Wow...is it merely the 7th week of school? It feels as if I should be going into the Christmas Holidays in about 2 weeks......not in the middle of October.
Alright, this week....Studying and Study Guides. But I do remember Mrs. Robinson saying something about us being able to explain something from past weeks right? Good!
How about I get what I don't understand out of the way right now?
1. STILL!! Optimization
2. Tangent Lines [[I get the general gist of it, just...not all of it]]
First Derivative Test (I'm sorry if everyone understands it, but I don't want to look dumb by explaining something I don't get at all wrong)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
Can someone check this for me? I don't know if I'm not in my right mind right now [no comments] or if I'm really failing at this? Or if I just do it and not think about it now?
f(x) = 1/2x - sinx
Find the extrema on the interval (0, 2pi)
f'(x) = 1/2 - cosx
1/2 - cosx = 0
-cosx = -1/2
cosx = 1/2
x=cos^-1(1/2)
x=pi/3 and 5pi/3
Because
| +
-------
| +
cos is positive on the first and forth quadrants
300 degrees in radians is 5pi/3
and
60 degrees in radians is pi/3
Also, for the powerpoint tomorrow:
1. If I downloaded fonts from the Internet, will they show up on Mrs. Robinson's computer?
2. Will someone be clicking the next slide button or does it have to be set on a timer?
Thanks!! =]
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Whenever you take the derivative of a function and solve for x, you are finding the critical points. When you set the points into intervals to plug back in, that is where you will find your max and mins and all that other stuff. The x values are just critical points. Hope this helps girl :)
ReplyDeleteyou said you don't know tangent lines up there.. so, to find the eqation of a tangent line
ReplyDeletefirst, take the derivative of the function, then plug in the x from the point that's given to you into the derivative you just found. Doing this gives you the slope, then you plug the point given and the slope you found into slope intercept form and you're done.