Function y = 1 / 2 (SiNx + cosx) - 1 / 2 [SiNx cosx], find its value range. [] represents the absolute value

Function y = 1 / 2 (SiNx + cosx) - 1 / 2 [SiNx cosx], find its value range. [] represents the absolute value

SiNx cosx = √ 2Sin (x - π / 4) makes sin (x - π / 4) ≥ 0. The solution is: 2K π + π / 4 ≤ x ≤ (2k + 1) π + π / 4, so when 2K π + π / 4 ≤ x ≤ (2k + 1) π + π / 4, SiNx cosx ≥ 0, when (2k + 1) π + π / 4 ≤ x ≤ (2k + 2) π + π / 4, SiNx cosx ≤ 0, so when 2K π + π / 4 ≤ x ≤ (2k + 1) π + π / 4

When the absolute value of X is ≤ π / 4, find the value range of function f (x) = square of cosx + SiNx

f(x)=cos ² x+sinx=1-sin ² x+sinx=-sin ² x+sinx+1=-(sinx-1/2) ²+ 5/4
|x|≤π/4
-π/4≤x≤π/4
-√2/2≤sinx≤√2/2
When SiNx = 1 / 2, f (x) has the maximum value, f (x) max = 5 / 4
When SiNx = - √ 2 / 2, f (x) has a minimum value, f (x) min = (1 - √ 2) / 2
The value range of the function is [(1 - √ 2) / 2,5 / 4]

Known function f (x) = x ^ 2 + ax INX (1) If the function FX is a subtractive function on [1,2], find the value range of A (2) Let g (x) = f (x) - x ^ 2 whether there is a real number A. when x ∈ [0. E], the minimum value of function GX is 3. If it exists, find the value of A. if it does not exist, explain the reason (3) When x ∈ (0,] indicates E ² x ²－ （5／2）x＞（x+1）㏑x

1、f'(x)=2x+a-1/x

Known function f (x) = 1 2(x−1)2+㏑x−ax+a． (1) If a = 3 2. Find the extreme value of function f (x); (2) If f (x) > 0 holds for any x ∈ (1,3), find the value range of A

(1) The definition domain of function f (x) is (0, + ∞)
f′（x）=x−1+1
x−a，
When a = 3
2, f '(x) = x + 1
x−5
2=2x2−5x+2
2x，
Let f '(x) = 0, and the solution is x = 1
2 or 2. List:
x (0，1
2) one
two (1
2，2) two （2，+∞）
f′（x） + 0 - 0 +
f（x） Monotonic increasing maximum value Monotonic decreasing Minimum Equal monotonic increasing
Function f (x) at x = 1
Get the maximum value f (1) at 2 places
2)＝−1
8−ln2，
The function f (x) obtains the minimum value f (2) = ln2-1 at x = 2
2；
（II）f′(x)＝x+1
X − (1 + a), when x ∈ (1,3), (x + 1)
x)∈(2，10
3)，
(i) When 1 + a ≤ 2, i.e. a ≤ 1, X ∈ (1,3), f '(x) > 0, the function f (x) is an increasing function in (1,3),
∀ x ∈ (1,3), f (x) > F (1) = 0 is constant;                      
(II) when 1 + a ≥ 10
3, i.e. a ≥ 7
3, when x ∈ (1,3), f '(x) ＜ 0, and the function f (x) is a subtractive function in (1,3),
∀ x ∈ (1,3), f (x) ＜ f (1) = 0 is always true, which does not agree with the question and should be omitted;
(III) when 2 < 1 + a < 10
3, i.e. 1 < a < 7
3, when x ∈ (1,3), f '(x) takes negative first, then 0, and finally positive. The function f (x) first decreases and then increases in (1,3), while f (1) = 0, ∀ x ∈ (1,3), f (x) > F (1) = 0 cannot be constant;
To sum up, the value range of a is (- ∞, 1)

Known function f (x) = ax + INX Let g (x) = x ^ 2-2x + 2, if there is x2 ∈ [0,1] for any x ∈ (0, + infinity), so that f (x) < g (x2) finds the range of A

First find the minimum value of G (x),
For any f (x)

Given the function f (x) = ax INX, a ∈ R, when a = 2, find the tangent equation of curve f [x] at point [1, f [1] If f [x] has an extreme value at x = 1, find the monotonically increasing interval of F [x]

When a = 2
f(x)=2x-Inx
f'(x)=2-1/x
So f (1) = 2-ln1 = 2
f'(1)=2-1/1=1
The tangent slope of curve f (x) at point (1, f (1)) is f '(1)
Therefore, the tangent equation is y-f (1) = f '(1) (x-1), that is, X-Y + 1 = 0