posted by Someone on .
A particle moves along the x-axis with position function s(t) = e^cos(x). How many times in the interval [0, 2π] is the velocity equal to 0?
1< My answer
More than 3
I don't really get this question will someone please explain it to me?
v(x) = -sin(x) e^cos(x)
since e^cos(x) is never zero, how many times in sin(x) = 0 in the interval?
Your answer is correct on the interval (0,2π), but not on [0,2π]
what is the difference between the two?
v(t) = 0 at 0,π,2π
If you exclude the endpoints of the interval, your answer is correct. But, we are working with the closed interval, including the endpoints. So, there are 3 points where v(t) = 0.
Better review interval notation.