Minimum of sum

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

Correct answer:

x =  1

Step-by-step explanation:

$$\begin{gathered} f(x)=x+1\mathrm{/}x \\ x>0 \\ {f}^{\mathrm{\prime }}(x)=1-1\mathrm{/}{x}^2 \\ 1-1\mathrm{/}{x}^2=0 \\ {x}^2-1=0 \\ {x}^2-1=0 \\ a=1;b=0;c=-1 \\ D=b^2-4ac=0^2-4\cdot 1\cdot (-1)=4 \\ D>0 \\ {x}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{\pm \sqrt{4}}{2}} \\ {x}_{1,2}={\displaystyle \frac{\pm 2}{2}} \\ {x}_{1,2}=\pm 1 \\ {x}_1=1 \\ {x}_2=-1 \\ \text{ Factored form of the equation: } \\ (x-1)(x+1)=0 \\ x=x_1=1 \end{gathered}$$

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