What is the minimum value of frac{1 - x + x^2 | vedclass

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(B) Let $y = \frac{1 - x + x^2}{1 + x + x^2}$.We can rewrite this as $y = \frac{(1 + x + x^2) - 2x}{1 + x + x^2} = 1 - \frac{2x}{1 + x + x^2}$.For $x

Vedclass · Accepted Answer

$1/3$

Vedclass · Answer

(B) Let $y = \frac{1 - x + x^2}{1 + x + x^2}$.We can rewrite this as $y = \frac{(1 + x + x^2) - 2x}{1 + x + x^2} = 1 - \frac{2x}{1 + x + x^2}$.For $x = 0$,$y = 1$.For $x 
eq 0$,$y = 1 - \frac{2}{\frac{1}{x} + 1 + x}$.By $AM$-$GM$ inequality,for $x > 0$,$\frac{1}{x} + x \geq 2$,so $\frac{1}{x} + 1 + x \geq 3$.Thus,$\frac{2}{\frac{1}{x} + 1 + x} \leq \frac{2}{3}$.Therefore,$y = 1 - \frac{2}{\frac{1}{x} + 1 + x} \geq 1 - \frac{2}{3} = \frac{1}{3}$.For $x  0$. Then $y = \frac{1 + t + t^2}{1 - t + t^2} = \frac{1}{\frac{1 - t + t^2}{1 + t + t^2}}$.Since the minimum value of the expression is $1/3$,the maximum value is $3$. Thus $y \geq 1/3$ for all $x$.The minimum value is $1/3$.

What is the minimum value of $\frac{1 - x + x^2}{1 + x + x^2}$?

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