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What is the smallest sum that can be obtained by adding a positive number and its reciprocal?
Wiki User
∙ 9y ago
let f(x) = x + 1/x ; To find the minimum of f(x), take the derivative and set equal zero, then solve for x.
f'(x) = 1 - 1/(x2) ; set equal to zero and solve for x --> 0 = 1 - 1/(x2) --> x = 1 & -1.
The question asked for a positive number so disregard x=-1.
So f(1) = 1 + 1 = 2. Is it a minimum or maximum, though? When the derivative = 0, the graph is changing from positive slope to negative[a maximum or relative max], or negative slope to positive [a minimum or relative min]. Let's find out.
When x < 1, we have f' = 1 - 1/(x2), where (x2) is smaller than 1, so [1/(x2)] is larger than 1, so the slope is negative (value is decreasing). When x > 1, (x2) is getting larger than 1, so [1/(x2)] is smaller than 1 and the slope is positive (increasing).
So at x =1 is a minimum [check by plugging in x = 1/2 and x = 3/2].
f(1/2) = 1/2 + 2 = 2.5, and f(3/2) = 3/2 + 2/3 = 9/6 + 4/6 = 13/6 = 2.1666 [both are greater than the sum of 2 at x=1]
So sum = 2 is a minimum.
Wiki User
∙ 9y ago
Georgian Rosales
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