Find a number in the interval (0,2) such that the sum of the number and its reciprocal is the absolute min.?

Answer 1

min y = x + 1/x
dy/dx = 1 - 1/x^2 = 0
1/x^2 = 1
x^2 = 1
in (0,2), x = 1 is the number where y = x + 1/x is minimum

Ans x=1

Answer 2

We wish to find a minimum of f(x) = x + 1/x in the interval (0, 2). This will occur at a point where f'(x) = 0, So differentiate and solve:

f'(x) = 1 - 1/x² = 0
x² = 1
x = ±1

But of course x must be positive to be in (0, 1), so x=1. This is a minimum, so we are done.

Answer 3

number 1 and its reciprocal 1/1 sum to 2, a minimum.