Known function. (1) If the tangents of curves y = f (x) and y=g(x) are parallel to each other at x = 1, find the distance between two parallel straight lines; (2) If f(x)

Test and analysis: (1) Because the tangents of curves Y = f (x) and y=g(x) are parallel to each other at x= 1, the derivatives of these two functions can be obtained respectively, and the slopes of the derivative functions at x = 1 are the same, that is, the values and two tangent equations can be obtained. Then we can get two tangent equations according to the distance formula between parallel lines.

(2) F (x) ≤ G (x)- 1 for any x > 0；; 0 is a constant, so a new function is constructed in x >; 0, find the range where the maximum value of the function meets the conditions.

(3) According to the conclusion (2), Dangdang

Problem analysis: (1), according to the meaning of the question: = "2;

The tangent of the curve y=f(x) at x= 1 is 2x-y-2 = 0,

The tangent equation of curve y=g(x) at x= 1 is 2x-y- 1 = 0. The distance between two straight lines is

(2) let H (x) = F (x)-G (x)+ 1, then

When ≤0, x>0, so

When > 0,

When, when,

So h(x) is a increasing function in theory and a decreasing function in theory.

∴h(x)≤

Because H (1) = 0, and ≠ 1 when ≠2 does not match. So = 2.

(3) When

Might as well set it to 0.

∴|h(x 1)-h(x ^ 2)|≥| x ^ 1-x ^ 2 |

Equivalent to h (x1)-h (x2) ≥ x2-x1,that is, h (x1)+x/≥ h (x2)+x2, so that h (x) = h (x)+x = lnx x.

∵ ? (x>0), ∴-2x2+x+≤ 0 in x > 0 is a constant, ∴≤ (2x2-x) min, x>0, (2x2-x) min =

∴ A ≤-, and A