Reflection #3

This week in calculus we had a multiple choice test. I thought i knew most of what i was doing, but today when i was making corrections i realized that i messed up on some really easy ones. Ill do a few examples of those.

Find the x values at which f(x) x/x^2+x is not continous

First you can take out an x and then you are left with x/x(x+1), because the x cancel you get 0 which is not continous. Next you take whats left and set = to 0. So you get x=-1 which has a removable discountinuity at 0.

Find the limit of approches 13 x^2 +13x/(x^2+169)(x+13) This was a very easy one that i made extremely complicated. All you you had to do was take out an x and you get x(x+13)/(x^2+169)(x+13), now plug in 13 for x and you should get 26.
They had a few others, but you get the genreal idea of how i turned easy problems into a disaster.

In section 1.5 we had the rules of lim x approches infiniti
1. degree at the top equal the degree at the bottom limit is the coefficent
2.degree top is greater than degree bottom limit is infiniti or negative infiniti
3. degree top less than degree bottom limit is 0
Example lim x approches infiniti 1/x=0

Find the symptote of the graph example: f(x)=1/2(x+1) 2(x+1)=0 you get x=-1

Friday we started learning about secant lines and the rules and formulas you follow for that.
We learned that a secant line is one that touches at two points while a tangent line only touches down at one point.
The formula for it is m= f(c+deltax)- f(c)/ deltax

The formula for a secant slope is lim f(x+deltax)/deltax
deltax approches 0

keywords to look for when finding dervatives are as y^1, f^1(x), dy/dx, d/dx, dx[y] slope of a tan line, slope of the curve at a point.

We will be working on this more through out the week, i really like doing this stuff though because its recalling the algebra.