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=1

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