Let f (x) = a (x + 1 / x) + 2lnx, G (x) = x ^ 2. If a > 0 and a is not equal to 2, the line L is tangent to the image of function f (x) and function g (x), and the equation of tangent L is obtained? It's really a talent. At that time, I was just like you, but I didn't expect to propose a 2x

The line L is tangent to the image of function f (x) and function g (x),
∴f'(x)=g'(x),
That is, a (1-1 / x ^ 2) + 2 / x = 2x,
a(x^2-1)+2x(1-x^2)=0,x>0,
X = 1 or x = A / 2
g(1)=1,g(a/2)=a^2/4,g'(1)=1,g'(a/2)=a,
a> 0 and a is not equal to 2,
The equation of L is y = x or Y-A ^ 2 / 4 = a (x-a / 2),
The latter is 4ax-4y-a ^ 2 = 0, where a satisfies f (A / 2) = g (A / 2), that is, a ^ 2 + 8 + 8ln (A / 2) = 0

Let f (x) = P (x-1x) - 2lnx, G (x) = X2, (I) if the line L is tangent to the image of function f (x), G (x) and tangent to the image of function f (x) at point (1, 0), find the value of real number p; (II) if f (x) is a monotone function in its domain, find the value range of real number P

(I) method 1: ∵ f '(x) = P + PX2 − 2x, ∵ f' (1) = 2p-2. Let a straight line and G (x) = x2 be tangent to the point m (x0, Y0) ∵ G '(x) = 2x, ∵ 2x0 = 2p-2, then the solution is ∵ x0 = P − 1, Y0 = (P − 1) 2, and the solution is p = 1 or P = 3 by substituting the straight line equation l into y = X2 to get 2 (p-1) (x-1) = 0, ∵ △ 4 (p-1) 2-8 (p-1) = 0, then the solution is p = 1 or P = 1= 3. (II) ∵ f '(x) = P + PX2 − 2x = PX2 − 2x + PX2.. ① if f (x) is a monotone increasing function, f' (x) ≥ 0 is constant at (0, + ∞), that is, px2-2x + P ≥ 0 is constant at (0, + ∞), that is, P ≥ 2xx2 + 1 = 2x + 1x is constant at (0, + ∞), and 2x + 1x ≤ 1, then f (x) is a monotone increasing function at (0, + ∞) when p ≥ 1; ② In order to make f (x) a monotone decreasing function, f '(x) < 0 must be constant at (0, + ∞), that is, P ≤ 2xx2 + 1, (0, + ∞), and 2xx2 + 1 ≥ 0, so p ≤ 0. When p ≤ 0, f (x) is a monotone decreasing function at (0, + ∞). In conclusion, if f (x) is a monotone function at (0, + ∞), then the value range of P is p ≥ 1 or P ≤ 0

Let f (x) = LNX, G (x) = f (x) + F '(x). (I) find the monotone interval and minimum value of G (x); (II) discuss the size relationship between G (x) and G (LX); (III) find the value range of a, so that G (a) - G (x) < 1A holds for any x > 0

(I) Let f (x) = LNX, G (x) = LNX + 1 x, G '(x) = X-1 x 2, let g' (x) = 0 get x = 1, when x ∈ (0,1), G '(x) < 0, so (0,1) is the monotone decreasing interval of G (x). When x ∈ (1, + ∞), G' (x) > 0, so (1, + ∞) is the monotone increasing interval of G (x), so x = 1 is the unique value point of G (x), and is the minimum point Let H (x) = g (x) - G (LX) = 2lnx-x + LX, then H '(x) = - (x-1) 2x2, when x = 1, H (1) = 0, i.e. g (x) = g (LX), when x ∈ (0,1) ∪ (1, + ∞), H' (1) < 0, therefore, H (x) monotonically decreases in (0, + ∞), when 0 < x < 1, H (x) > H (1) = 0, i.e. g (x) > G (LX), when x ∈ (0,1, + ∞), H '(x) > G When x ＞ 1, H (x) ＜ H (1) = 0, that is g (x) ＜ g (1x). (III) from (I), we know that the minimum value of G (x) is 1, so g (a) - G (x) ＜ 1a, for any x ＞ 0, holds ⇔ g (a) - 1 ＜ 1a, that is, ina ＜ 1, thus 0 ＜ a ＜ E

F (x) = LNX let g (x) = f (x) + F '(x) g (x) = LNX + 1 / X ① g (1 / x) = - LNX + X ② g (1 / x) = f (1 / x) + F' (1 / x) = - lnx-1 / X ① which pair

