1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
EVT: a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval
Rolle's: Let F be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number "c" in (a,b) such that f(c)=0.
MVT: If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f '(c)=f(b)-f(a)/b-a
Rolle's: Let F be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number "c" in (a,b) such that f(c)=0.
MVT: If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f '(c)=f(b)-f(a)/b-a
And i still have trouble with integrals
If they are asking you to take the derivative of the integral, you basically plug the top bound where x is and then take the derivative as usal. Whenever you see a +1 at the bottom, most likely the integral is tan inverse. When the top is the derivative of the bottom, then the integral is probably ln |u|. When you have an e in the problem, your u is whatever e is raised to. Anything else you most likely have to use substitution. Hope this helps!
ReplyDeleteall u do is plug the top bound in and take the normal der.
ReplyDeleteyou can even plug in integrals into your calculator to get an easy answer! :D
ReplyDeleteIntegrals are pretty easy...basically think of it as the anti derivative. Like, for instance...
ReplyDeleteThe derivative of x^3 is 3x^2, right?
So the integral of 3x^2 is what? x^3.
Once you get the mindset of it being an anti derivative, things will become way easier...at least they did for me.