The real number $x$ when added to its inverse gives the minimum value of the sum at $x$ equal to:

Let $f(x)=x+\frac{1}{x}$.To find critical points, $f^{\prime }(x)=1-\frac{1}{x^2}=0⟹x^2=1⟹x=\pm 1$.For $x=1$, $f(1)=2$. $f^{\prime \prime }(x)=\frac{2}{x^3}>0$, so $x=1$ is a minimum.For $x=-1$, $f(-1)=-2$. $f^{\prime \prime }(x)<0$, so $x=-1$ is a maximum.Thus the minimum value occurs at $x=1$.Note: Usually, "minimum value" in this context refers to positive $x$ unless specified.