Minimum of sum

Find a positive number that the sum of the number and its inverted value was minimal.

Correct result:

x =  1

Solution:

f(x)=x+1/x x>0 f′(x)=1−1/x2 1−1/x2=0 x2−1=0  x2−1=0  a=1;b=0;c=−1 D=b2−4ac=02−4⋅1⋅(−1)=4 D>0  x1,2=−b±D2a=±42 x1,2=±22 x1,2=±1 x1=1 x2=−1   Factored form of the equation:  (x−1)(x+1)=0 x=x1=1f(x)=x + 1/x \ \\ x>0 \ \\ f'(x)=1 -1/x^2 \ \\ 1 -1/x^2=0 \ \\ x^2 - 1=0 \ \\ \ \\ x^2 -1=0 \ \\ \ \\ a=1; b=0; c=-1 \ \\ D=b^2 - 4ac=0^2 - 4\cdot 1 \cdot (-1)=4 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ \pm \sqrt{ 4 } }{ 2 } \ \\ x_{1,2}=\dfrac{ \pm 2 }{ 2 } \ \\ x_{1,2}=\pm 1 \ \\ x_{1}=1 \ \\ x_{2}=-1 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -1) (x +1)=0 \ \\ x=x_{1}=1

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