What is max and min value of A

Question

A= sin²x + cos⁴x

Answer 1

It may be worthwhile in general to confirm via the second derivative test as such:

 

Re-writing dA/dx as -sin(2x)cos(2x)= -sin(4x)/2, we can easily compute the second derivative: -2cos(4x)

 

At x=0, the second derivative is <0, and thus this is indeed a maximum-generating point

At x= pi/4, the second derivative is >0 and thus this is indeed a minimum-generating point

 

(no boundary value conditions stated)

Answer 2

A=sin²x + (1-sin²x)²

A= sin²x + 1 -2sin²x + sin⁴x

A = sin⁴x - sin²x + 1

 

Now to find the max and min vale of A we will find dA/dx and equal it to 0

 

dA / dx = 4sin³xcosx - 2sinxcosx = 0

 

2sinxcox (2sin²x - 1) = 0

Sin2x (2sin²x-1 ) = 0

 

So Either Sin2x=0 or 2sin²x-1 = 0

 

if sin2x=0, we have x=0,90,180 degree and so on

if 2sin²x-1 = 0, we have x=45, 135, 225 degree and so on

 

So  when x = 0 degree , A = 1

when x = 45 degree, A = 3/4

 

Max and min value of A is 1 and 3/4 respectively