For all real x,the minimum value of frac{1-x+ | vedclass

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(B) Let $y = \frac{1-x+x^{2}}{1+x+x^{2}}$.We can rewrite the expression as $y = \frac{(x^{2}+x+1) - 2x}{x^{2}+x+1} = 1 - \frac{2x}{x^{2}+x+1}$.To find

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$1/3$

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(B) Let $y = \frac{1-x+x^{2}}{1+x+x^{2}}$.We can rewrite the expression as $y = \frac{(x^{2}+x+1) - 2x}{x^{2}+x+1} = 1 - \frac{2x}{x^{2}+x+1}$.To find the minimum value of $y$,we need to maximize the term $\frac{2x}{x^{2}+x+1}$.Let $f(x) = \frac{x}{x^{2}+x+1}$.Taking the derivative with respect to $x$ and setting it to $0$:$f'(x) = \frac{(x^{2}+x+1)(1) - x(2x+1)}{(x^{2}+x+1)^{2}} = \frac{x^{2}+x+1-2x^{2}-x}{(x^{2}+x+1)^{2}} = \frac{1-x^{2}}{(x^{2}+x+1)^{2}}$.Setting $f'(x) = 0$ gives $1-x^{2} = 0$,so $x = 1$ or $x = -1$.For $x = 1$,$f(1) = \frac{1}{1+1+1} = \frac{1}{3}$.For $x = -1$,$f(-1) = \frac{-1}{1-1+1} = -1$.Since we want to minimize $y = 1 - 2f(x)$,we choose the maximum value of $f(x)$,which is $1/3$.Thus,the minimum value of $y$ is $1 - 2(1/3) = 1 - 2/3 = 1/3$.

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

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