Question

Find a positive number for which the sum of it and its reciprocal is the smallest​ (least) possible.

Answer 1

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:

$V(x) = x + \dfrac{1}{x}$

First, we differentiate V(x) with respect to x, to get,

$\frac{d(V(x))}{dx} = \frac{d(x+\frac{1}{x})}{dx} = 1-\dfrac{1}{x^2}$

Equating the first derivative to zero, we get,

$\frac{d(V(x))}{dx} = 0\\\\1-\dfrac{1}{x^2}= 0$

Solving, we get,

$x^2 = 1\\x= \pm 1$

Since x is a positive number x = 1.

Again differentiation V(x), with respect to x, we get,

$\frac{d^2(V(x))}{dx^2} = \dfrac{2}{x^3}$

At x = 1

$\frac{d^2(V(x))}{dx^2} > 0$

Thus, by double derivative test minima occurs for V(x) at x = 1.

Thus, smallest possible sum of a number and its reciprocal is

$V(1) = 1 + \dfrac{1}{1} = 2$

Thus, 1 is the positive number for which the sum of it and its reciprocal is the smallest.