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(B) Let $y = \frac{x^2 - x + 1}{x^2 + x + 1}$.To find the extrema,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{(x^2 + x + 1)(2x - 1

Vedclass · Accepted Answer

$3, \frac{1}{3}$

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(B) Let $y = \frac{x^2 - x + 1}{x^2 + x + 1}$.To find the extrema,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = \frac{(x^2 + x + 1)(2x - 1) - (x^2 - x + 1)(2x + 1)}{(x^2 + x + 1)^2}$$= \frac{(2x^3 + 2x^2 + 2x - x^2 - x - 1) - (2x^3 - 2x^2 + 2x + x^2 - x + 1)}{(x^2 + x + 1)^2}$$= \frac{(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1)}{(x^2 + x + 1)^2}$$= \frac{2x^2 - 2}{(x^2 + x + 1)^2}$.Setting $\frac{dy}{dx} = 0$,we get $2x^2 - 2 = 0$,which implies $x^2 = 1$,so $x = 1$ or $x = -1$.For $x = 1$,$y = \frac{1 - 1 + 1}{1 + 1 + 1} = \frac{1}{3}$.For $x = -1$,$y = \frac{1 + 1 + 1}{1 - 1 + 1} = \frac{3}{1} = 3$.Thus,the greatest value is $3$ and the least value is $\frac{1}{3}$.

If $x$ is real,then the greatest and least values of $\frac{x^2 - x + 1}{x^2 + x + 1}$ are

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