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

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(C) Let $f(x) = \frac{1-x+x^2}{1+x+x^2}$.Using the quotient rule,$f'(x) = \frac{(1+x+x^2)(-1+2x) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2}$.Expanding the numera

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$\frac{1}{3}$

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(C) Let $f(x) = \frac{1-x+x^2}{1+x+x^2}$.Using the quotient rule,$f'(x) = \frac{(1+x+x^2)(-1+2x) - (1-x+x^2)(1+2x)}{(1+x+x^2)^2}$.Expanding the numerator: $(1+x+x^2)(-1+2x) = -1+2x-x+2x^2-x^2+2x^3 = 2x^3+x^2+x-1$.$(1-x+x^2)(1+2x) = 1+2x-x-2x^2+x^2+2x^3 = 2x^3-x^2+x+1$.Subtracting these: $(2x^3+x^2+x-1) - (2x^3-x^2+x+1) = 2x^2-2$.Thus,$f'(x) = \frac{2x^2-2}{(1+x+x^2)^2}$.Setting $f'(x) = 0$ gives $2x^2-2 = 0$,so $x^2 = 1$,which means $x = 1$ or $x = -1$.Evaluating $f(x)$ at these points:$f(1) = \frac{1-1+1}{1+1+1} = \frac{1}{3}$.$f(-1) = \frac{1-(-1)+(-1)^2}{1+(-1)+(-1)^2} = \frac{1+1+1}{1-1+1} = \frac{3}{1} = 3$.Therefore,the minimum value of $f(x)$ is $\frac{1}{3}$.

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

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