How do you find the intervals of increasing and decreasing using the first derivative given #y=x-2cosx#?

Question

Paisley Johnson · Accepted Answer

The intervals of increasing are #(-1/6pi+2kpi, 7/6pi+2kpi)#
The intervals of decreasing are #(7/6pi+2kpi, 11/6pi+2kpi)#, #AA k in ZZ# Determine the initial derivative. #y=x-2cosx# #dy/dx=1+2sinx# The crucial moments are when #dy/dx=0# #1+2sinx=0# #sinx=-1/2# #x in (-1/6pi+2kpi) uu (7/6pi+2kpi)#, #AA k in ZZ# We build a sign chart in the interval # x in [-1/6pi, 19/6pi]# #color(white)(aaaa)##x##color(white)(aaaa)##-1/6pi##color(white)(aaaaaaa)##7/6pi##color(white)(aaaaa)##11/6pi##color(white)(aaaa)##19/6pi# #color(white)(aaaa)##dy/dx##color(white)(aaaaa)##0##color(white)(aaaa)##+##color(white)(aa)##0##color(white)(aaa)##-##color(white)(aa)##0##color(white)(aaaa)##+# #color(white)(aaaa)##y##color(white)(aaaaaaa)####color(white)(aaaa)##↗##color(white)(aa)####color(white)(aaa)##↘##color(white)(aa)####color(white)(aaaa)##↗# Consequently, The intervals of increasing are #(-1/6pi+2kpi, 7/6pi+2kpi)# The intervals of decreasing are #(7/6pi+2kpi, 11/6pi+2kpi)# #AA k in ZZ# graph{x-2cosx[-11.2, -4.82, 17.09, -14.95]}

Emilia Aguilar · Answer

undefined To find the intervals of increasing and decreasing using the first derivative $y'=x+2\sin(x)$, follow these steps:

1. Set the first derivative equal to zero to find critical points: $x + 2\sin(x) = 0$.
2. Solve for $x$ to find the critical points.
3. Test the intervals between critical points and at the endpoints of the domain.
4. Determine where the first derivative is positive or negative in each interval to identify intervals of increasing and decreasing.

Note: The domain of $y = x - 2\cos(x)$ is all real numbers.