Answer:
--- Objective function
Interval = ![\{x:x=1\}](https://tex.z-dn.net/?f=%5C%7Bx%3Ax%3D1%5C%7D)
Step-by-step explanation:
Given
Represent the number with x
The required sum can be represented as:
![x + \frac{1}{x}](https://tex.z-dn.net/?f=x%20%2B%20%5Cfrac%7B1%7D%7Bx%7D)
Hence, the objective function is:
![S(x) = x + \frac{1}{x}](https://tex.z-dn.net/?f=S%28x%29%20%3D%20x%20%2B%20%5Cfrac%7B1%7D%7Bx%7D)
To get the the interval, we start by differentiating w.r.t x
<em>Using first principle, this gives:</em>
![S'(x) = 1 - \frac{1}{x^2}](https://tex.z-dn.net/?f=S%27%28x%29%20%3D%201%20-%20%5Cfrac%7B1%7D%7Bx%5E2%7D)
Equate S'(x) to 0 in order to solve for x
![0 = 1 - \frac{1}{x^2}](https://tex.z-dn.net/?f=0%20%3D%201%20-%20%5Cfrac%7B1%7D%7Bx%5E2%7D)
Subtract 1 from both sides
![0 -1 = 1 -1 - \frac{1}{x^2}](https://tex.z-dn.net/?f=0%20-1%20%3D%201%20-1%20-%20%5Cfrac%7B1%7D%7Bx%5E2%7D)
![-1 = - \frac{1}{x^2}](https://tex.z-dn.net/?f=-1%20%3D%20-%20%5Cfrac%7B1%7D%7Bx%5E2%7D)
Multiply both sides by -1
![1 = \frac{1}{x^2}](https://tex.z-dn.net/?f=1%20%3D%20%5Cfrac%7B1%7D%7Bx%5E2%7D)
Cross Multiply
![x^2 * 1 = 1](https://tex.z-dn.net/?f=x%5E2%20%2A%201%20%3D%201)
![x^2 = 1](https://tex.z-dn.net/?f=x%5E2%20%20%3D%201)
Take positive square root of both sides because x is positive
![\sqrt{x^2} = \sqrt{1](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2%7D%20%3D%20%5Csqrt%7B1)
![x = 1](https://tex.z-dn.net/?f=x%20%3D%201)
Representing x using interval notation, we have
Interval = ![\{x:x=1\}](https://tex.z-dn.net/?f=%5C%7Bx%3Ax%3D1%5C%7D)
To get the smallest sum, we substitute 1 for x in ![S(x) = x + \frac{1}{x}](https://tex.z-dn.net/?f=S%28x%29%20%3D%20x%20%2B%20%5Cfrac%7B1%7D%7Bx%7D)
![S(1) = 1 + \frac{1}{1}](https://tex.z-dn.net/?f=S%281%29%20%3D%201%20%2B%20%5Cfrac%7B1%7D%7B1%7D)
![S(1) = 1 + 1](https://tex.z-dn.net/?f=S%281%29%20%3D%201%20%2B%201)
![S(1) = 2](https://tex.z-dn.net/?f=S%281%29%20%3D%202)
<em>Hence, the smallest sum is 2</em>