Sunday, March 7, 2010

post 29

here go teh blogs. again.


The limit rules are:
1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2) If the highest exponent is on the top then the limit is infinity.
3) But if the highest exponent is on the bottom then the limit is 0.

LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals

Tangent lines:

The problem will give you a function and an x value. Sometimes they may give you a y value; if not then you plug the x value into the original function and solve for y to get the y value. Next, you take the derivative of the function and plug in x to get the slope. After that, you plug everything into point-slope form.

First derivative test:

For the first derivative test, you are solving for max and mins and may be trying to see where the graph is increasing and decreasing. You take the derivative of the function and and set it equal to zero and solve for the x values (critical points). Then you set those points up into intervals between negative infinity and infinity. Then, you plug in numbers between those intervals to see if it is positive or negative.

Second derivative test:

For the second derivative test, you are solving to see whether the graph is concave up, concave down, or where there is a point of inflection in the graph. You take the derivative of the function twice and set it equal to zero and solve for the x values. You set those values up into intervals between negative infinity and infinity. You then plug in numbers between those intervals to see if it is positive or negative. If it is positive, it is concave up. If it is negative it is concave down. Where there is a change in concavity, there is a point of inflection.

im not good at e integration

3 comments:

  1. The integral of e^u is just e^u + c
    So if you are integrating e^3x, the integral would be e^3x, but you have to get rid of the three by adding a 1/3 in the front.
    The final answer would be 1/3 e^3x + c

    It works the same no matter what e is raised to.
    The integral of e^7x is 1/7 e^7x + c.

    Don't forget your +c.

    ReplyDelete
  2. Also if you need to get rid of an e or an ln use the other b/c they cancel each other out.

    ReplyDelete
  3. When you integrate e, the answer is always itself +c but remember that if there's a number like 4 inside of the integral you have to account for it outside of the integral by placing a 1/4 on the ouside of the sign.

    ReplyDelete