Given that f (x) = LNX + (1 / x) (x > 0), G (x) = lnx-x (x > 0), prove that when x > 0, XLN (1 + 1 / x)

Given that f (x) = LNX + (1 / x) (x > 0), G (x) = lnx-x (x > 0), prove that when x > 0, XLN (1 + 1 / x)

We only need to prove that 1 / (x + 1) g (0) = 0 when x > 0
So ln (1 + T) > t / (1 + T) 1 / x > 0, ln (1 + 1 / x) > x / 1 + X

d(x*ln(x^2+1))=_____ d(lnx)
How to solve it

Left = (LN (x ^ 2 + 1) + X * (1 / x ^ 2 + 1) * 2x) DX = (x * ln (x ^ 2 + 1) + 2x ^ 3 / (x ^ 2 + 1)
)*(1/x)dx,
Right d (LNX) = (1 / x) DX,
So we should fill in X * ln (x ^ 2 + 1) + 2x ^ 3 / (x ^ 2 + 1)

How to find the derivative of x ^ x (LNX + 1), that is, the second derivative of x ^ x

f’=（lnx+1）^2*x^[x(lnx+1）]
f‘’=x^[x(lnx+1）-1]*2（lnx+1）+（lnx+1）^4*x^[x(lnx+1）]

Finding the derivative of (1 / x ^ 2) 2lnx

[(1/x^2) *2lnx] '
=(1/x^2)' *2lnx + (1/x^2) *(2lnx)'
obviously
(1/x^2)'= -2/x^3
(2lnx)'=2/x
therefore
[(1/x^2) *2lnx] '
= -4lnx /x^3 + 2/x^3

Given the function f (x) = ax LNX + B, the minimum value 0 is obtained at x = 1,
(1) Find the value of real number Q, B;
(2) For any real number x > 1, the inequality K

Answer: (1) f (x) = ax LNX + B, x > 0 derivation: F '(x) = A-1 / X derivation again: F' '(x) = 1 / x ^ 2 > 0 so: F (x) has a minimum value which must satisfy the condition that f' (x) = A-1 / x = 0A = 1 / XX = 1, then: a = 1 / x = 1 / 1 = 1F (x) = ax LNX + B = x-lnx + BF (1) = 1-ln1 + B = 01-0 + B = 0b = - 1 so: a = 1

F (x) = LNX / (x-1 / x)

The most value of F (x (x) is the most value of F (x) = (xlnx) / (x-lnx) / (x-lnx) / (x (x-lnx) / (x (x) (x-lnx) / (x (x) (x (x) (x-lnx) / (x (x) = (xlnxx) / (x (x) / (x-lnx) / (x-lnx) / (x (x) / (x \\\\\\; (x \\\\\\\\\\\\\\\\\\\\\\\\ා178; + 1) LNX

The minimum value of the function y = | LNX + 4 | + 2 is

y = | lnx+4 | + 2
lnx ∈（-∞,+∞）
lnx+4 ∈（-∞,+∞）
| lnx+4 | ∈【0,+∞）
| lnx+4 | + 2 ∈【2,+∞）
Minimum 2

The minimum point of function y = | (LNX) + 4 | + 2 is. The answer is x = e ^ (- 4)

LNX + 4 = 0 gives x = e ^ (- 4),
According to (- ∞, e ^ (- 4)), (e ^ (- 4), + ∞), the absolute value is removed
Get the left end minus function, right end increasing function, so change the point to the minimum point

Let f (x) = lnx-p (x-1), P ∈ R. (1) when p = 1, find the monotone interval of F (x); (2) Let G (x) = XF (x) + P (2x2-x-1), for any x ≥ 1, G (x) ≤ 0 holds, find the value range of P

(1) When p = 1, f (x) = ln & nbsp; X - (x-1), f ′ (x) = 1 X-1, Let f ′ (x) > 0, X ∈ (0, 1), so the monotone increasing interval of F (x) is (0, 1), and the monotone decreasing interval of F (x) is (1, + ∞); Let f ′ (x) < 0, then x ∈ (1, + ∞), so the monotone decreasing interval of F (x) is (1, + ∞)

Monotone interval of FX = LNX / X

The domain of F (x) is x > 0
f‘(x)=(1/x*x-lnx*1)/x^2
=(1-lnx)/x^2
Let f '(x) > = 0
1-lnx>=0
lnx