Given f (x) = LNX, if f '(x0) = 2, then x0

f'(x)=1/x=2
x=1/2
That is, x0 = 1 / 2

It is known that f (x) = LX LNX has a zero point x0 in the interval (1,2). If we use dichotomy to find the approximate value of x0 (accuracy 0.1), we need to divide the interval equally ()
A. 3B. 4C. 5D. 6

Every time the interval is divided equally, the length of the interval becomes half of the original length, and the length of the original interval is 1, so that 1 × (12) n ≤ 0.1, the minimum natural number n is 4, so B

The function f (x) = LNX - Px + 1 proves: (2ln2 / 2 ^ 2) + (2ln3 / 3 ^ 2) + +2lnn/n^2
=2

Let P = 1
f(x)=lnx-x+1,x>=1
f'(x)=(1-x)/x1
Then f (x) decreases monotonically on x > 1, and f (x) can be continuous on x = 1
f(x)1,lnx-x+11
Lnx1
Let's take n & # 178; (> 1) to replace the above formula X
lnn²

Let f (x) = x ^ 2-ax, G (x) = LNX. Let H (x) = f (x) + G (x) have two extreme points x1, X2, and 0

From the meaning of the title, we know that h (x) = x ^ 2-ax + LNX
So h '(x) = 2x-a + 1 / X=
There are two extreme points x1, x2,
Therefore, if △ > 0, then a ^ 2 > 8 and 2x1 - A + 1 / X1 = 0, 2x2 - A + 1 / x2 = 0
The solution is X1 = √ [(A / 4) ^ 2-1 / 2] + A / 4. X2 = - √ [(A / 4) ^ 2-1 / 2] + A / 4
Let H (x) = H (x1) - H (x2) - 3 / 4 + LN2
Take the above X1 x2 into it to simplify, and then use a ^ 2 > 8
That's the way of thinking. Simplification can reduce a lot of things. It's not very complicated. Come on!

Given f (x) = LNX, if f '(x0) = 2, then x0