Find the variation range of the slope of the tangent of the function y = sin (π / 3-2x)

Find the variation range of the slope of the tangent of the function y = sin (π / 3-2x)

y=sin(π/3-2x)
∴ y'=cos(π/3-2x)*(-2)∈[-2,2]
That is, the variation range of the slope of the tangent of the function y = sin (π / 3-2x) is [- 2,2]

Function f (x) = 2x-b / (x-1) ^ 2. It is known that the tangent slope of the image of this function at x = 2 is 2 (1) Find the analytical formula of function f (x) (2) Set a

(1) F (x) = (2x-b) / (x-1) ^ 2, f '(x) = (2b-2-2x) / (x-1) ^ 3 the tangent slope of the image of the function at x = 2 is 2, f' (2) = (2b-2-4) / (2-1) ^ 3 = 2, B = 4f (x) = (2X-4) / (x-1) ^ 2, (2) x belongs to [2,4], f '(x) = (6-2x) / (x-1) ^ 3 = 0, x = 3. It is easy to judge the sign of the derivative function on both sides, x

Let the function FX = x + ax ^ 2 + blnx, the curve y = FX passes through P (1.0), and the tangent slope at point P is 2, which proves that FX ≤ 2x-2

fx=x+ax^2+blnx
Bring in x = 1 and y = 0
A = - 1 if 1 + a = 0
Derivation
F '(x) = 1 + 2aX + B / X brings in x = 1
1+2a+b=2
So B = 3
f(x)=x-x ²+ 3lnx
Let g (x) = X-X ²+ 3lnx-2x+2
=-x ²- x+3lnx+2
Derivation
g'(x)=-2x-1+3/x
=(-2x ²- x+3)/x
=-(2x+3)(x-1)/x=0
Get x = 1
G (x) gets the maximum value at x = 1
g(1)=0
So g (1) ≤ 0
That is, f (x) ≤ 2x-2

Let the function f (x) = x + AX2 + blnx, the curve y = f (x) passes through P (1,0), and the tangent rate at point P is 2 (I) find the values of a and B; (II) proof: F (x) ≤ 2x-2

（Ⅰ）f'（x）=1+2ax+b
x，
From known conditions:
f(1)＝0
f/(1)＝2 ， Namely
1+a＝0
1+2a+b＝2
Solution: a = - 1, B = 3
(II) the domain of F (x) is (0, + ∞). From (I), f (x) = x-x2 + 3lnx,
Let g (x) = f (x) - (2x-2) = 2-x-x2 + 3lnx, then
g/(x)＝−1−2x+3
x=−(x−1)(2x+3)
x
At that time, 0 < x < 1, G '(x) > 0; When x > 1, G '(x) < 0
Therefore, it increases monotonically on (0,1) and decreases monotonically on (1, + ∞)
‡ g (x) gets the maximum value at x = 1, G (1) = 0
That is, when x > 0, the function g (x) ≤ 0
‡ f (x) ≤ 2x-2 is constant on (0, + ∞)

The tangent slope of the image with function y = (2x-1) ^ 3 at point (O, - 1) is

The derivative of Y is equal to 6 (2x-1) ^ 2
Replace 0 with 6

If the function f (x) = the x power of E * SiNx, the inclination angle of the tangent of the function at point (4, f (4)) is

f(x)=e^x*sinx
f'(x)=e^x*sinx+e^x*cosx=e^x*(sinx+cosx)
So k = f '(4) = e ^ 4 * (SiN4 + Cos4) = e ^ 4 * √ 2Sin (4 + π / 4) ＜ 0 (because 4 + π / 4 is the fourth quadrant angle)
K = Tan θ
therefore θ It must be an obtuse angle