Given that the function y = (1 + LNX) / X is tangent to y = KX, find the value of real number K

Given that the function y = (1 + LNX) / X is tangent to y = KX, find the value of real number K

Let a, B be the tangent point
Then at the tangent point, y '= [(1 + LNX) / x]' = - LNX / X & # 178; = k, i.e., Ka & # 178; + LNA = 0
And B = Ka, B = (1 + LNA) / A
Three equations are solved simultaneously: a = 1 / √ e, B = √ E / 2, k = E / 2
So the linear equation is y = ex / 2, and the tangent coordinates are (1 / √ e, √ E / 2)
The value of the real number k is e / 2
derivatives,,,
Finding the maximum and minimum of the function y = 3-4sinx-cos & # 178; X
∵ cosx ^ 2 = 1-sinx ^ 2 ∵ y = 3-4sinx-cosx ^ 2 = 3-4sinx - (1-sinx ^ 2) = SiNx ^ 2-4sinx + 2 = (sinx-2) ^ 2-2 and - 1 ≤ SiNx ≤ 1 ∵ - 3 ≤ sinx-2 ≤ - 1 ∵ 1 ≤ (sin-2) ^ 2 ≤ 9 ∵ when SiNx = 1, y has the minimum value and y = - 1; when SiNx = - 1, y has the maximum value and y = 7
If the line y = KX is the tangent of y = LNX, then the value of K is equal to?
This is the derivative. The derivative of y = LNX is y '= 1 / x, so k = 1 / X
unsolvable
1/e
The minimum value of the function y = cos2x SiNx is,
t=sinx,|t|
The number of solutions of the equation lnx-2tx = 0 about X is discussed
Let f (x) = lnx-2tx, then f '(x) = 1 / x-2t (x > 0)
When t0 (x
y=cos2x+sinx(-π/6
y=cos2x+sinx
》y=1-2sin^2(x)+sinx
》=-2（sin^2(x)-1/2sinx+1/16-1/16)+1
》=-2(sinx-1/4)^2+9/8
Because - π / 6
The maximum value is 3 / 2 and the minimum value is - 1 / 2?
Analysis: Let f (x) = lnx-kx-1, the problem of the equation KX + 1 = LNX having solution is transformed into the problem of the function f (x) having zero point, then the monotonicity and extremum of the function f (x) are studied by using the derivative, and the range of K that makes the function have zero point is found
Let f (x) = lnx-kx-1
Then f ′ (x) = 1 - KX / X,
（x＞0）
If K ≤ 0, then f ′ (x) > 0, f (x) is an increasing function on (0, + ∞). When ∵ x → 0, f (x) → - ∞, f (x) has and has only one zero point, that is, the equation KX + 1 = LNX has a solution
According to the above ideas, why there must be a solution >, zero point is not 1 / K, K is less than 0. And step by step, then when x is greater than 0, there is no solution, do not draw
This is to judge whether f (x) = 0 has a solution by F '(x)
Yes, the solution of F '(x) = 0 is x = 1 / K,
When K0
In other words, in the domain of definition, f '(x) = (1-kx) / x > 0,
If the derivative is greater than 0, the original function f (x) increases monotonically
And f (x) increases monotonically from negative infinity to positive infinity, so there must be and only one solution
It is known that x belongs to (3 factions / 4,3 factions / 2). If the maximum value of function f (x) = cos2x SiNx + B + 1 is 9 / 8, try to find its minimum value
The minimum value is 0
If y = KX and y = LNX have a common point, then the value range of K is
kx=lnx
k=(1/x)lnx
Let f (x) = (LNX) / X
f'(x)=(1-lnx)/x^2
F '(x) is positive on (0. E) and negative on (E, + ∞)
We can get the value range of F (x) is (- ∞, 1 / E], which is the value range of K
This last line is not very clear
The results show that k = (LNX) / x, x > 0
The value of K is required to make the above equation solvable
So the value of K should be in the range of F (x)
The range of F (x) is (- ∞, 1 / E], so this is also the range of K
Given that x ∈ [π / 2,3 π / 2], if the maximum value of function f (x) = cos2x SiNx + B + 1 is 9 / 8 (1), find the value of B; (2) find the minimum value of F (x), then x takes the value
First of all, from the double angle formula cos2x = 1-2 * (SiNx) ^ 2, we can get f (x) = - 2 (SiNx) ^ 2-sinx + B + 2. From the range of X, we can get that the range of SiNx is [- 1,1] 1). From the form of F (x) and the image of SiNx, we can see that when SiNx gets the maximum, f (x) gets the minimum, and when SiNx gets the minimum, f (x) is the maximum, and SiNx is the minimum
Firstly, from the formula cos2x = 1-2 * (SiNx) ^ 2, f (x) = - 2 (SiNx) ^ 2-sinx + B + 2 is obtained. From the range of X, the range of SiNx is [- 1, 1]