P point is on the curve y 1 2ex times, Q is on y ln 2x , what is the minimum value of PQ

Solution: The function y=1 2e x and the function y=ln(2x) are inverse functions of each other, and the image is symmetric with respect to y=x.

The distance from the point p(x,1 2 e x) on the function y=1 2e x to the straight line y=x is d=|1/2e^x-x|/ √2

Let g(x)=1 2 e x-x, (x 0) then g (x)=1 2 e x-1

From g (x) = 1 2 e x-1 0 we get x ln2 and g (x) = 12 e x-1 0 gives 0 x ln2

The function g(x) decreases monotonically at (0,ln2) and increases monotonically at [ln2,+.

When x=ln2, the function g(x)min=1-ln2

dmin=(1-ln2)/ √2

From the image symmetry with respect to y=x: |pq|The minimum value is 2dmin=2 (1-ln2).

Let the coordinates of point P (a, (1 2)e a) and point Q coordinates (b, ln(2b)).

pq=（a-b）^2 + 1/2)e^a-ln(2b))^2=f(a,b)

f'a(a,b)=2(a-b)+2((1/2)e^a-ln(2b))(1/2)e^a=0

f'b(a,b)= -2(a-b)-2((1/2)e^a-ln(2b))/b=0

The addition of the two equations yields ((1 2)e a-ln(2b))(1 2)e a-1 b)=0;

At the same time, a=b is obtained;

Thus pq=(1, 2)e-a-ln(2a)=g(a).

g'(a)=(1/2)e^a -1/a=0; a=。

Bring in the formula to calculate the minimum distance of pq.

y=ln(x-1) and y=e x+1 are inverse functions of each other, i.e., they are symmetrical with respect to y=x, sketched:

y=-x intersects them, and when the tangent slope at the intersection is equal to 1, the pure distance between the intersection points is smallest.

So y'=1/(x-1)=1 y'=e x=1 yields: x=2 x=0

So |pq|The minimum value of prudence is (2 2 + 2 2) and q is (0 2) and the minimum value of caution is (2 2 + 2 2) = 2 2

The friend above miscalculated - you ended up turning x 2 into x ......The floor theme is also mistyped p on y=x 2+2 so that the straight line where p,q is an inverse function of each other, about y=x symmetry, only require a point to the shortest distance of y=x and multiply it by 2 to set q(x, (x-2)) y=x d=|x-√(x-2)|2, and then the numerator is written by the friend above in the exact flat way, ( x-2)-1 2) 2+7 4 2+7 4 2=7* 2 8 2d=7* 2 4 is what is sought.

[Note: A conclusion:.]

Let p, q be two moving points on two disjoint curves.

When the normals of two curves that cross these two points coincide, |pq|Least. Solution: p(a, (e a) 2,) q(b, ln(2b)) is easy to know

The normal equation for the curve y=(e x) 2 at the point p is: y=[-2 (e a)]x+[2a (e a)]+e a) 2].

The normal equation for the curve y=ln(2x) at the point q is: y=-cx+c +ln(2c)

From the above conclusions, the comparison can be obtained:

c=2/(e^a)

c +ln(2c)=[2a (e a)]+e a) 2] solves: c=1, a=ln2

p(ln2, 1), q(1,ln2)∴|pq|min=√[(1-ln2)²+1-ln2)²]=(1-ln2)√2.

The two curves are inverse functions of each other That is to say, with respect to y=x symmetry, make the tangent of the two curves and parallel to y=x, and the distance between the two tangents is the minimum.

Make the tangents of the two curves and parallel to y=x, and the distance between the two tangents is the minimum value. This statement is wrong.

The derivatives are y=1 2e x y=1 x

When equal, the slope is not equal to 1

Should be used |pq|Find |when it is perpendicular to y=xpq|minimum value.

Let the coordinates of point P (a,(1 2)e a) and point coordinate of point Q (b,ln(2b))Pq=(a-b) 2

1/2)e^a-ln(2b))^2=f(a,b)f'a(a,b)=2(a-b)+2((1/2)e^a-ln(2b))(1/2)e^a=0

f'b(a,b)=

2(a-b)-2((1 2)e a-ln(2b)) b=0 is added to obtain ((1 2)e a-ln(2b))(1 2)e a-1 b)=0;

At the same time, a=b is obtained;

Thus pq=(1, 2)e, a-ln(2a)=g(a)g'(a)=(1/2)e^a

1/a=0;

a=。Bring in the formula to calculate the minimum distance of pq.