After this week in calculus i realized that it's not getting much easier. We mainly worked with the first and second derivative tests again this week and justifications.
A justification is when you pretty much explain everything that you did in the problem. You first explain that you took the first derivative test then you explain all the internal steps, like possible points of inflection, actual points of inflection, concavity, max/min, and things along those lines.
We also learned how to find the absolute max or min in a problem.
1. Take the first derivative
2. Find your critical values
3. Plug in for x to find y-values
4. Find your endpoints
5. Plug in for x to find y-values
6. Highest y-value is your absolute max, lowest y-value is your absolute min
We also learned about differntiability. This is one of the things I did not understand. Another thing I did not understand was horizontal tangent. I tried going over these two things at the study group today, but it still doesn't completely make sense to me. if anyone knows an easy and simple explanation, that would be nice :)
I think calc will start to get easier as the weeks continue because it is only our fifth week. What probably confused me the most this week were those extremely hard take home tests and study guides for the test. They kind of made me nervous for the test tomorrow, but i've been studying all weekend so I think it will turn out okay.
Subscribe to:
Post Comments (Atom)
Ok, so for the differentiability, which you might understand better after today, if you have an absolute value, a piecewise function, or a cusp, then it's not differentiable.
ReplyDeleteand to find equation of the horizontal tangent, you find the slope (by taking the derivative of the given function) and if they give you a point, just plug it into the slope intercept formula. So if you have the point (3,2) and f(x)=2x+1, your equation would be y-2=2(x-3).
Hoped that helped somewhat?!
if you want to find the equation for a horizontal tangent find the slope, and if you have a point plug it into the slope intercept formula and there you go.
ReplyDeletea horizontal tangent is just a line perpendicular to the function at a certain point. first you find the slope that you want it to be paralel to, then you set it equal to the derivative and solve for x, then you plug that into the original function to find your y. then you plug into point slope to find the equation of the tangent line.
ReplyDeletei'm pretty sure that's how you do it. hope this helped.