Y = derivative of sin (3x - π / 6)

Y = derivative of sin (3x - π / 6)

Y '= cos (3x - π / 6) * 3 = 3cos (3x - π / 6) can be obtained by deriving the composite function
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High school mathematics - understanding of determining the period of function: About f (x + T) = 1 / F (x), f (x + T) = - f (x) How to apply the relationship of these functions in the topic? What are the characteristics? Better ~)

okay
This is simple
In fact, there are rules
My summary is
Cycle = |2t|
This is a rule
Generally, I don't do much
If you meet, just use this directly

The minimum positive period of high school mathematical function f (x) = (1 + √ 3tanx) cosx

f(x)=(1+（√3）tanx)cosx
=cosx+√3tanxcosx
=cosx+√3sinx
=2(cosxcos60°+sinxsin60°）
=2cos(x-π/3)
Because the period of cosx is 2 π, the minimum positive period of the function is 2 π

Given the function f (x) = ax-1-lnx (a ∈ R). (I) discuss the number of extreme points of function f (x) in the definition domain; (II) if the function f (x (II) if the function f (x) obtains the extreme value at x = 1 and is constant for ∀ x ∈ (0, + ∞), f (x) ≥ bx-2, find the value range of real number B; (III) when E-1 < y < x, try to prove that e ^ (X-Y) > {ln (x + 1)} / {ln (y + 1)}

The domain of function definition is x > 0, the derivative of function f (x) is f '(x) = A-1 / x, and the extreme point is f' (x) = 0 = A-1 / x, that is, x = 1 / a (1). Discussion: when a ≤ 0 and f '(x) 0, f (x) obtains the extreme value at x = 1 / A, that is, the number of extreme points is 1 (2) the function obtains the extreme value at x = 1, then a = 1, f (x) = x-1-lnxf (x) ≥ bx-2 is constant

Given that the function f (x) = x (LNX ax) has two extreme points, the value range of real number a is? My idea is to take the derivative to get lnx-2ax + 1 = 0,

According to the relationship between the extreme point and the derivative function, that is, the derivative function of this function passes through the x-axis twice in the definition domain
After the derivation of the original function, f '(x) = lnx-2ax + 1 means that if the derivative function = 0, the structural equation lnx-2ax + 1 = 0 has two different solutions
In addition, G (x) = lnx-2ax + 1 g '(x) = 1 / x-2a, let g' (x) = 0, get x = 1 / 2a, and the definition domain is x ∈ (0, positive infinity)
1. When a is less than or 0, it is obvious that G '(x) is greater than 0. At this time, G (x) = lnx-2ax + 1 monotonically increases. It is impossible to cross the X axis twice, which is not true!
2. When a is greater than 0, G (x) increases at (0,1 / 2a) and decreases at (1 / 2a, positive infinity). When x approaches 0 and x approaches positive infinity, G (x) approaches negative infinity. Therefore, to make g (x) have two different solutions, only G (1 / 2a) is greater than 0, that is, ln (1 / 2a) > 0
Combined with the above a is greater than 0, it can be solved that a belongs to (0,1 / 2)

F (x) = LNX + x ^ 2 + ax, 9 (1) if x = 1 / 2, f (x) obtains the extreme value, then the value of a is (2) if f (x) is an increasing function in its definition field, find the value of A

(1) First find the derivative f '(x) = 1 / x + 2x + A, replace x = 1 / 2, make f' (x) = 0, and find a = - 3
(2) First find the derivative f '(x) = 1 / x + 2x + A, whose definition field is x greater than 0, and then make f' (x) greater than 0 to get a greater than - (1 + 2x) ²)/ x. Let H (x) = - (1 + 2x) ²)/ x. When the maximum value of H (x) is x = root 2 / 2, 2 root 2 is obtained. Therefore, the value of a is greater than 2 root 2