Answer:
1 is the positive number for which the sum of it and its reciprocal is the smallest.
Step-by-step explanation:
Let x be the positive number.
Then, the sum of number and its reciprocal is given by:
[tex]V(x) = x + \dfrac{1}{x}[/tex]
First, we differentiate V(x) with respect to x, to get,
[tex]\frac{d(V(x))}{dx} = \frac{d(x+\frac{1}{x})}{dx} = 1-\dfrac{1}{x^2}[/tex]
Equating the first derivative to zero, we get,
[tex]\frac{d(V(x))}{dx} = 0\\\\1-\dfrac{1}{x^2}= 0[/tex]
Solving, we get,
[tex]x^2 = 1\\x= \pm 1[/tex]
Since x is a positive number x = 1.
Again differentiation V(x), with respect to x, we get,
[tex]\frac{d^2(V(x))}{dx^2} = \dfrac{2}{x^3}[/tex]
At x = 1
[tex]\frac{d^2(V(x))}{dx^2} > 0[/tex]
Thus, by double derivative test minima occurs for V(x) at x = 1.
Thus, smallest possible sum of a number and its reciprocal is
[tex]V(1) = 1 + \dfrac{1}{1} = 2[/tex]
Thus, 1 is the positive number for which the sum of it and its reciprocal is the smallest.
