Question

The positive real number x when added to its reciprocal gives the minimum value of the sum when, x = __________________.

Answer 1

Let S(x) be the sum of the positive real number x and its reciprocal.

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

Differentiating both sides with respect to x, we get

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

For maxima or minima,

$S\text{'}\left(x\right)=0$

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

$\Rightarrow x^2=1$

⇒ x = −1 or x = 1

Now,

$S\text{'}\text{'}\left(x\right)=\frac{2}{x^3}$

At x = −1, we have

$S\text{'}\text{'}\left(1\right)=\frac{2}{{\left(-1\right)}^3}=-2<0$

So, x = −1 is the point of local maximum of S(x).

At x = 1, we have

$S\text{'}\text{'}\left(1\right)=\frac{2}{{\left(1\right)}^3}=2>0$

So, x = 1 is the point of local minimum of S(x).

Therefore, S(x) is minimum when x =1.

Thus, the sum of positive real number x and its reciprocal is minimum when x = 1.


The positive real number x when added to its reciprocal gives the minimum value of the sum when, x = ___1___.