For all real $x$, the minimum value of the function $f(x)=\frac{1-x+x^2}{1+x+x^2}$ is
For all real $x$, the minimum value of the function $f(x)=\frac{1-x+x^2}{1+x+x^2}$ is
$\frac{1}{3}$
0
3
1
Solution
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 $x=1, f(x)=\frac{1}{3}$ and when $x=-1, f(x)=3$
Hence minimum value of $f(x)$ is $\frac{1}{3}$.