What positive number added to its reciprocal gives the minimum sum?

11 Answers

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  • 1 month ago

    1 added to its reciprocal gives the minimum sum.

  • 1 month ago

    if i want to go to space then is this question necessary?

  • Amy
    Lv 7
    1 month ago

    y = x + 1/x

    dy/dx = 1 - 1/x^2

    0 = 1 - 1/x^2

    1/x^2 = 1

    x^2 = 1

    x = ±1

    x = 1 gives the minimum sum over positive x

    (and x = -1 gives the maximum sum over negative x)

  • 1 month ago

    Let y=x+1/x

    y'=1-1/x^2

    y"=2/x^3

    y'=0=>

    x=+/-1

    y"(1)=2>0

    =>

    x=1 is the number required.

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  • 1 month ago

    one                                                    .

  • Philip
    Lv 6
    1 month ago

    Put required positive number = x;

    Adding gives f(x) = x+1/x = x + x^(-1);

    We seek x such that f(x) is a minimum;

    f'(x) = 1-x^(-2);

    f''(x) = 3x^(-3), > 0 for all x > 0, ---> any extremum reached at any x > 0  will be a rel

    min.;

    f'(x) = [x^(-2)](x^2-1) = [x^(-2)](x-1)(x+1);

    f'(x) = 0 for x = 1;

    f(1) = 2;

    Then f(x) reaches a rel min of 2 at x = 1  

  • Ian H
    Lv 7
    1 month ago

    S(x) = x + 1/x The curve indicates the minimum for positive x at x = 1

    https://www.wolframalpha.com/input/?i=S%28x%29+%3D...

    dS/dx = 1 – 1/x^2 = 0 for turning points

    Selecting positive, min at (1, 2)

  • 1 month ago

    The smallest candidate is n=1, since 0 has no reciprocal.  The sum is 1 + 1/1 = 2.

    Every other value of n is greater than or equal to that sum *before* adding 1/n, and every reciprocal is greater than 0, so just add inequalities:

        n >= 2               .

       1/n > 0

       n + 1/n > 2

    So no whole number n greater than 1 can have an n+1/n sum less than 1+1/1.

  • Jeremy
    Lv 6
    1 month ago

    n = unknown number

    S(n) = n + 1/n <=== "S(n) means sum as a function of n."

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    First derivative: S'(n) = (S(n))' = (n + 1/n)' = 1 - 1/n^2

    We must solve the equation S'(n) = 0, in order to find the stationary (or turning) points.

    S'(n) = 0 ===> 1 - 1/n^2 = 0 ===> (n^2 - 1)/n^2 = 0

    It has to be: n^2 - 1 = 0 ===> n^2 = 1 ===> n = ±√1 = ±1

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    Second derivative test

    S''(n) = (S'(n))' = (1 - 1/n^2)' = ( 0 - (-2n/n^4) ) = 2n/n^4 = 2/n^3

    If n = -1 ==> S''(-1) = 2/(-1)^3 = -2 ==> S''(-1) < 0 (MAXIMUM POINT)

    If n = 1 ==> S''(1) = 2/1^3 = 2 ==> S''(1) > 0 (MINIMUM POINT)

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    If n = 1, the sum of n to its reciprocal is minimum and equal to:

    S(1) = 1 + 1/1 = 1 + 1 = 2

  • 1 month ago

    You start with the lowest positive number because any number bigger than the lowest positive number will give a bigger number. 

    If x < y and x,y > 0 then (x + 1/x) < (y + 1/y).  So it’s either 1 or 2 depending on whether or not 1/1 is a reciprocal. 

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