Minimum of sum

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

Result

x =  1

Solution:

f(x)=x+1/x x>0 f′(x)=1−1/x2 1−1/x2=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=(x−1)(x+1)=0x=x1=1f(x) = x + 1/x \ \\ x>0 \ \\ f'(x) = 1 -1/x^2 \ \\ 1 -1/x^2 = 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=(x -1) (x +1) = 0x = x_{ 1 } = 1

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