the function is not contiunuous. 

2. Intermediate Value Theorem

Example that gives a solution:

y=6x+9 on the interval [-3, 0]

y=6(-3)+9 → y=-18+9 → y=-9

y=6(0)+9 → y=0+9 → y=9

Since f is continuous on [-3, 0] and f(-3) = -9 < 0 < 9 = f(0), then there exists c in [-3, 0] such that f(c) = 0 (Because one solution is positive and one is negative and the function is continuous, it must cross the x-axis within the interval.)

Example that does not give a solution:

y=x^3+5 on interval [2, 5]

y=(2)^3+5 → y=8+5 → y=13

y=(5)^3+5 → y=125+5 → y=130

Since f is continuous on [2, 5] and f(2) = 13 > 0 < 130 = f(5), then it cannot be concluded that there exists c in [2, 5] such that f(c) = 0 (Because both solutions are positive, we can not determine if the function crosses the x-axis within the given interval.)

3. Derivatives

Definition: A function which gives the slope of a curve or the slope of the line tangent to a function.

    Types of derivatives (example problem: f(x)=5x-5 at x=5):

The difference quotient. (H approaches 0.)
The limit of ((5(x+h) -5) - (5x-5))/h as h approaches 0

(5x+5h -5 -5x +5)/h = 5h/h = 5

The derivative that is found using the limit as x approaches a of the slope formula.

The Limit of ((5x-5)-(5(5)-5))/x-5 as x approaches 5

(5x-5)-(25-5)/x-5=(5x-5)-20/x-5=5x-25/x-5
=5(x-5)/x-5=5

    The hardest part of finding the derivative is remembering to completely distribute when it is necessary.

4. Difference Between Instantaneous Velocity and Average Velocity

Instantaneous velocity is the slope of the tangent line at one point in a function. The average velocity on the other hand, is over an interval in a function.

Blog #3

1. The limit of a function is the behavior before and after a specific point.

2. You evaluate limits by plugging in the number or value that you are testing into the given function. Certain values do not always work and you will end up with an undefined equation. (Divide by zero).When this happens the equation must be rationalized. We dealt with square root functions, absolute value functions, quadratic functions, cubic functions, and lines.

Examples of functions that don't immediately work: 


Square Root Function: Evaluate the limit as x approaches 0 from the equation
The first step is to rationalize the equation by multiplying by the square root of h + 16, +4. Multiply both the numerator and the denominator. The numerator should end up being h and the denominator should end up being h times the square root of h+16, + 4. The next step would be to divide by h, leaving you with 1 over the square root of h+16, +4. Then plug in the desired value (0). The answer should be 1/8. 



Absolute Value Function: Evaluate the limit as X approaches -1 from the equation:
The first step is to plug the value (-1) in for x without the absolute value sign and solve. This answer should be one. The second step is to make what is inside the absolute value equation opposite and solve. It should be -X-1 over X+1. The answer for this is -1. Because the two answers are different, the limit does not exist for this problem.

Quadratic Function: Evaluate the limit as x approaches 5 from the equation
The first step is to factor the numerator of the equation. In this case, the factors are (X-1) and (X-5). Because there is (X-5) on the numerator and denominator, they can be cancelled. The final equation should be X-1 over 1. The final step is to plug in the desired value (5) and solve. The answer should be 4.

Cubic Function: Evaluate the limit as x approaches 0 from the equation
First, triple foil the numerator of the equation. Once the numerator is triple foiled, every number in the numerator should have an h that can be divided out. Divide by h. After this, plug your value (0) into the equation. The answer should be 12.

3. https://www.desmos.com/calculator/vyl6gaur84
 Limits do not exist at -4, -2 and 2 because the behavior on both sides of these points is not the same.
:'( <Sad Face. I'm a failure.

4. This function has an infinite limit at zero because because we can get "arbitrarily large for sufficiently close." Meaning that it goes on forever and as we get closer to zero, the y-value gets significantly larger.

Calculus Attempt 2

Descartes
Rational Root Theorem
Synthetic Division
Graphing and End Behavior
Triangles and Trig Values

Descartes, Rational Root Theorem, Synthetic Division, Graphing and End Behavior, Triangles and Trig Values

Blog Assignment 1

1.) The hardest topic in math so far was solving word problems. This topic was hard because it is sometimes difficult to figure out what the word problem is asking.
2.) Solving Systems of equations is the most fun because it is challenging and I enjoy a challenge.
3.) I would like to become a chemical engineer and live like a boss.
4.) I would like to be accepted into Texas A&M University. I can accomplish this goal by turning in my applications, meeting deadlines, and being awesome.
5.) I would like to be an All-State track athlete two years in a row. I can accomplish this goal by training hard, listening to my coaches, and being awesome.
6.) A function is a set of ordered pairs that do not have a repeating x-value.


Function Picture Example



Function Example: 4x^2+2x+2