The positive real number x when

Solution:

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

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

Differentiating both sides with respect to x, we get

$S^{\prime}(x)=1-\frac{1}{x^{2}}$

For maxima or minima,

$S^{\prime}(x)=0$

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

$\Rightarrow x^{2}=1$

$\Rightarrow x=-1$ or $x=1$

Now,

$S^{\prime \prime}(x)=\frac{2}{x^{3}}$

At $x=-1$, we have

$S^{\prime \prime}(1)=\frac{2}{(-1)^{3}}=-2<0$

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

At x = 1, we have

$S^{\prime \prime}(1)=\frac{2}{(1)^{3}}=2>0$

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___.