For all real $x$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

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Aicharya · Accepted Answer

$\begin{aligned} & f(x)=\frac{1-x+x^2}{1+x+x^2} \ 	herefore \quad & f^{\prime}(x)=\frac{\left(1+x+x^2ight)(-1+2 x)-\left(1-x+x^2ight)(1+2 x)}{\left(1+x+x^2ight)^2} \ = & \frac{\left(-1+2 x-x+2 x^2-x^2+2 x^3ight)}{\left(1+x+x^2ight)^2} \ = & \frac{-2+2 x^2}{\left(1+x+x^2ight)^2}\end{aligned}$ If $\mathrm{f}^{\prime}(x)=0$, then $\frac{-2+2 x^2}{\left(1+x+x^2ight)^2}=0 \Rightarrow x^2=1$ $\Rightarrow x= \pm 1$ $	herefore \quad \mathrm{f}(x)$ at $x=1$ is $\frac{1}{3}$ and $\mathrm{f}(x)$ at $x=-1$ is 1 , $	herefore \quad$ Minimum value of $\mathrm{f}(x)$ is $\frac{1}{3}$.

For all real $x$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

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