Sum of the maximum and minimum values of y=2cos^2 x + 3sin^2 x + 5

To begin with, we can simplify y using the Pythagorean identity:

y = 2 cos²(x) + 3 sin²(x) + 5

y = 2 (cos²(x) + sin²(x)) + sin²(x) + 5

y = 2 + sin²(x) + 5

y = sin²(x) + 7

Next, we can further rewrite this using the half angle identity for sine:

y = (1 - cos(2x))/2 + 7

y = 15/2 - 1/2 cos(2x)

Now, since cos(x) is bounded between -1 and 1, we have

max(y) = 15/2 - 1/2×(-1) = 15/2 + 1/2 = 8

and

min(y) = 15/2 - 1/2×1 = 15/2 - 1/2 = 7

Then the sum of the maximum and minimum is 8 + 7 = 15.

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)