Question #a8660

1 Answer

There are two maximum points
#(pi/6, 5/4)=(0.523599, 1.25)" " "# and #((5pi)/6, 5/4)=(2.61799, 1.25)#
There is one minimum point #(pi/2, 1)=(1.57, 1)" "#

Explanation:

Let the given by #y=sin x+cos^2 x#

Determine the first derivative #dy/dx# then equate to zero, that is #dy/dx=0#

Let us begin

from the given

#y=sin x+cos^2 x=sin x+ (cos x)^2#

#d/dx(y)=d/dx(sin x) + d/dx(cos x)^2#

#dy/dx=cos x * dx/dx+ 2*(cos x)^((2-1))*d/dx(cos x)#

#dy/dx=cos x * 1 +2*(cos x)^1*(-sin x)*dx/dx#

#dy/dx=cos x-2*sin x*cos x*1#

#dy/dx=cos x-2*sin x*cos x#

Equate #dy/dx=0#

#cos x-2*sin x*cos x=0#

solve by factoring

#cos x(1-2 sin x)=0#

Equate each factor to zero

#cos x=0" " "# the first factor
#arccos(cos x)=arccos 0#

#x=pi/2#

find #y#, using the original equation
#y=sin x+cos^2 x#
#y=sin (pi/2)+cos^2 (pi/2)#
#y=1+(0)^2#
#y=1#

solution #(pi/2, 1)=(1.57, 1)" "#the Minimum Point
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#1-2 sin x=0" " " "# the second factor

#2*sin x=1#

#sin x=1/2#

#arcsin(sin x)=arcsin (1/2)#

#x=pi/6# also #x=(5pi)/6#

find #y#, using #x=pi/6# in the original equation

#y=sin x+cos^2 x#
#y=sin (pi/6)+cos^2 (pi/6)#

#y=1/2+(sqrt3/2)^2#

#y=1/2+3/4#
#y=5/4#

solution #(pi/6, 5/4)=(0.523599, 1.25)" " "#the Maximum Point

the other Maximum Point is at #((5pi)/6, 5/4)=(2.61799, 1.25)#

because #sin (pi/6)=sin ((5pi)/6)#. That is why there are two maximum points.
Kindly see the graph and locate the critical points
graph{y=sin x+(cos x)^2 [-1, 5, -1, 1.5]}

God bless....I hope the explanation is useful.