For what positive number(s) is the sum of the number and its reciprocal the smallest? Respond to question with full detail, and possibly a graph to better illustrate solution.  

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degeneratecircle's profile pic

degeneratecircle | High School Teacher | (Level 2) Associate Educator

Posted on

You may be interested in a non-calculus solution:

Let `x` be a positive number, so that `sqrt(x)` is defined. Since

`0<=(sqrt(x)-1/sqrt(x))^2=x+1/x-2,` we have

`2<=x+1/x,` so the sum of a positive number `x` and its reciprocal is never less than 2. It is equal to `2` if and only if `sqrt(x)-1/sqrt(x)=0,` which holds if and only if `x=1.` ``

Thus, for positive `x,` `x+1/x` achieves its minimum value of `2` when `x=1.`

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lfryerda | High School Teacher | (Level 2) Educator

Posted on

To answer this question, let x be a positive real number.  Then we want to minimize the function

`f(x)=x+1/x`   rewrite as powers

`f(x)=x+x^{-1}`   now take derivative to minimize


at some point we want to test the critical point, so take the derivative again

`f''(x)=2x^{-3} = 2/x^3`

Note that since the second derivative is always positive for any `x>0` , then any critical point must be a minimum.

Rearrange the first derivative so we can set it to zero.

`f'(x)=1-1/x^2={x^2-1}/x^2`  now factor


The critical points of this function are where the first derivative is zero or does not exist.  In this case, where `x=1` , `x=-1` and `x=0` .  Since the problem states that `x>0` , the critical point we want is `x=1` .

The number `x=1` , minimizes the sum of the number and its reciprocal.

A graph of the function is provided.  Note the function is a minimum at the blue dot where `x=1` .