For all real \(\mathrm{x}\), the minimum value of the function \(f(x)=\frac{1-x+x^2}{1+x+x^2}\) is

We have \(f(x)=\frac{1-x+x^2}{1+x+x^2}\)
\(
f^{\prime}(x)=\frac{\left(1+x+x^2\right)(2 x-1)-\left(1-x+x^2\right)(2 x+1)}{\left(1+x+x^2\right)^2}
\)
\(
=\frac{\left(2 x+2 x^2+2 x^3-x-1-x^2\right)-\left(2 x-2 x^2+2 x^3+1-x+x^2\right)}{\left(1+x+x^2\right)^2}
\)
\(
=\frac{\left(x+x^2+2 x^3-1\right)-\left(x-x^2+2 x^3+1\right)}{\left(1+x+x^2\right)^2}
\)
\(=\frac{2 x^2-2}{\left(1+x+x^2\right)^2}\) and when \(f^{\prime}(x)=0\), we get
\(
2\left(\mathrm{x}^2-1\right)=0 \Rightarrow \mathrm{x}= \pm 1
\)
When \(\mathrm{x}=1, \mathrm{f}(\mathrm{x})=\frac{1}{3}\) and when \(\mathrm{x}=-1, \mathrm{f}(\mathrm{x})=3\)
Hence minimum value of \(f(x)\) is \(\frac{1}{3}\).