If x is real, then the minimum value of frac{ | 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 values, set $f'(x) = 0$:$(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, we check the second derivative $f''(x)$ or test values.At $x = 1$, $f(1) = \frac{1-1+1}{1+1+1} = \frac{1}{3}$.At $x = -1$, $f(-1) = \frac{1+1+1}{1-1+1} = 3$.Thus, 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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