If the point P is on the curve y = ln (x-1) and the point q is on the curve y = e ^ x + 1, then the minimum value of | PQ |

If the point P is on the curve y = ln (x-1) and the point q is on the curve y = e ^ x + 1, then the minimum value of | PQ |

Y = ln (x-1) and y = e ^ x + 1 are reciprocal functions, that is, they are symmetric with respect to y = X
When y = - x intersects them, and the tangent slope at the intersection is equal to 1, the distance between the intersections is the smallest
So y '= 1 / (x-1) = 1, y' = e ^ x = 1
The solution is: x = 2, x = 0
So when | PQ | is the smallest, P is (2,0), q is (0,2)
That is, the minimum value is √ (2 ^ 2 + 2 ^ 2) = 2 √ 2

Let point P move on the circle (x + 1) 2 + (Y-1) 2 = 1, and point Q move on the curve xy = 1, then find the minimum value of PQ
To explain in detail, I know to find the minimum value from the center of the circle to xy = 1, and then reduce the radius, but I won't find the minimum value. Just write out the process of finding the minimum value

If Q (x, 1 / x) is Q (x, 1 / x), then: MQ ^ 2 = (x + 1) ^ 2 + ((1 / x) - 1) ^ 2 = x ^ 2 + (1 / x ^ 2) + 2x - (2 / x) + 2 = (x - (1 / x)) ^ 2 + 2 (x - (1 / x)) + 4 = (x - (1 / x) + 1) ^ 2 + 3 > = 3mqmin = radical, minimum value of 3pq = MQ min - r = (radical 3) - 1