The monotone increasing interval of function y = x-ln (5 + x ^ 2) is_____ . The answer is (- ∞, + ∞) How to calculate? If you take the derivative of Y, you should get - 2x / (5 + x ^ 2), so that the derivative is greater than 0, and you get X

The monotone increasing interval of function y = x-ln (5 + x ^ 2) is_____ . The answer is (- ∞, + ∞) How to calculate? If you take the derivative of Y, you should get - 2x / (5 + x ^ 2), so that the derivative is greater than 0, and you get X

y' = 1 - 2x/(5 + x²) = (x² - 2x + 5)/(x² + 5)
x² + 5 > 0
x² - 2x + 5 = (x - 1)² + 4 > 0
The function is incremented in the domain (- ∞, + ∞)
You left out the first X

Given the function FX = 2cos & # 178; X + 2 √ 3sinxcosx + A, and when x ∈ [0, π / 6], the minimum value of FX is 2. (1) find a, monotonically increasing interval. (2) keep the ordinate of each point in the image unchanged, shorten the abscissa to 1 / 2 times of the original, and then shift the image to the right π / 6 units, get the function GX, find the sum of all roots of equation GX = 2 in the interval [0, π / 2]

This question is very simple, but some symbols will not be printed directly send pictures!

Given the function f (x) = (X & # 178; - 2x + a) / x, X ∈ (0,2], where constant a > 0, find the minimum value of function f (x)
If it doesn't work for me

∵X∈（0,2],
∴f（x）=x-2+a/x
∵ x + A / x > = 2 radical a
Ψ f (x) > = 2 radical A-2
That is, the minimum value is: 2 radical A-2
Sorry, I can't type root