Answer:
[tex]S(x) = x + \frac{1}{x}[/tex] --- Objective function
Interval = [tex]\{x:x=1\}[/tex]
Step-by-step explanation:
Given
Represent the number with x
The required sum can be represented as:
[tex]x + \frac{1}{x}[/tex]
Hence, the objective function is:
[tex]S(x) = x + \frac{1}{x}[/tex]
To get the the interval, we start by differentiating w.r.t x
Using first principle, this gives:
[tex]S'(x) = 1 - \frac{1}{x^2}[/tex]
Equate S'(x) to 0 in order to solve for x
[tex]0 = 1 - \frac{1}{x^2}[/tex]
Subtract 1 from both sides
[tex]0 -1 = 1 -1 - \frac{1}{x^2}[/tex]
[tex]-1 = - \frac{1}{x^2}[/tex]
Multiply both sides by -1
[tex]1 = \frac{1}{x^2}[/tex]
Cross Multiply
[tex]x^2 * 1 = 1[/tex]
[tex]x^2 = 1[/tex]
Take positive square root of both sides because x is positive
[tex]\sqrt{x^2} = \sqrt{1[/tex]
[tex]x = 1[/tex]
Representing x using interval notation, we have
Interval = [tex]\{x:x=1\}[/tex]
To get the smallest sum, we substitute 1 for x in [tex]S(x) = x + \frac{1}{x}[/tex]
[tex]S(1) = 1 + \frac{1}{1}[/tex]
[tex]S(1) = 1 + 1[/tex]
[tex]S(1) = 2[/tex]
Hence, the smallest sum is 2
