Question

Find a positive number for which the sum of it and its reciprocal is the smallest​ (least) possible. Let x be the number and let S be the sum of the number and its reciprocal. Write the objective function in terms of x. ​S(x) nothing The interval of interest of the objective function is nothing. ​(Simplify your answer. Type your answer in interval​ notation.) The number is 0.

Answer 1

Answer:

$S(x) = x + \frac{1}{x}$ --- Objective function

Interval = $\{x:x=1\}$

Step-by-step explanation:

Given

Represent the number with x

The required sum can be represented as:

$x + \frac{1}{x}$

Hence, the objective function is:

$S(x) = x + \frac{1}{x}$

To get the the interval, we start by differentiating w.r.t x

Using first principle, this gives:

$S'(x) = 1 - \frac{1}{x^2}$

Equate S'(x) to 0 in order to solve for x

$0 = 1 - \frac{1}{x^2}$

Subtract 1 from both sides

$0 -1 = 1 -1 - \frac{1}{x^2}$

$-1 = - \frac{1}{x^2}$

Multiply both sides by -1

$1 = \frac{1}{x^2}$

Cross Multiply

$x^2 * 1 = 1$

$x^2 = 1$

Take positive square root of both sides because x is positive

$\sqrt{x^2} = \sqrt{1$

$x = 1$

Representing x using interval notation, we have

Interval = $\{x:x=1\}$

To get the smallest sum, we substitute 1 for x in $S(x) = x + \frac{1}{x}$

$S(1) = 1 + \frac{1}{1}$

$S(1) = 1 + 1$

$S(1) = 2$

Hence, the smallest sum is 2