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
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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.
