What is min & max value of 2 sin^2x + 3 cos^2x?

Question

What is min & max value of 2 sin^2x + 3 cos^2x ?

Answer 1

sin²x = 1 - cos²x



=> 2(1 - cos²x) + 3cos²x



i.e. 2 + cos²x.......(1)



cos2x = cos²x - sin²x



=> cos2x = 2cos²x - 1



so, cos²x = (cos2x + 1)/2



From (1) => 5/2 + (1/2)cos2x



Now, cos2x minimum is -1 and cos2x maximum is 1



=> max 5/2 + 1/2 = 3 and min 5/2 - 1/2 = 2



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Answer 2

2 sin^2x + 3 cos^2x = 2 + cos^2x

cos^2x has a minimum of 0 and a maximum of 1, so the answer is minimum: 2, maximum: 3

Answer 3

d/dx [2 sin^2x + 3 cos^2x] = -2*(sin x)*cos(x) = -sin(2x)



Zeros of the derivative are every pi/2.



If x = 0 we have f (0) = 3, for x = pi/2 we have f(pi/2) = 2



for x = pi we have f(pi) = 3, for x = 3pi/2 we have f(3pi/2) = 2



It would appear that 2 <= f(x) <= 3