### Smarter Than The Average Bear

My title doesn't have anything to do with my scribe but I just had to say it. I felt that we say this often that it stucked in my mind now. I don't know if that's good but hey, it wouldn't hurt me.

To start of today we as usual had to do questions on the board related to what we've talked about yesterday. The first question was about calculating the average rate of change(slope) between two points values of the function. Then we had to approximate the intanteneous rate of change of the values. The second question is again finding the rate of change but by using the derivative of a function f at x=a denoted by f'(a), is

f'(a) = lim [ f(a+h) - f(a)] / h

h-> 0

For question one and two nobody had any problem with because its just a review from yesterday's class. For number three, i didn't quite understand it. It is very new to me and I had no idea answering the problem. Mr. K then explained what it is about. The question was;

A function f has a derivative f ' (x) = X^3 - 3x + 1 use your calculator to graph f'

To get the next question we have to remember that to get the equation we need the slope and a point in the line and use the point-slope formula which is y - y1 = m( x - x1 )

for this question the answer is y - 5 = 3(x + 1) . Mr. K said we can leave it like this because we might make little mistakes and get the answer incorrect.

The next question we looked at is about absolute functions. We learned today that you can't find the derivative of a corner or a cusps because it doesn't respect the definition that if we get the slope of the secant line on either side of a point it should be coming to a same number. In corners, the left side and right side of the point are different.

We also learned the nderiv and dy/dx function of our calculator. both calculates the slope but just in a different way. dy/dx must have a given graph to be used. you select a point in the graph then "calc dy/dx" then there you are with the slope. The nderiv one is a bit complicated. you put nderiv then type the equation comma the variable which is x then you put the point where you want to calculate the derivative and then there it is.

Lastly, then again we found out that our calculator is a liar and stupid because when we use the dy/dx capability of our calculator it still calculated the a derivative in corners. It is like this because mainly it calculates the symmetrical difference quotient of the point but doesnt know the restriction that there is no tangent at that point. Well we just have to remember always that we are smarter than the average bear.

To start of today we as usual had to do questions on the board related to what we've talked about yesterday. The first question was about calculating the average rate of change(slope) between two points values of the function. Then we had to approximate the intanteneous rate of change of the values. The second question is again finding the rate of change but by using the derivative of a function f at x=a denoted by f'(a), is

f'(a) = lim [ f(a+h) - f(a)] / h

h-> 0

For question one and two nobody had any problem with because its just a review from yesterday's class. For number three, i didn't quite understand it. It is very new to me and I had no idea answering the problem. Mr. K then explained what it is about. The question was;

A function f has a derivative f ' (x) = X^3 - 3x + 1 use your calculator to graph f'

- Where on the graph of f will the tangent lines be horizontal
- Suppose f(-1) = 5 write the equation of the tangent line to f when x= -1

To get the next question we have to remember that to get the equation we need the slope and a point in the line and use the point-slope formula which is y - y1 = m( x - x1 )

for this question the answer is y - 5 = 3(x + 1) . Mr. K said we can leave it like this because we might make little mistakes and get the answer incorrect.

The next question we looked at is about absolute functions. We learned today that you can't find the derivative of a corner or a cusps because it doesn't respect the definition that if we get the slope of the secant line on either side of a point it should be coming to a same number. In corners, the left side and right side of the point are different.

We also learned the nderiv and dy/dx function of our calculator. both calculates the slope but just in a different way. dy/dx must have a given graph to be used. you select a point in the graph then "calc dy/dx" then there you are with the slope. The nderiv one is a bit complicated. you put nderiv then type the equation comma the variable which is x then you put the point where you want to calculate the derivative and then there it is.

Lastly, then again we found out that our calculator is a liar and stupid because when we use the dy/dx capability of our calculator it still calculated the a derivative in corners. It is like this because mainly it calculates the symmetrical difference quotient of the point but doesnt know the restriction that there is no tangent at that point. Well we just have to remember always that we are smarter than the average bear.

## 1 Comments:

At 10:42 PM CDT, Mr. Kuropatwa said…

That was an outstanding summary Jay! You really nailed down all the little details that are so important to remember and keep track of.

You really are "smarter than the average bear!" ;-)

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