Known function y = 2sin1 2X, please (1) The maximum, minimum and minimum positive period of function y; (2) Monotonically increasing interval of function y

Known function y = 2sin1 2X, please (1) The maximum, minimum and minimum positive period of function y; (2) Monotonically increasing interval of function y

(1) According to the properties of sine function, − 1 ≤ sin1
2x≤1
∴-2≤y≤2
The maximum value of the function is 2 and the minimum value is - 2,
T=2π
one
2=4π
(2) Order − 1
2π+2kπ≤1
2x≤1
2π+2kπ，k∈Z
∴4kπ-π≤x≤4kπ+π，k∈Z
The monotonically increasing interval of the function is [4K π - π, 4K π + π] (K ∈ z)

Let the function f (x) = ax - (a + 1) ln (x + 1), where a ≥ - 1, find the monotone interval of F (x)

It is known that the definition domain of function f (x) is (- 1, + ∞), and f '(x) = (AX-1) / (x + 1) (a ≥ - 1),
(1) When - 1 ≤ a ≤ 0, f '(x) < 0, the function f (x) decreases monotonically on (- 1, + ∞),
(2) When a > 0, x = 1 / A is obtained from F '(x) = 0
When x ∈ (- 1,1 / a), f '(x) < 0, the function f (x) decreases monotonically on (- 1,1 / a)
When x ∈ (1 / A, + ∞), f '(x) > 0, the function f (x) increases monotonically on (1 / A, + ∞)
in summary:
When - 1 ≤ a ≤ 0, the function f (x) decreases monotonically on (- 1, + ∞)
When a > 0, function f (x) decreases monotonically on (- 1,1 / a) and function f (x) increases monotonically on (1 / A, + ∞)

Known function f (x) = ln (2-x) + ax (1) Let the tangent of curve y = f (x) at point (1, f (1)) be L. if the straight line L is tangent to circle (x + 1) 2 + y2 = 1, find the value of a; (2) Find the monotone interval (a ∈ R) of function f (x)

(1) From the meaning of the question, f '(x) = a + 1x − 2, substituting x = 1 into f (x) to get: F (1) = a, then the tangent point coordinate is (1, a), substituting x = 1 into the derivative function to get: F' (1) = A-1, then the tangent slope k = A-1, so the tangent equation L is: Y-A = (A-1) (x-1), that is, (A-1) X-Y + 1 = 0, and circle

Let the domain of the function f (x) = ln (x2 ax + 2) be a (1) If 2 ∈ a, - 2 ∉ a, find the range of real number a; (2) If the domain of function y = f (x) is r, find the value range of real number a

(I) from the meaning of the question
4−2a+2＞0
4+2a+2≤0 ， (2 points)
So a ≤ - 3
Therefore, the range of real number a is (- ∞, - 3]. (4 points)
(II) from the meaning of the question, it is obtained that x2 + ax + 2 > 0 is constant on R,
Then △ = a2-8 ＜ 0, (6 points)
Solution - 2
2＜a＜2
2. (7 points)
Therefore, the range of real number a is (- 2)
2，2
2) (8 points)

Known function f (x) = ln (AX + 1) + (1-x) / (1 + x), x > = 0, where a > 0, (1) find the monotonic interval of F (x) (2) if the minimum value of F (x) is 1, find the value range of A

f′(x)=[a/(a+1)]-[2/(1+x) ²]
=(ax ²+ a-2)/(ax+1)(1+x) ²
∵x≥0
a>0
∴ax+1>0
① When a ≥ 2
F '(x) > 0 on interval (0, + ∞)
② When 0 √ [(2-A) / a]
By F '(x)

It is known that the function f (x) = ln (1 + ax) - x2 a > 0 belongs to the monotone interval of finding f (x) in [01]

f’(x)=-2x+（a／（1+ax））＝-（2ax＾2+2x-a）／（1+ax）
From F '(x) ≥ 0
（-1-√（1+2a＾2））／2a≤x≤（-1+√（1+2a＾2））／2a
And because x ∈ (0,1)
Therefore, the monotonic increasing interval is (0, (- 1 + √ (1 + 2A ＾ 2)) / 2A]
The monotonic decreasing interval is [(- 1 + √ (1 + 2A ＾ 2)) / 2a, 1]