Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
f(t) = 2 cos t + sin 2t
[0, ( pi)/2]
so i know the deriv is -2(sint)+cos(2t), but i don't understand how to find a value which makes this zero
Use calculus to find the absolute maximum and minimum values of the function. (Round all answers to three decimal places.)
f(x) = x + 2cos(x)
[0, 2]
again, I know the derivative is 1+-2sinx, but i don't understand how to get a value for x which makes the equation equal to zero.
but i ammm sooooo confused
the first problem, i don't even understand a little
why is the deriv of sin(2t)) = 2cos2t? the deriv of sinx=cosx. right?
and the second problem, i understand that i can only use pi/6 or 5pi/6
and i know that f" =-2cosx
do i just plug in these values?
I am extremely lost
In the unit circle, is positive in the first and second quadrants, that means that two solutions are
(Quadrant 1)
and
(Quadrant 2).
But are these the only solutions? Of course not. It depends on how many times you go around the unit circle. So adding an integer multiple of specifies that you have gone around the unit circle a certain number of times.