The first is right
The second is that the wrong g (1 / x) = f '(1 / x) * (1 / x)' = - f '(1 / x) / x ^ 2 is not equal to (1 / x) + F' (1 / x)

F (x) = x ^ 3-3ax, G (x) = LNX, (1) when a = 1, find the minimum value of F (x) in the interval [- 2,2]
(2) If the image of F (x) in the interval [1,2] is always above the image of G (x), the value range of the real number a is obtained
(3) An analytic formula for finding the maximum f (a) of F (x) in the interval [- 1,1]

(1) If a = 1, the function f (x) = x ^ 3-3ax is continuous in the interval [- 2,2], so it can be derived, f '(x) = 3x ^ 2-3a = 3 (x ^ 2-1), the stationary point of F (x) is x = ± 1, when x = 1, f (x) = 2, when x = - 1, f (x) = 2, and when x = - 2, f (x) = 2, when x = 2, f (x) = 2, so f (x)

Given the function f (x) = x-lnx.1 find the monotone interval of F (x); 2 find the minimum value of F (x) in the interval [1 / 2,2]

1.f(x)=x-lnx
Derivation: F '(x) = 1-1 / x = 0, x = 1
When x ∈ (0,1), f '(x)

Let f (x) = x + 1 / (x-1). It is proved that the tangent of any point on the curve y = f (x) and the area of the triangle surrounded by the straight line x = 1 and y = x are fixed values

f'(x)=1-1/(x-1)².
Let the tangent at x = t be y = f '(T) (x-t) + F (T)
That is y = [1-1 / (t-1) & # 178;] (x-t) + T + 1 / (t-1) = x-x / (t-1) & # 178; + T / (t-1) & # 178; + 1 / (t-1)
The intersection of x = 1 and x = 1
y=1-1/(t-1)²+t/(t-1)²+1/(t-1)=1+(-1+t+t-1)/(t-1)²=1+(2t-2)/(t-1)²=1+2/(t-1),
That is, the intersection point is (1,1 + 2 / (t-1)),
The intersection of y = x and y = x: substitute y = x into x = x-x / (t-1) &# 178; + T / (t-1) &# 178; + 1 / (t-1) has x = (t-1) &# 178; [T / (t-1) &# 178; + 1 / (t-1)] = t + T-1 = 2t-1,
The intersection point is at (2t-1,2t-1),
The intersection of y = x and x = 1 is at (1,1),
So the area of the triangle s = 1 / 2 × [1 + 2 / (t-1) - 1] × [2t-1-1] | = 1 / 2 × [2 / (t-1) × 2 (t-1) | = 2 is a fixed value

(2) Let the line l be the tangent of the curve y = f (x), and prove that the area of the triangle surrounded by the line L and the line x = 1 and the line y = x is a fixed value
Let f (x) = ax + 1 / (x + b) (a, B are integers), the tangent equation of curve y = f (x) at point (2, f (2)) is y = 3, (1) find the analytic expression of F (x); the number of words is not enough, so only the most important problem is written!

F (x) '= A-1 / (x + b) ^ 2F (2)' = A-1 / (2 + b) ^ 2 = 0A, B are integers, so 1 / (2 + b) ^ 2 = 1, otherwise it is impossible to satisfy the problem, so B + 2 = + - 1, B = - 1 or B = - 3A = 1, and f (2) = 3, so 3 = 2 + 1 / (2 + b), B = - 1, so f (x) = 4x-1 / (x-1) 2

What is the derivative of e to the - x power
My question is: is the result - e ^ - X or (e ^ x) ^ - 1, and then - (e ^ x) ^ - 2?

Analysis,
(e^-x)'=-e^（-x）
f(x)=[e^x]^(-1),
Let t = e ^ X,
Then f (T) = T ^ (- 1), f '(T) = - 1 / T & # 178;
f'(x)=f'(t)*t'=-1/t²*e^x=-1/e^(x)=-e^(-x),
The derivation of compound function is one layer at a time. First, the outer layer and then the inner layer

Derivative of e to the power of π

The π power of E is a constant
So derivative = 0

Let f (x) = a (x + 1 / x) + 2lnx, G (x) = x ^ 2. If a > 0 and a is not equal to 2, the line L is tangent to the image of function f (x) and function g (x), and the equation of tangent L is obtained? It's really a talent. At that time, I was just like you, but I didn't expect to propose a 2x