If x is real,then the minimum value of frac{x | vedclass

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(A) Let $f(x) = \frac{x^2-x+1}{x^2+x+1}$ ... $(i)$On differentiating with respect to $x$,we get$f'(x) = \frac{(x^2+x+1)(2x-1) - (x^2-x+1)(2x+1)}{(x^2+

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$\frac{1}{3}$

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(A) Let $f(x) = \frac{x^2-x+1}{x^2+x+1}$ ... $(i)$On differentiating with respect to $x$,we get$f'(x) = \frac{(x^2+x+1)(2x-1) - (x^2-x+1)(2x+1)}{(x^2+x+1)^2}$For maximum or minimum,put $f'(x) = 0$$(x^2+x+1)(2x-1) - (x^2-x+1)(2x+1) = 0$$(2x^3 - x^2 + 2x^2 - x + 2x - 1) - (2x^3 + x^2 - 2x^2 - x + 2x + 1) = 0$$(2x^3 + x^2 + x - 1) - (2x^3 - x^2 + x + 1) = 0$$2x^2 - 2 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$Now,$f'(x) = \frac{2x^2-2}{(x^2+x+1)^2}$Again differentiating,we get $f''(x) = \frac{(x^2+x+1)^2(4x) - (2x^2-2) \cdot 2(x^2+x+1)(2x+1)}{(x^2+x+1)^4}$At $x = 1$,$f''(1) = \frac{(1+1+1)^2(4) - 0}{(1+1+1)^4} = \frac{9 	imes 4}{81} = \frac{36}{81} > 0$Therefore,the function has a minimum at $x = 1$.Putting $x = 1$ in equation $(i)$,we get$f(1) = \frac{1^2-1+1}{1^2+1+1} = \frac{1}{3}$$	herefore$ The minimum value is $\frac{1}{3}$.

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

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