The minimum value of function y = x + 2x + 2 / x + 1 (x ＞ - 1)

The minimum value of function y = x + 2x + 2 / x + 1 (x ＞ - 1)

y=x^2+2x+2/x+1
=(x^2+1)+(2x+2/x)
>=(x ^ 2 + 1) + 2 * root (2x * 2 / x)
=x^2+1+4=x^2+5
If and only if 2x = 2 / x, take equality, that is, x ^ 2 = 4
The minimum value is
y=x^2+5=9

Finding the minimum value of function f (x) = x-2x
Any X1 < x2
F (x1) - f (x2) = (x1-2x) - (x2-2x)
=
=

=(x1²-x2²）-2（x1-x2）
=（x1+x2-2）（x1-x2）
When X1 + x2-2 > 0, it increases
x1+x2-2

Let X be greater than - 1, find the minimum value of (2x ^ 2 + 2x + 1) of the function f (x) = (x + 1). (x + 1) is the denominator!
It is about solving inequality, that is, a ^ 2 + B ^ 2 is greater than or equal to 2Ab, or a + B is greater than or equal to 2 times the root sign ab

F (x) = (2x ^ 2 + 2x + 1) / (x + 1) = (2x ^ 2 + 4x + 2 - 2x-2 + 1) / (x + 1) = [2 (x + 1) ^ 2 + 1 - 2 (x + 1)] / (x + 1) = 2 (x + 1) + 1 / (x + 1) - 2 because x > - 1, x + 1 > 0 from the basic inequality, we get 2 (x + 1) + 1 / (x + 1) > = 2 radical sign 2 under 2 (x + 1) * 1 / (x + 1) = 2 radical sign 2, so f (x) > = 2 radical sign 2 - 2 is the minimum value

Given that the definition field of the function f (x) = 2asin (2x Pie / 3) + B is [0, Pie / 2], the maximum value of the function is one and the minimum value is 5, find a and B

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